Physics I.

Gyorgy HarsGabor Dobos

2013.04.30

Contents

1 Kinematics of a particle - Gyorgy Hars 31.1 Rectilinear motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 Uniform Rectilinear Motion . . . . . . . . . . . . . . . . . . . . . 41.1.2 Uniformly Accelerated Rectilinear Motion . . . . . . . . . . . . . 41.1.3 Harmonic oscillatory motion . . . . . . . . . . . . . . . . . . . . . 5

1.2 Curvilinear motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2.1 Projectile motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.2.2 Circular motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.3 Areal velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Dynamics of a Particle - Gyorgy Hars 142.1 Inertial system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.2 The mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.3 Linear momentum (p) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.4 Equation of motion: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.5 The concept of weight . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.6 The concept of work in physics . . . . . . . . . . . . . . . . . . . . . . . 202.7 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.8 Theorem of Work (Kinetic energy) . . . . . . . . . . . . . . . . . . . . . 232.9 Potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.10 Conservation of the mechanical energy . . . . . . . . . . . . . . . . . . . 272.11 Energy relations at harmonic oscillatory motion . . . . . . . . . . . . . . 282.12 Angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.13 Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.14 Central force field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3 Dynamics of system of particles - Gyorgy Hars 343.1 Momentum in system of particles . . . . . . . . . . . . . . . . . . . . . . 34

3.1.1 Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.1.2 Missile motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.2 Angular momentum in system of particles . . . . . . . . . . . . . . . . . 42

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3.2.1 The skew rotator . . . . . . . . . . . . . . . . . . . . . . . . . . . 443.2.2 The pirouette dancer (The symmetrical rotator) . . . . . . . . . . 46

3.3 Discussion of the total kinetic energy in the system of particles . . . . . 48

4 Dynamics of rigid body - Gyorgy Hars 504.1 Moment of inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504.2 Equation of motion of the rigid body: . . . . . . . . . . . . . . . . . . . . 56

4.2.1 Demonstration example 1. . . . . . . . . . . . . . . . . . . . . . . 574.2.2 Demonstration example 2. . . . . . . . . . . . . . . . . . . . . . . 58

4.3 Kinetic energy of the rigid body . . . . . . . . . . . . . . . . . . . . . . . 60

5 Non-inertial (accelerating) reference frames - Gyorgy Hars 625.1 Coordinate system with translational acceleration . . . . . . . . . . . . . 625.2 Coordinate system in uniform rotation . . . . . . . . . . . . . . . . . . . 65

5.2.1 Earth as a rotating coordinate system . . . . . . . . . . . . . . . . 67

6 Oscillatory Motion - Gabor Dobos 776.1 The simple harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . 77

6.1.1 Complex representation of oscillatory motion . . . . . . . . . . . . 796.1.2 Velocity and acceleration in oscillatory motion . . . . . . . . . . . 80

6.2 Motion of a body attached to a spring . . . . . . . . . . . . . . . . . . . 806.3 Simple pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 846.4 Energy in simple harmonic motion . . . . . . . . . . . . . . . . . . . . . 866.5 Damped oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 876.6 Forced oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 906.7 Superposition of simple harmonic oscillations . . . . . . . . . . . . . . . . 93

6.7.1 Same frequency, same direction . . . . . . . . . . . . . . . . . . . 936.7.2 Different frequency, same direction . . . . . . . . . . . . . . . . . 956.7.3 Lissajous figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 966.7.4 Fourier analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

7 Waves - Gabor Dobos 1007.1 Sine wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1017.2 Transverse wave on a string . . . . . . . . . . . . . . . . . . . . . . . . . 1037.3 Energy transport by mechanical waves . . . . . . . . . . . . . . . . . . . 1057.4 Group velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1077.5 Wave packets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1097.6 Standing waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1117.7 The Doppler Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

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8 First law of thermodynamics and related subjects - Gyorgy Hars 1178.1 Ideal gas equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1178.2 The internal energy of the gas (U) . . . . . . . . . . . . . . . . . . . . . 1188.3 The p-V diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1218.4 Expansion work of the gas . . . . . . . . . . . . . . . . . . . . . . . . . . 1218.5 First law of thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . 122

8.5.1 Isochoric process . . . . . . . . . . . . . . . . . . . . . . . . . . . 1238.5.2 Isobaric process . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1238.5.3 Isothermal process . . . . . . . . . . . . . . . . . . . . . . . . . . 1248.5.4 Adiabatic process . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

8.6 Summary of the molar heat capacitances . . . . . . . . . . . . . . . . . . 1288.7 The Carnot cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

9 The entropy and the second law of thermodynamics - Gyorgy Hars 1329.1 The entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1329.2 The isentropic process . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1369.3 The microphysical meaning of entropy . . . . . . . . . . . . . . . . . . . 1369.4 Gay-Lussac experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

9.4.1 Phenomenological approach . . . . . . . . . . . . . . . . . . . . . 1389.4.2 Statistical approach . . . . . . . . . . . . . . . . . . . . . . . . . . 139

9.5 The Boltzmann equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 1419.6 Approximate formula (lnn! ≈ n lnn− n) a sketch of proof: . . . . . . . . 1419.7 Equalization process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142

9.7.1 Equalization between gaseous components . . . . . . . . . . . . . 1439.7.2 Equalization of non-gaseous materials without phase transition . . 1489.7.3 Ice cubes in the water . . . . . . . . . . . . . . . . . . . . . . . . 149

9.8 The second law of thermodynamics . . . . . . . . . . . . . . . . . . . . . 151

3

Introduction - Gyorgy Hars

Present work is the summary of the lectures held by the author at Budapest Universityof Technology and Economics. Long verbal explanations are not involved in the text,only some hints which make the reader to recall the lecture. Refer here to the book:Alonso/Finn Fundamental University Physics, Volume I where more details can be found.

Physical quantities are product of a measuring number and the physical unit. In con-trast to mathematics, the accuracy or in other words the precision is always a secondaryparameter of each physical quantity. Accuracy is determined by the number of valuabledigits of the measuring number. Because of this 1500 m and 1.5 km are not equivalentin terms of accuracy. They have 1 m and 100 m absolute errors respectively. The oftenused term relative error is the ratio of the absolute error over the nominal value. Thesmaller is the relative error the higher the accuracy of the measurement. When makingoperations with physical quantities, remember that the result may not be more accuratethan the worst of the factors involved. For instance, when dividing 3.2165 m with 2.1s to find the speed of some particle, the result 1.5316667 m/s is physically incorrect.Correctly it may contain only two valuable digits, just like the time data, so the correctresult is 1.5 m/s.

The physical quantities are classified as fundamental quantities and derived quan-tities. The fundamental quantities and their units are defined by standard or in otherwords etalon. The etalons are stored in relevant institute in Paris. The fundamentalquantities are the length, the time and the mass. The corresponding units are meter(m), second (s) and kilogram (kg) respectively. These three fundamental quantities aresufficient to build up the mechanics. The derived quantities are all other quantities whichare the result of some kind of mathematical operations. To describe electric phenomenathe fourth fundamental quantity has been introduced. This is ampere (A) the unit ofelectric current. This will be used extensively in Physics 2, when dealing with electricity.

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Chapter 1

Kinematics of a particle - GyorgyHars

Kinematics deals with the description of motion, without any respect to the cause of themotion. Strictly speaking there is no mass involved in the theory, so force and relatedquantities do not show up. The fundamental quantities involved are the length and thetime only.

To describe the motion one needs a reference frame. Practically it is the Cartesiancoordinate system with x, y, z coordinates, and corresponding i, j, k unit vectors.

The particle is a physical model. This is a point like mass, so it lacks of any extension.

1.1 Rectilinear motion

(Egyenes vonalu mozgas)The motion of the particle takes place in a straight line in rectilinear motion. This

means that the best mathematical description is one of the axes of the Cartesian coor-dinate system. So the position of the particle is described by x(t) function.

The velocity of the particle is the first derivative of the position function. The every-day concept of speed is the absolute value of the velocity vector. Therefore the speed isalways a nonnegative number, while the velocity can also be a negative number.

v(t) = limt=0

∆x

∆t=dx

dt

[ms

](1.1)

The opposite direction operation recovers the position time function from the velocityvs. time function. Here x0 is the initial value of the position in t = 0 moment, t

′denotes

5

the integration parameter from zero to t time.

x(t) = x0 +

t∫0

v(t,)dt, (1.2)

The acceleration of the particle is the first derivative of the velocity vs. time function,thus it is the second derivative of the position vs. time function.

a(t) = limt=0

∆v

∆t=dv

dt=d2x

dt2

[ms2

](1.3)

The opposite direction operation recovers the velocity time function from the accelerationvs. time function. Here v0 is the initial value of the position in t = 0 moment, t

′denotes

the integration parameter from zero to t time.

v(t) = v0 +

t∫0

a(t,)dt, (1.4)

1.1.1 Uniform Rectilinear Motion

Here the acceleration of the particle is zero. The above formulas transform to the fol-lowing special cases. a = 0, v = v0, x = x0+ vt.

Figure 1.1: Uniform Rectilinear Motion

1.1.2 Uniformly Accelerated Rectilinear Motion

Here the acceleration of the particle is constant. The above formulas transform to thefollowing special cases. a = const, v = v0 + at x = x0 + v0t+ a

2t2

Typical example is the free fall, where the acceleration is a = g = 9.81 m/s2.

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Figure 1.2: Uniformly Accelerated Rectilinear Motion

1.1.3 Harmonic oscillatory motion

The trajectory of the harmonic oscillation is straight line, so this is a special rectilinearmotion. First let us consider a particle in uniform circular motion.

Figure 1.3: Harmonic oscillatory motion

The two coordinates in the Cartesian coordinate system are as follows:

x = A cos(ωt+ φ) y = A sin(ωt+ φ) (1.5)

If the uniform circular motion is projected to one of its coordinates, the motion of theprojected point is “harmonic oscillatory motion”. We choose the x coordinate.

x(t) = A cos(ωt+ φ) (1.6)

The displacement at oscillatory motion is called excursion. The sum in the parenthesisis called the “phase”. The multiplier of time is called angular frequency, and additiveϕ constant is the initial phase. The multiplier in front is called the “amplitude”. Thevelocity of the oscillation is the derivative of the displacement function.

dx(t)

dt= v(t) = −Aω sin(ωt+ φ) (1.7)

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The multiplier of the trigonometric term is called the “velocity amplitude” (vmax).

vmax = Aω (1.8)

The acceleration is the derivative of the velocity:

dv(t)

dt= a(t) = −Aω2 cos(ωt+ φ) (1.9)

If one compares the displacement and the acceleration functions the relation below canreadily found:

a(t) = −ω2x(t) (1.10)

Accordingly, the acceleration is always opposite phase position relative to the displace-ment.

In the kinematics of the harmonic oscillations it is very much helpful to go back to theorigin of the oscillatory motion and contemplate the phenomena as projected componentof a uniform circular motion. This way one gets rid of the trigonometric formalism andthe original problem could have a far easier geometric interpretation. Best example forthat if we want to find out the resultant oscillation of two identical frequency harmonicoscillations with different amplitudes and different initial phases. In pure trigonometryapproach this is a tedious work, while in the circle diagram this is a simple geometryproblem, actually a cosine theorem application in the most ordinary case.

1.2 Curvilinear motion

(Gorbervonalu mozgas)The motion of the particle is described by an arbitrary r(t) vector scalar function,

where i, j, k are the unit vectors of the coordinate system.

r(t) = x(t)i + y(t)j + z(t)k (1.11)

The velocity of the particle is the first derivative of the position function.

v(t) = limt=0

∆r

∆t=dx

dti +

dy

dtj +

dz

dtk (1.12)

The velocity vector is tangential to the trajectory of the particle always.The vector of acceleration is the derivative of the velocity vector. The vector of

acceleration can be decomposed as parallel and normal direction to the velocity.

a(t) = limt=0

∆v

∆t=dvxdt

i +dvydt

j +dvzdt

k =d2x

dt2i +

d2y

dt2j +

d2z

dt2k (1.13)

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Figure 1.4: Curvilinear motion

The parallel component of the acceleration (called tangential acceleration) is theconsequence of the variation in the absolute value of the velocity. In other words this iscaused by the variation of the speed. The normal component of the acceleration (calledcentripetal acceleration) is the consequence of the change in the direction of the velocityvector.

If one drives a car on the road, speeding up or slowing down causes the tangentialacceleration to be directed parallel or opposite with the velocity, respectively. By turningthe steering wheel, centripetal acceleration will emerge. The direction of the centripetalacceleration points in the direction of the virtual center of the bend.

1.2.1 Projectile motion

(Hajıtas)In the model of the description the following conditions will be used:Projectile is a particle,Gravity field is homogeneous,Rotation of the Earth, does not take part,No drag due to air friction will be considered.In real artillery situation the phenomenon is much more complex. This is far beyond

the present scope.

Figure 1.5: Projectile motion

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The projectile is fired from the origin of the Cartesian coordinate system. The motionis characterized by the initial velocity v0 and the angle of the velocity α relative to thehorizontal direction. The motion will take place in the vertical plane, which contains thevelocity vector. The motion is the superposition of a uniform horizontal rectilinear mo-tion, a uniform vertical rectilinear motion and a free fall. Thus the velocity componentsare as follows:

vx = v0 cosα (1.14)

vy = v0 sinα− gt (1.15)

The corresponding position coordinates are the integrated formulas with zero initialcondition.

x =

t∫0

vx(t′)dt

′= v0t cosα (1.16)

y =

t∫0

vy(t′)dt

′= v0t sinα− g

2t2 (1.17)

Two critical parameters are needed to find out. These are the height of the trajectory(h) and the horizontal flight distance (d). First, the rise time should be calculated. Therise time τrise is the time when the vertical velocity component vanishes. Accordinglyvy =0 condition should be met. From the equation the following results:

τrise =v0 sinα

g(1.18)

The height of the trajectory shows up as a vertical coordinate just in rise time moment.

h = y(t = τrise) = v0τrise sinα− g

2τ 2rise (1.19)

By substituting the formula of τrise into the equation above, the height of the trajectoryresults:

h =v2

0 sin2 α

g− g

2

v20 sin2 α

g2(1.20)

Accordingly:

h =v2

0 sin2 α

2g(1.21)

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The rise and the fall part of the motion last the same duration, due to the symmetry ofthe motion. Because of this, the total flight time of the motion is twice longer than therise time alone. The horizontal flight distance (d) can be calculated as the horizontal xcoordinate at double rise time moment.

d = x(t = 2τrise) = v02v0 sinα

gcosα =

v20

g2 sinα cosα (1.22)

By using elementary trigonometry, the final formula of horizontal flight distance results:

d =v2

0

2sin 2α (1.23)

This clearly shows that the projectile flies the furthest if the angle of the shot is 45degrees.

1.2.2 Circular motion

In circular motion, the particle moves on a circular plane trajectory. To describe theposition of the particle polar coordinates are used. The origin of the polar coordinatesystem is the center of the motion. The only variable parameter is the angular positionϕ(t) since the radial position is constant.

Figure 1.6: Circular motion

The derivative of the angular position is the angular velocity ω.

ω(t) = limt=0

∆φ

∆t=dφ

dt

[1

s

](1.24)

Up to this moment it looks as if the angular velocity were a scalar number. But thisis not the case. The angular velocity is a vector in fact, because it should contain theinformation about the rotational axis as well. By definition, the angular velocity vector

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ω is as follows: The absolute value of the ω is the derivative of the angular position aswritten above. The direction of the ω is perpendicular, or in other words, normal to theplane of the rotation, and the direction results as a right hand screw rotation. This lattermeans that by turning a usual right hand screw in the direction of the circular motion,the screw will proceed in the direction of the ω vector. Just an example: If the circularmotion takes place in the plane of this paper and the rotation is going clockwise, the ωwill be directed into the paper. Counter clockwise rotation will obviously result in a ωvector pointing upward, away from the paper.

With the help of ω vector number of calculation will be much easier to carry out. Forexample finding out the velocity vector of the particle is as easy as that:

v = ω × r (1.25)

This velocity vector is sometimes called “circumferential velocity” however this notationis redundant, since the velocity vector is always tangential to the trajectory. The crossproduct of vectors in mathematics has a clear definition. By turning the first factor ( ω)into the second one (r) the corresponding turning direction defines the direction of thevelocity vector by the right hand screw rule. The absolute value of the velocity is theproduct of the individual absolute values, multiplied with the sine of the angle betweenthe vectors.

Before going into further details, let us state three mathematical statements. Leta(t) and b(t) are two time dependent vectors and λ(t) a time dependent scalar. Thenthe following differentiation rules apply:

d

dt(a(t)× b(t)) =

da(t)

dt× b(t) + a(t)× db(t)

dt(1.26)

d

dt(a(t)b(t)) =

da(t)

dtb(t) + a(t)

db(t)

dt(1.27)

d

dt(λ(t)a(t)) =

dλ(t)

dta(t) + λ(t)

da(t)

dt(1.28)

These formulas make it possible to use the same differentiation rules among the vectorproducts, just like among the ordinary product functions. End this is true both the crossproduct and the dot product operations. The proof of these rules, are quite straightfor-ward. The vectors should be written by components, and the match of the two sidesshould be verified.

Using the ω vector is a powerful means. This way the acceleration vector of theparticle can be determined with a relative ease.

a =dv(t)

dt=

d

dt(ω × r) =

dω

dt× r + ω × dr

dt(1.29)

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The derivative of ω vector is called the vector of angular acceleration β. This is theresult of the variation in the angular velocity either due to spinning faster or slower orby changing the axis of the rotation.

dω

dt= β

[1

s2

](1.30)

Last term is the derivative of the position vector. This is the velocity, which can bewritten as above wit the help of ω vector. So ultimately the acceleration vector can besummarized.

a = β × r + ω × (ω × r) (1.31)

The above formula consists of two major terms. The first term is called tangentialacceleration. In case of plane motion, this is parallel or opposite to the velocity andit is the consequence of speeding up or slowing down, as explained in the earlier partof this chapter. The second term is called the centripetal or normal acceleration. Thiscomponent points toward the center of the rotation. The centripetal acceleration is theconsequence of the direction variation of the velocity vector. The absolute values of thesecomponents can readily be expressed.

atan = βr acpt = rω2 =v2

r(1.32)

There are two special kinds of circular motion, the uniform and the uniformly acceleratingcircular motion.

Uniform circular motion:

In here the angular velocity is constant. The angle or rotation can be expressed accord-ingly:

φ(t) = ωt+ φ0 (1.33)

Since the angular acceleration is zero, no tangential acceleration will emerge. Howeverthere will be a constant magnitude centripetal acceleration, with an ever changing direc-tion, pointing always to the center.

acpt = rω2 =v2

r(1.34)

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Uniformly accelerating circular motion

In here the angular acceleration is constant. The corresponding formulas are analogousto that of uniformly accelerating rectilinear motion, explained earlier in this chapter.

ω(t) = βt+ ω0 (1.35)

φ(t) =β

2t2 + ω0t+ φ0 (1.36)

The magnitude of the tangential acceleration is constant and parallel with the velocityvector.

atan = βr = const (1.37)

The magnitude of the centripetal component shows quadratic dependence in time.

acpt = rω2 = r(βt+ ω0)2 (1.38)

1.2.3 Areal velocity

(Teruleti sebesseg)

Figure 1.7: Areal velocity

Let us consider particle travelling on its trajectory. If one draws a line between theorigin of the coordinate system and the particle, this line is called the “radius vector”.The vector of areal velocity is the ratio of the area swept by the radius vector over time.The crosshatched triangle on the figure above is the absolute value of the infinitesimalvariation (dA) of the swept area vector.

dA =1

2r× dr (1.39)

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dA =1

2r× vdt (1.40)

dA

dt=

1

2r× v

[m2

s

](1.41)

Areal velocity will be used in the study of planetary motion later in this book.

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Chapter 2

Dynamics of a Particle - GyorgyHars

(Tomegpont dinamikaja)Dynamics deals with the cause of motion. So in dynamics a new major quantity

shows up. This is the mass of the particle (m). The concept of force and other relatedquantities will be treated as well. In this chapter only one piece of particle will be thesubject of the discussion, in the next chapter however the system of particles will betreated.

2.1 Inertial system

In kinematics any kind of coordinate system could be used, there was no restriction inthis respect. In dynamics however, a dedicated special coordinate system is used mostly.This is called inertial system. The inertial system is defined as a coordinate system inwhich the law of inertia is true. The law of inertia or Newton’s first law says that themotion state of a free particle is constant. This means that if it was standstill it stayedstandstill, if it was moving with a certain velocity vector, it continues its motion withthe same velocity. So the major role of Newton’s first law is the definition of the inertialsystem. Other Newton’s laws use the inertial system as a frame of reference further on.The best approximation of the inertial system is a free falling coordinate system. Inpractice this can be a space craft orbiting the Earth, since the orbiting space craft is inconstant free fall.

The inertial systems are local. This means that the point of the experimentation andits relative proximity belongs to a dedicated inertial system. An example explains thisstatement: Imagine that we are on a huge spacecraft circularly orbiting the Earth, sowe are in inertial system. Now a small shuttle craft is ejected mechanically from thespacecraft without any rocket engine operation. The shuttle craft also orbits the Earth

16

on a different trajectory and departs relatively far from the mother ship. Observing theevents from the inertial system of the mother ship the shuttle supposed to keep its originalejection velocity and supposed to depart uniformly to the infinity. Much rather insteadthe shuttle craft also orbits the Earth and after a half circle it returns to the mother shipon its own. So the law of inertia is true in the close proximity of the experiment only. Ifone goes too far the law of inertia looses validity.

On the surface of the Earth we are not in inertial system. Partly because we ex-perience weight, which is the gravity force attracting the objects toward the center,partly because the Earth is rotating, which rotation causes numerous other effects. Eventhough in most cases phenomena on the face of our planet can be described in inertialsystem, by ignoring the rotation related effects, and by considering the gravity a separateinteraction.

2.2 The mass

Mass is a dual face quantity. Mass plays role in the interaction with the gravity field.This type of mass called gravitational mass and this is something like gravitational chargein the Newton’s gravitational law.

F = Gm1m2

r2(2.1)

Here the m1 and m2 are the gravitational masses, r is the distance between the objects, Fis the resulting force and G is the gravitational constant (6.67x10−11 m3kg−1s−2). Whensomebody measures the body weight with a bathroom scale he actually measures thegravitational mass.

Other major feature is that the mass shows resistance against the accelerating ef-fect. This resistance is characterized by the inertial mass. It has been discovered laterthat these fundamentally different features can be related to the same origin, and sothe two types of mass are equivalent. Therefore the distinction between them becameunnecessary.

This equivalency makes the free falling objects drop with the same acceleration. Thegravity force is proportional with the gravitational mass, which force should be equal withthe acceleration times the inertial mass. So if the ratio of these masses were different,then the free fall would happen with different acceleration for different materials. Thisis harshly against the experience, so mass will be referred without any attribute later inthis book.

2.3 Linear momentum (p)

(Impulzus, mozgasmennyiseg, lendulet)

17

By definition the linear momentum is the product of the mass and the velocity.Therefore linear momentum is a vector quantity.

p = mv

[kgm

s

](2.2)

Newton’s second law: This law is the definition of force (F).The force exerted to a particle is equal to the time derivative of the linear momentum.

The unit of force is Newton (N).

F =dp

dt

[kgm

s2

]= N (2.3)

Conclusion 1.If the force equals to zero, then the linear momentum is constant. This is in agreement

with the law of inertia. However it is worth mentioning, that it is only true in inertialsystem. Which means that on an accelerating train or in a spinning centrifuge it is notvalid.

Conclusion 2.According to the fundamental theorem of calculus, the time integral of force results

in the variation of the linear momentum:

p2 − p1 =

t2∫t1

F(t)dt (2.4)

The right hand side is called impulse (erolokes).Conclusion 3.The well-known form of the Newton second law can be readily expressed:

F =dp

dt=d(mv)

dt= m

dv

dt= ma (2.5)

Or briefly:

F = ma (2.6)

Newton third law: (Action reaction principle)When two particles interact, the force on one particle is equal value and opposite

direction to the force of the other particle.

18

Figure 2.1: Action reaction principle

2.4 Equation of motion:

The particle is affected by numerous forces. The sum of these forces, cause the acceler-ation of the particle. This leads to a second order ordinary differential equation. This iscalled the equation of motion: ∑

i

Fi(r, t,v) = md2r

dt2(2.7)

In principle the forces may be the function of position, time and velocity.

Example 1 for the equation of motion :Attenuated oscillation:(csillapodo rezges)A particle is hanging on a spring in water in vertical position. The particle is deflected

to a higher position, and left alone to oscillate. Describe the motion by solving theequation of motion. Ignore the buoyant force. The motion will take place in the verticalline. The position is denoted y(t) which is positive upside direction.

The forces affecting the particle are as follows:

Fspring = −Dy Fdrag = −kdydt

Fgrav = −mg (2.8)

Here D is the direction constant of the spring in N/m units, k is the drag coefficient andg is 9.81 m/s2 .Accordingly the equation of motion can be written:

−kdy(t)

dt−Dy(t)−mg = m

d2y(t)

dt2(2.9)

Ordering it to the form of a differential equation:

d2y(t)

dt2+k

m

dy(t)

dt+D

my(t) = g (2.10)

19

Let us introduce β for the attenuation coefficient with the following definition:

β =k

2m(2.11)

d2y(t)

dt2+ 2β

dy(t)

dt+D

my(t) = g (2.12)

The mathematical method for solving this differential equation is beyond the scope ofthis chapter. The solution below can be verified by substitution:

y(t) = Ae−βt cosωt− mg

D(2.13)

Here A is the original value of the deflection, ω0 is called the Thomson angular frequencyand ω is the angular frequency of the attenuated oscillation with the following definitions:

ω0 =

√D

mω =

√ω2

0 − β2 (2.14)

Figure 2.2

Example 2 for the equation of motion: Conical pendulum(kupinga)

20

Figure 2.3: Conical pendulum

The conical pendulum circulates in horizontal plane with ω angular frequency. Theangle of the rope α relative to the vertical direction is the unknown parameter to bedetermined. The coordinate system is an inertial system with horizontal and verticalaxes, with the particle in the origin. There are two forces affecting the particle, gravityforce (mg) and the tension of the rope (K). The equation of the motion is a vectorequation in two dimensions so two scalar equations are used.

← K sinα = macp (2.15)

↓ mg −K cosα = 0 (2.16)

In addition the centripetal acceleration can be expressed readily:

acp = lω2 sinα (2.17)

After substitution K = mlω2 results.By means of this result the cosine of the angular position is determined:

cosα =g

lω2(2.18)

21

2.5 The concept of weight

Let us place a bathroom scale on the floor of an elevator. The normal force (N) isdisplayed by the scale that is transferred to the object.

Figure 2.4: The concept of weight

The positive reference direction is pointing down. The following equation of motioncan be written:

↓ mg −N = ma (2.19)

Here the acceleration of the elevator is denoted (a). Let us express the normal forceindicated by the scale:

N = m(g − a) (2.20)

If the elevator does not accelerate (in most cases it is standstill) the scale shows theforce which is considered the weight of the object in general. (N =mg). This force isjust enough to compensate the gravity force, so the object does not accelerate. However,when the elevator accelerates up or down, the indicated value is increased or decreased,respectively. This also explains that in a freefalling coordinate system, where a = g, theweight vanishes. Similarly zero gravity shows up on the orbiting spacecraft, which is alsoin constant freefall.

2.6 The concept of work in physics

The concept of work in general is very broad. Besides physics, it is used in economy, alsoused as “spiritual work”. Concerning the physical concept, the amount of work is not

22

too much related, how much tiredness is suffered by the person who actually made thiswork. For example, if somebody is standing with fifty kilogram sack on his back for anhour without any motion, surely becomes very tired. Furthermore if this person walkson a horizontal surface during this time, he gets tired even more. Physical work has notbeen done in either case.

In high school the following definition was learnt. “The work equals the product offorce and the projected displacement”. This is obviously true, but only for homogeneousforce field and straight finite displacement. In equation: ∆W = F∆r. Here we used themathematical concept of dot product, which results in a scalar number, and the productof the two absolute values is multiplied with the cosine of the angle.

In general case when the related force field F(r) is not homogeneous and the displace-ment is not straight, the above finite concept is not applicable. We have to introduce theinfinitesimal contribution of work (dW=F(r)dr). The amount of work made betweentwo positions is the sum or in other words integral of dW contributions. The physicalunit is Newton meter (Nm) which is called Joule (J).

W =

r2∫r1

F(r)dr

[J = Nm =

kgm2

s2

](2.21)

Figure 2.5: The concept of work

There is a special case when the force is the function of one variable F (x) only, and itsdirection is parallel with the x direction. The above definition simplifies to the following:

W =

x2∫x1

F (x)dx

[J = Nm =

kgm2

s2

](2.22)

23

Figure 2.6: The work is the area under the curve

In this special case the work done between two positions is displayed by the areaunder the F (x) curve.

2.7 Power

(Teljesıtmeny)The power (P ) is associated with the time needed to carry out a certain amount of

work. In mathematics, this is the time derivative of the work done. The physical unit isJoule per second which is called Watt (W ).

P =dW

dt=

d

dt(Fdr)

[W =

kgm2

s3

](2.23)

Provided the force does not depend directly on time, the above formula can be trans-formed:

P =dW

dt=

d

dt(Fdr) = Fv (2.24)

So the instantaneous power is the dot product of the force and the actual velocity vector.

24

2.8 Theorem of Work (Kinetic energy)

Munkatetel (Mozgasi energia)Kinetic energy is the kind of energy which is associated with the mechanical motion of

some object. In high school the following simplified argument was presented to calculateit:

Figure 2.7: Velocity vs. time function by the effect of constant force

A particle with mass (m) is affected by constant force. Initially the particle is stand-still. The acceleration is constant, thus the v(t) graph is a sloppy line through the origin.After (t) time passed, the displacement (s) shows up as the area under the v(t) curve.Its shape is a right angle triangle.

s =vt

2(2.25)

The acceleration is the slope of the v(t) line.

a =v

t(2.26)

Let us multiply the above equation with the mass of the particle:

ma =mv

t(2.27)

The left hand side equals the force affecting the particle.

F =mv

t(2.28)

25

We also know that the work done in this simple case is:

W = Fs (2.29)

So let us substitute the related formulas. Time cancels out:

W =mv

t

vt

2=

1

2mv2 (2.30)

This is the work done on the particle which generated the kinetic energy.The above argument is not general enough, due to the simplified conditions used.

The general argument is presented below:Let us start with Newton’s second law:

F = md2r

dt2(2.31)

The work done in general is as follows:

W =

r2∫r1

Fdr (2.32)

Substitute first to the second formula:

W =

r2∫r1

md2r

dt2dr (2.33)

Switch the limits of the integration to the related time moments t1 and t2.

W = m

t2∫t1

(d2r

dt2dr

dt)dt (2.34)

Take a closer look at the formulas in the parenthesis. In here the product of the firstand the second derivative of some function are present.

The following rule is known in mathematics:

d

dx

(1

2f 2(x)

)= f(x)

df(x)

dx(2.35)

Using this formula for the last expression of work:

W = m

t2∫t1

d

dt

[1

2

(dr

dt

)2]dt = m

t2∫t1

d

[1

2

(dr

dt

)2]

=m

2

t2∫t1

d[v2]

(2.36)

26

By integrating the variations of the v2, the total variation will be the result:

m

2

t2∫t1

d[v2]

=1

2mv2

2 −1

2mv2

1 (2.37)

W =

r2∫r1

Fdr =1

2mv2

2 −1

2mv2

1 = Ekin2 − Ekin1 (2.38)

Thus: The work done on a particle equals the variation of the kinetic energy. This is thetheorem of work.

Note there is no any restriction to the kind of force. So the force is not required tobe conservative, which concept will be presented later in this chapter. This can be evensliding friction, drag or whatever other type of force.

The kinetic energy is accordingly:

Ekin =1

2mv2 (2.39)

2.9 Potential energy

(Helyzeti energia)Potential energy is the kind of energy which is associated with the position of some

object in a force field. Force field is a vector-vector function in which the force vectorF depends on the position vector r. In terms of mathematics the force field F(r) isdescribed as follows:

F(r) = X(x, y, z)i + Y (x, y, z)j +X(x, y, z)k (2.40)

where i, j, k are the unit vectors of the coordinate system.Take a particle and move it slowly in the F(r) force field from position 1 to position

2 on two alternative paths.Let us calculate the amount of work done on each path. The force exerted to the

particle by my hand is just opposite of the force field -F(r). If it was not the case,the particle would accelerate. The moving is thought to happen quasi-statically withoutacceleration.

Let us calculate my work for the two alternate paths:

W1 =

r2∫r1

(−F)dr

path1

W2 =

r2∫r1

(−F)dr

path2

(2.41)

27

Figure 2.8: Integration on two alternative paths

In general case W1 and W2 are not equal. However, in some special cases they maybe equal for any two paths. Imagine that our force field is such, that W1 and W2 areequal. In this case a closed loop path can be made which starts with path 1 and returnsto the starting point on path 2. Since the opposite direction passage turns W2 to itsnegative, ultimately the closed loop path will result in zero value. That special forcefield where the integral is zero for any closed loop is considered CONSERVATIVE forcefield. In formula: ∮

F(r)dr = 0 (2.42)

At conservative force field, one has to choose a reference point. All other destinationpoints can be characterized with the amount of the work done against the force field toreach the destination point. This work is considered the potential energy (Epot) of thepoint relative to the reference point:

Epot(r) =

r∫ref

(−F(r,))dr, = −r∫

ref

F(r,)dr, (2.43)

The reference point can be chosen arbitrarily, however it is worth considering the practicalaspects of the problem.

Due to the fact that the reference point is arbitrary, the value of the potential energyis also indefinite since direct physical meaning can only be associated to the variation ofthe potential energy. In other words, the individual potential energy values of any two

28

points can be altered by changing the reference point, but the difference of the potentialenergy values does not change.

Now the work done against the forces of the field between r1 and r2 points can beexpressed:

r2∫r1

(−F(r))dr =

ref∫r1

(−F(r))dr +

r2∫ref

(−F(r))dr = −r1∫

ref

(−F(r))dr +

r2∫ref

(−F(r))dr

(2.44)

The last two integrals are the potential energies of r1 and r2 points respectively.

r2∫r1

(−F(r))dr = Epot(r2)− Epot(r1) (2.45)

2.10 Conservation of the mechanical energy

(Mechanikai energia megmaradasa)Mechanical energy consists of kinetic and potential energy by definition. Earlier in

this chapter the theorem of work was stated. Work done on a particle equals the variationof its kinetic energy. In addition F(r) could be any kind of force.

r2∫r1

F(r)dr = Ekin2 − Ekin1 (2.46)

Later the potential energy has been treated.

r2∫r1

(−F(r))dr = Epot(r2)− Epot(r1) (2.47)

Let us switch the sign of the above equation:

r2∫r1

F(r)dr = Epot(r1)− Epot(r2) (2.48)

At potential energy however conservative force field is required. This means that the socalled dissipative interactions are excluded, such as the sliding friction and the drag. Letus make the right hand sides of the relevant equations equal.

Epot(r1)− Epot(r2) = Ekin2 − Ekin1 (2.49)

29

Ordering the equation:

Epot(r1) + Ekin1 = Epot(r2) + Ekin2 (2.50)

Using the conservation of mechanical energy requires conservative force, because this isthe more stringent condition.

Ultimately let us declare again clearly the conservation of mechanical energy: Inconservative system the sum of the kinetic and potential energy is constant in time.Accordingly, these two types of energy transform to each other during the motion, butthe overall value is unchanged. In contrast to this when dissipative interaction emergesin the system, the total mechanical energy gradually decreases by heat loss.

In this chapter the concept of work end energy have been used extensively. To improveclarity, the following statement needs to be declared: Work is associated to some kindof process or action. Energy on the other hand is associated to some kind of state of asystem, when not necessarily happens anything, but the capacitance to generate actionis present.

2.11 Energy relations at harmonic oscillatory mo-

tion

The equation of motion of the harmonic oscillation is as follows:

F (t) = −Dx(t) (2.51)

Here Dis direction coefficient of the spring on which a particle with mass m oscillates.In the chapter of kinematics the harmonic oscillatory motion has been introduced,

and the basic formulae have all been derived. The following relation was recovered:

a(t) = −ω2x(t) (2.52)

Let us multiply it with mass:

ma(t) = −mω2x(t) (2.53)

The left hand side of the equation is the force affecting the particle.

F (t) = −mω2x(t) (2.54)

By comparing the two expressions of the force one can conclude as follows:

D = mω2 ω =√

Dm

30

The harmonic oscillatory motion is a conservative process. This means that the totalmechanical energy (the sum of kinetic and the potential energy) should be constant.

Etot =1

2Dx2 +

1

2mv2 = const (2.55)

Let us verify the above statement with the concrete formulas of displacement and velocity:

Etot =1

2DA2 cos2(ωt+ φ) +

1

2mω2A2 sin2(ωt+ φ) = const (2.56)

Now we can proceed on two alternate tracks by substituting the direction coefficient intothe equation and using the most basic trigonometric relation:

Etot =1

2DA2 cos2(ωt+ φ) +

1

2DA2 sin2(ωt+ φ) =

1

2DA2 (2.57)

Or alternatively:

Etot =1

2mω2A2 cos2(ωt+ φ) +

1

2mω2A2 sin2(ωt+ φ) =

1

2mω2A2 (2.58)

By using the velocity amplitude (vmax) defined in the chapter of kinematics one canconclude as follows:

Etot =1

2mv2

max (2.59)

Ultimately we found two alternate formulae for the total mechanical energy. Theseformulae prove that the process is truly conservative, and the total energy may show upeither as potential or kinetic energy. In amplitude position the total energy is storedin the spring as potential (elastic) energy, at zero excursion position the total energy iskinetic energy.

Figure 2.9: Energy relations of the oscillatory motion

In the figure above the energy relations are displayed. The motion takes place underthe solid horizontal line of total energy.

31

2.12 Angular momentum

(Impulzus nyomatek, perdulet)By definition the angular momentum of the particle is the cross product of the position

vector and the linear momentum.

L = r× p

[kgm2

s

](2.60)

2.13 Torque

(Forgato nyomatek)By definition the torque (M) is the cross product of the position vector and the force

affecting the particle.

M = r× F

[kgm2

s2

](2.61)

Let us consider the situation when r and p and F are in the plane of the sheet. Accordingto the definition, both the angular momentum and the torque are normal to the sheet.

Figure 2.10: Both the angular momentum and the torque point into the paper

If the vectors depend on time, one can determine the derivative of the product:

dL

dt=dr

dt× p + r× dp

dt(2.62)

Since drdt

= v and dpdt

= F the above equation can be transformed:

dL

dt= v × p + r× F (2.63)

The first term on the right cancels out because v and p vectors are parallel. Therefore:

dL

dt= r× F (2.64)

32

The product on the right hand side is the torque. Ultimately one can conclude:

dL

dt= M (2.65)

In words: The time derivative of the angular momentum of some particle equals thetorque affecting this particle. (Obviously the reference point of both L and M must bethe same.)

This formula is analogous to that of Newton’s second law, expressed with the linearmomentum. By means of the fundamental theorem of calculus, this formula can beintegrated.

L2 − L1 =

t2∫t1

M(t)dt (2.66)

In words:The variation of the angular momentums is the time integral of the torque affecting

the particle. This integral is called the angular impulse. (Nyomatek lokes)

2.14 Central force field

(Centralis eroter)If the force is collinear with the position vector and the magnitude depends on the

distance alone, then the force field is considered central force field:

F = k(r)r (2.67)

here k is a scalar number which may depend only on the distance from the center.As it has already been calculated:

dL

dt= r× F (2.68)

Let us substitute the central force field:

dL

dt= r× k(r)r (2.69)

The cross product is zero because of the collinear arrangement:

dL

dt= 0 (2.70)

Accordingly, in central force field the angular momentum is constant in time: (L= const)It has the important conclusion. Planets, moons or spacecrafts which orbit their central

33

body in the space also move in the central force field of gravity. Therefore the angularmomentum referred to the central body is constant.

In the chapter of kinematics the concept of areal velocity was introduced in general.Accordingly:

dA

dt=

1

2r× v (2.71)

On the other hand, the angular velocity is:

L = r×mv (2.72)

By combining these two last equations:

dA

dt=

1

2mL = const (2.73)

Ultimately the areal velocity is constant in the central force field.Planetary motion: A meteorite is orbiting the sun on an ellipse trajectory. The

ellipse trajectory is the consequence of the Newton’s gravitational law. The constantareal velocity will make the meteorite travel faster when close to the sun and slowerwhen it is far away. The crosshatched areas in the figure below are equal. So, the motionis far not uniform.

Figure 2.11: Planetary motion with constant areal velocity

34

35

Chapter 3

Dynamics of system of particles -Gyorgy Hars

(Tomegpont rendszer dinamikaja)

3.1 Momentum in system of particles

The subject of analysis will be the system of particles. The system of particles in practicemay consist of several particles (mass points). Each of the particles may travel arbitrarilyin 3D space. The particles may exert force to each other (internal force) and may beaffected by forces originating in the environment (external force).

In mathematical calculations however it is worth reducing the number of particlesto two particles. This way, calculations become much easier without loosing generality.The physical meaning behind the equations becomes even more apparent. At the end ofthe argument the result will be stated in full generality for any number of particles.

Figure 3.1: System of particles

36

The center of mass is the weighted average of the position vectors.

m1r1 +m2r2

m1 +m2

= rc (3.1)

Its time derivative is the velocity of the center of mass.

m1v1 +m2v2

m1 +m2

= vc ptot =∑i

pi = vc

∑i

mi (3.2)

The numerator is the total momentum of the system of particles. So the total momentumcan be expressed as the product of the velocity of the center of mass multiplied by thetotal mass.

Let us make one more time derivation:

m1a1 +m2a2

m1 +m2

= ac (3.3)

Accordingly:

m1a1 +m2a2 = ac

∑i

mi (3.4)

Now consider the Newton equation for m1 and m2:

m1a1 = F1 + F12 m2a2 = F2 + F21 (3.5)

Internal forces show up with double subscript. By substituting the forces to the aboveequation:

(F1 + F12) + (F2 + F21) = ac

∑i

mi (3.6)

Here we have to take into account the fact, that the internal forces show up in pairs andthey are opposite of each other. F12 =-F21. So they cancel out and only the externalforces remain. ∑

i

Fiext = ac

∑i

mi (3.7)

In words: The sum of the external forces accelerates the center of mass. Internal forcesdo not affect the acceleration of the center of mass. This is the theorem of momentum.

If on the other hand the sum of the external forces is zero, the acceleration of thecenter of mass becomes also zero, or in other words, the velocity of the center of mass isconstant. If the velocity of the center of mass is constant, then the total momentum ofthe system of particles will also be constant.

37

So all together, let us state the conservation of momentum: In an isolated mechanicalsystem (in here the sum of the external forces is zero) the total momentum of the systemof particles is constant.

ptot = Const (3.8)

This law can also be used in coordinate components. So if the system of particles ismounted on a little rail cart, and external force parallel with the rail does not affect thesystem, then that component of the total momentum will be constant which is parallelwith the rail. In terms of other directions no any law applies.

3.1.1 Collisions

(Utkozesek)At commonly happening collisions the conservation of momentum is valid because

the system of the two colliding particles represents an isolated mechanical system. Thereare two specific types of collisions, the inelastic and elastic collision. The distinction isbased on the kinetic energy variation during the process.

Inelastic collisions:

(Rugalmatlan utkozes)The two colliding particles get stuck together. The kinetic energy of the system

is partly dissipated. Substantial amount of heat can be generated. Let us write theconservation of momentum:

m1v1 +m2v2 = (m1 +m2)u (3.9)

The velocity after collision (u) results:

m1v1 +m2v2

m1 +m2

= u (3.10)

The “lost” mechanical energy, which has been dissipated to heat, is the difference of thetotal kinetic energy before and after the collision:

∆Eloss =1

2mv2

1 +1

2mv2

2 −1

2(m1 +m2)

(m1v1 +m2v2

m1 +m2

)2

(3.11)

Elastic collision:

(Rugalmas utkozes)Word “elastic” means that the mechanical energy is conserved. Thus, both the mo-

mentum and the kinetic energy are conserved. After collision the particles get separated

38

with different velocities. The velocities before and after the collision are denoted withv1 v2 and u1 u2 respectively. The conservation of momentum follows:

m1v1 +m2v2 = m1u1 +m2u2 (3.12)

The conservation of mechanical energy is also valid. Here the total mechanical energy iskinetic energy since no potential energy is involved.

1

2m1v

21 +

1

2m1v

22 =

1

2m1u

21 +

1

2m1u

22 (3.13)

Let us group terms with subscript 1 to the left and terms with subscript 2 to the righthand side for two equations above.

m1v1 −m1u1 = m2u2 −m2v2 (3.14)

1

2m1v

21 −

1

2m1u

21 =

1

2m2u

22 −

1

2m2v

22 (3.15)

Now factor out m1 and m2 from the equations, multiply the kinetic energy equation withtwo and use the equivalency for the difference of squares:

m1(v1 − u1) = m2(u2 − v2) (3.16)

m1(v1 − u1)(v1 + u1) = m2(u2 − v2)(u2 + v2) (3.17)

Up to this point of discussion the 3D vector equations above are fully valid. Among dotproducts, division operation is impossible. This is due to the fact that reverse directionof the operation is ambiguous.

From this point, the mathematical argument is confined to the central collision only.At central collision the velocities before the collision are parallel with the line between thecenters of the particle. This way the collision process takes place in a single line, and thevelocities before and after the collision will all be 1D vectors in the line of the collision.The 1D vectors are practically plus, minus or zero numbers, and the dot product betweenthese vectors is basically product between real numbers. So from the above equationsthe vector notation will be omitted. Accordingly any division can readily be carried out.

m1(v1 − u1) = m2(u2 − v2) (3.18)

m1(v1 − u1)(v1 + u1) = m2(u2 − v2)(u2 + v2) (3.19)

Let us divide the last equation with the former one:

v1 + u1 = v2 + u2 (3.20)

39

Now group the v terms to the left and u terms to the right hand side:

v1 − v2 = u2 − u1 (3.21)

Multiply the equation with m2:

m2v1 −m2v2 = m2u2 −m2u1 (3.22)

The original equation for momentum conservation is simplified for 1D central collision:

m1v1 +m2v2 = m1u1 +m2u2 (3.23)

Let us subtract the former equitation from the last one. Here m2u2 term will cancel out:

v1(m1 −m2) + 2m2v2 = u1(m1 +m2) (3.24)

Thus u1 can be expressed:

v1m1 −m2

m1 +m2

+2m2

m1 +m2

v2 = u1 (3.25)

Due to symmetry, formula for u2 can be easily derived by switching the subscripts 1 and2.

v2m2 −m1

m1 +m2

+2m1

m1 +m2

v1 = u2 (3.26)

The above formulas are not simple enough to provide plausible results. For this purposesome special cases will be treated separately:

Discussion 1: What if m1 = m2 = m is the case.Basically the masses are equal. Then the final result simplifies to:v2 = u1 and v1 = u2

Accordingly the particles swap their velocities. If on the other hand one of theparticle had zero velocity originally (v1 =0), and the other particle slammed into it withv2velocity. Then :

v2 = u1 and u2 = 0This means that the standing particle will start travelling with the velocity of the

moving particle, and the originally moving particle will stop.

Discussion 2: What if m2 is far larger than m1.−v1 + v2 = u1 and v2 = u2

This is the case when a ball bounces back from the face of the incoming bus. Thevelocity of bus does not change (v2 = u2), and the velocity of ball is reflected plus thespeed of the buss is added.

40

Discussion 3: Billiard ball collision:In this game, the balls are equal in mass, but the collisions are not necessarily central.

Consider the situation when one ball is standing and an equal weight ball collides to itin a skew elastic collision. Let us go back to the original equation with vectors.

The momentum conservation for the present case:

mv = mu1 +mu2 (3.27)

The mechanical energy conservation for the present case:

1

2mv2

1 =1

2mu2

1 +1

2mu2

2 (3.28)

After some obvious mathematical simplifications:

v = u1 + u2 v21 = u2

1 + u22 (3.29)

Figure 3.2: Billiard ball collision

First equation means that the vectors create a closed triangle. The second equationshows that the created triangle is a right angle triangle, since the Pythagoras theorem istrue only then. As a summary, one can say that the balls travel 90 degree angle relativeto each after collision in billiard game.

Ballistic pendulum

(Ballisztikus inga)This is a pendulum with some heavy sand bag on the end of some meter long rope.

The rope is hung on a high fix point, letting the pendulum swing. A simple indicatormechanism shows the highest angular excursion.

41

Figure 3.3: Ballistic pendulum

The pendulum is left to get quiet and hang vertically. Then the gun is fired, thebullet penetrates into the sandbag and get stuck in it. The pendulum starts to swing.The first highest angular excursion is detected. From the above information, the speedof the bullet can be found.

The whole process consists of two steps. In step 1 the bullet collides with the sandbag.Up to this point, conservation momentum is valid but the mechanical energy is notconserving quantity, due to the inelastic collision. After collision in step 2, there are nomore dissipative effects, so conservation of mechanical energy is true. The two relevantequations are as follows:

mv = (m+M)u (3.30)

1

2(m+M)u2 = (m+M)gl(1− cosφ) (3.31)

Here m and M are the mass of the bullet and the sandbag respectively. The v and u arethe speed of the bullet and the speed of the sandbag respectively. The l is the length ofthe rope and g is the gravity acceleration.

By eliminating u from the equations one can readily express the incoming speed ofthe bullet:

m

m+Mv = u (3.32)

1

2(

m

m+Mv)2 = gl(1− cosφ) (3.33)

v =m+M

m

√2gl(1− cosφ) (3.34)

This is an excellent example how careful one must be. If wrongly the whole process isassumed to be conservative, the resulting bullet speed will be some ten meters per secondwhich is roughly hundred times smaller than the real result.

42

3.1.2 Missile motion

(Raketa mozgas)Jet propulsion is the fundamental basis of the missile motion. This is based on the

conservation of momentum. If one tries to hold the garden hose when sprinkling thegarden, one will experience a recoil type force, which is pushing back. This force iscalled “thrust”, and this drives the missiles, aircrafts and jet-skis.

Figure 3.4: Missile motion

The missile ejects mass in continuous flow with the ejection speed (u) relative to themissile. The rate with which the mass is ejected is denoted (µ) and measured in kg/s.The infinitesimal ejected momentum will provide the impulse to the missile:

udm = Fdt (3.35)

udm

dt= F (3.36)

uµ = F (3.37)

So the product of the speed and ejection rate determines the thrust (F ). During themissile motion the thrust is a constant force. As the missile progresses the overall massis continuously reduced by burning the fuel. The equation of motion is as follows:

F = m(t)a (3.38)

Here m(t) is the reducing mass and m0 is the initial mass:

m(t) = (m0 − µt) (3.39)

uµ = (m0 − µt)a (3.40)

The acceleration can be expressed:

uµ

m0 − µt= a(t) (3.41)

43

In order to find out the velocity time function, the above formula needs to be integrated:

v(t) =

t∫0

a(t,)dt, =

t∫0

uµ

m0 − µt,dt, (3.42)

After the integration the velocity function is revealed:

v(t) = u lnm0

m0 − µt(3.43)

The final formula shows that approaching the m0/µ time the speed grows to the infinity.This value can not be reached since there must be a payload on the missile.

3.2 Angular momentum in system of particles

(Tomegpont rendszer impulzus nyomateka)

Figure 3.5: Angular momentum in system of particles

Consider the total angular momentum of a system of particles.

L1 = r2 ×m1v1 L2 = r2 ×m1v2 (3.44)

This is the sum of the angular momentums of the individual particles:

L1 + L2 = Ltot (3.45)

Check out the time derivative oft this equation:

dL1

dt+dL2

dt=dLtot

dt(3.46)

44

In Chapter 2 it has been shown that the derivative of angular momentum is the torqueaffecting the particle. Accordingly the above equation is transformed:

M1 + M2 =dLtot

dt(3.47)

Based on the definition of torque following formulas are true:

M1 = r1 × F1 + r1 × F12 M2 = r2 × F2 + r2 × F21 (3.48)

Let us substitute them to the equation above:

r1 × F1 + r1 × F12 + r2 × F2 + r2 × F21 =dLtot

dt(3.49)

Here we have to use that F12 = -F21 which is the consequence of action reactionprinciple.

r1 × F1 + r1 × F12 + r2 × F2 − r2 × F12 =dLtot

dt(3.50)

Now regroup the left hand side:

r1 × F1 + r2 × F2 + (r1 − r2)× F12 =dLtot

dt(3.51)

If now one looks at the figure above, the fact is readily apparent that the r1-r2 vectorand F12 force vector are collinear vectors, thus their cross product is zero. Therefore thetorques of the internal forces cancels out.

r1 × F1 + r2 × F2 =dLtot

dt(3.52)

The left hand side terms are all the torques of the external forces. So all together thegeneralized statement is as follows:∑

i

Miext =dLtot

dt(3.53)

In words:In system of particles, the time derivative of the total angular momentum is the sum

of the external torques. This statement is called the theorem of angular momentum.Therefore the internal torques are ineffective in terms of total angular momentum.

If on the other hand the total external torque is zero, then the total angular momen-tum is constant. This is the conservation of angular momentum.

Ltot = Const (3.54)

45

In summary: In a system of particles where the total external torque is zero, the totalangular momentum is constant, or in other words it is a conserving quantity.

This law can also be used in coordinate components. So if the system of particlesis mounted on a bearing, and external torque parallel with the axis of the bearing doesnot affect the system, then that component of the angular momentum will be constantwhich is parallel with the axis of the bearing. In terms of other directions no any lawapplies.

3.2.1 The skew rotator

(Ferdeszogu forgas)Consider the figure below. Two equal masses are placed on the ends of a weightless

rod. The center of mass is mounted on a vertical axis, which is rotating freely in twobearings. The angle of the fixture is intentionally not ninety degrees, but a skew acuteangle. The system rotates with a uniform angular velocity. The job is to find out thedeviational torque which emerges, due to the rotation of asymmetric structure.

Figure 3.6: The skew rotator

The origin of the coordinate system is the center of mass. The coordinate system isnot rotating together with the mechanical structure and it is considered inertial system.Gravity cancels out from the discussion, since the center of mass is supported by theaxis, and the gravity does not affect torque to the system. The mechanical setup is inthe plane of the figure. The two position vectors of the particles are (r) and (–r).

Momentums of the particles are p1 and p2. They are normal to the paper sheet.

p1 = m(ω × r) p2 = −m(ω × r) (3.55)

The corresponding L1 and L2 angular momentums are equal, because of the twice neg-ative multiplication:

L1 = r× p1 = r×m(ω × r) L2 = −r× p2 = −r×m(ω × (−r)) (3.56)

46

So the total angular momentum is the sum of these two:

Ltot = 2r×m(ω × r) (3.57)

Now we use the theorem of angular momentum:∑i

Miext =dLtot

dt(3.58)

∑i

Miext =d

dt(2r×m(ω × r)) = 2

dr

dt×m(ω × r) + 2r×m(ω × dr

dt) (3.59)

Now take it into consideration that the derivative of the position is the velocity whichcan be expressed by means of angular velocity vector:

dr

dt= v = ω × r (3.60)

After substitution:∑i

Miext = 2(ω × r)×m(ω × r) + 2r×m(ω × (ω × r)) (3.61)

The first term on the right hand side is zero, because this is a cross product of collinearvectors. The final result comes up immediately.∑

i

Miext = 2r×m(ω × (ω × r)) (3.62)

After following the numerous cross products in terms of direction one can conclude, thatthe torque is pointing out of the paper sheet. The direction of torque is rotating togetherwith the mechanical structure and always perpendicular to its plane. This is deviationaltorque and this emerges because the angular momentum vector is constantly changing,not in absolute value but in direction. This effect is very detrimental to the bearingsdue to the load that it generates. There are cases however when such effect does notshow up. When the angular momentum vector is parallel with angular velocity vectorno deviational torque will emerge. These are called principal axes. In general there arethree perpendicular directions of principal axes.

A new interesting aspect:Imagine that this experiment is carried out on a spacecraft orbiting the Earth. Sud-

denly the bearing and the mechanical axis disappear. How will the mass-rod-mass struc-ture move after this?

Since no external torque affects the system, the angular momentum will be constant.But now the angular velocity vector starts to go around on the surface of a virtual cone.The symmetry axis of such virtual cone is just the angular momentum vector. This kindof motion is called precession.

47

3.2.2 The pirouette dancer (The symmetrical rotator)

(A piruett tancos)In conjunction with the previous section this section could be called as “symmetrical

rotator”. The setup fundamentally similar, the major difference is that the mass-rod-mass system is positioned perpendicularly to the rotation axis.

Figure 3.7: Symmetrical rotator

p1 = m(ω × r) p2 = −m(ω × r) (3.63)

Similarly to the previous section total angular momentum is:

Ltot = 2r×m(ω × r) (3.64)

In present case however the position vectors and the angular velocity vector are normalto each other. Therefore the direction of the angular momentum and the angular velocitywill be both parallel with the rotation axis. Since the direction of the vectors is clear, itis enough to deal with the absolute values only.

Ltot = 2mr2ω (3.65)

Notice: The emerging 2mr 2 quantity is the so called the moment of inertia. Find detailsin Chapter 4.

Let us see what happens when the pirouette dancer pulls his arms in. Then r2 ≤ r1

is the case.The conserving quantity is the angular momentum.

2mr21ω1 = 2mr2

2ω2 (3.66)

48

From here:

ω1

(r1

r2

)2

= ω2 (3.67)

Accordingly, the dancer is spinning much faster.Let us check out how the kinetic energy changed during the pirouette. Clearly the

difference between the final state and the initial state should be calculated.

∆Ekin = 21

2m(r2ω2)2 − 2

1

2m(r1ω1)2 = m(r2

2ω22 − r2

1ω21) (3.68)

Now let us substitute ω2:

∆Ekin = m

(r2

2ω21

(r1

r2

)4

− r21ω

21

)(3.69)

∆Ekin = mr21ω

21

((r1

r2

)2

− 1

)(3.70)

The kinetic energy increased. This only could happen due to the work done by thedancer. The force of the dancer is internal force, so the kinetic energy of the system ofparticles can be changed by internal forces too. In order to calculate the work done, thefirst task is to find out the force function. The force against which the dancer pulls hisarm is the centrifugal force. The centrifugal force is an inertial force which emerges inspinning coordinate system only. See details in chapter 5.

Let us find out the angular velocity as a function of the arbitrary position:

ω1

(r1

r

)2

= ω(r) (3.71)

The formula of centrifugal force in present case is:

Fcf (r) = mrω2 = mrω21

(r1

r

)4

= mω21r

41

1

r3(3.72)

The work carried out by the dancer is the integral of the dancer’s force, which is justopposite of the centrifugal force.

Wdancer = 2

r2∫r1

(−Fcf )dr = −2mω21r

41

r2∫r1

dr

r3(3.73)

49

Now let us study the integral alone. Use the fundamental theorem of calculus:

r2∫r1

dr

r3=

[−1

2r−2

]r=r2r=r1

=1

2

(1

r21

− 1

r22

)(3.74)

After substitution:

Wdancer = mω21r

41

(1

r22

− 1

r21

)= mω2

1r21

((r1

r2

)2

− 1

)(3.75)

This result completely matches the growth of the kinetic energy calculated earlier, so theincrease of the kinetic energy is the consequence of the work done by the dancer.

3.3 Discussion of the total kinetic energy in the sys-

tem of particles

In contrast to the earlier habit the system of particles will be treated in general up to npieces of particles.

Particle positions are defined by ri position vectors relative to the origin of the coordi-nate system. The center of mass has the position vector rc. Now, we introduce the centerof mass as a new coordinate origin. In this new coordinate system the correspondingposition vectors will be denoted ρi.

Figure 3.8: Kinetic energy in the system of particles

Therefore the following equation is true:

ri = rc + ρi (3.76)

After derivation similar rule is found among the velocities:

vi = vc + υi (3.77)

50

Here vi and vc are the velocity of i-th particle and the center of mass respectively in theoriginal coordinate system. . The υi is the velocity of the i-th particle in the center ofmass coordinate system.

Take a look at the total kinetic energy of the system in the laboratory coordinatesystem:∑

Ekin =n∑i=1

(1

2miv

2i

)=

n∑i=1

(1

2mi(vc + υi)

2

)=

n∑i=1

(1

2mi(v

2c + υ2

i + 2vcυi)2

)(3.78)

Let us discuss the individual terms separately:∑Ekin =

n∑i=1

(1

2miv

2c

)+

n∑i=1

(1

2miυ

2i

)+

n∑i=1

(mivcυi) (3.79)

In the first term the velocity of center of mass can be factored out:

n∑i=1

(1

2miv

2c

)=

1

2

(n∑i=1

mi

)v2c =

1

2mtotv

2c (3.80)

The first energy term is associated with the velocity of the center of mass in the laboratorycoordinate system. The second term is associated with the velocities relative to the centerof mass.

The third term gives zero result. The proof is as follows: From the third term vc canbe factored out:

n∑i=1

(mivcυi) = vc

n∑i=1

(miυi) = 0 (3.81)

The right hand side summa is the total momentum in the center of mass coordinatesystem. The total momentum is the product of the total mass multiplied with thevelocity of the center of mass. But in the center of mass coordinate system the velocityof center of mass is obviously zero. QED.

Accordingly the total kinetic energy of the system of particles can be divided into twoparts: Kinetic energy associated to the velocity of the center of mass in the laboratorycoordinate system, and kinetic energy associated to the velocities in the center of masscoordinate system. In formulas:∑

Ekin =1

2mtotv

2c +

n∑i=1

(1

2miυ

2i

)(3.82)

The first term can be changed external force only, due to the theorem of momentum.The second term can be changed by any type of forces.

51

Chapter 4

Dynamics of rigid body - GyorgyHars

(Merev testek dinamikaja)

4.1 Moment of inertia

(Tehetetlensegi nyomatek)Up to this point only the motion of particles was discussed. The rigid bodies however

are extended objects. The kinematics of extended objects contains major distinctionsin terms of motion. Translation means that the points of the object travel the sametrajectory except for a shift, by which all the trajectories cover each other. The rotationhowever contains circular trajectories with different radii.

Rigid body is a special system of particles. Here the positions of the particles arefixed relative to each other. In addition the geometrical shape of the body is constant,and it is independent of the mechanical load.

In present discussion a majorly simplified theory will be treated. The simplificationsare as follows:

• The origin of the coordinate system will be the center of mass of the rigid body.

• The rotation will take place around principal axis, thus the angular momentumand the angular velocity vectors are parallel.

Let us consider the mass-rod-mass structure in chapter 3. The angular momentum ingeneral case can be expressed as follows:

Ltot = 2r×m(ω × r) (4.1)

52

In present case however the position vectors and the angular velocity vector are normalto each other. Therefore the direction of the angular momentum and the angular velocitywill be parallel. If two vectors are parallel then a scalar multiplier can be found betweenthem. (In the general case tensor describes the relation.)

Ltot = 2mr2ω = θpairω (4.2)

The scalar multiplier is called the “moment of inertia”. This is denoted with the Greekletter θ. So for a symmetrical pair of particles the moment of inertia is: θpair = 2mr2.

For one single piece of particle this value is: θparticle = mr2.Let us calculate the moment of inertia for a large diameter thin wheel with the radius

r.

θwheel =

∫r2dm (4.3)

Since the radius is constant this can be factored out.

θwheel = r2

∫dm = mr2 (4.4)

Here m means the total mass of the wheel.Now consider an arbitrary rotationally symmetric object. Imagine as this object

consisted of several wheels, each of them with dm mass.

Figure 4.1: Moment of inertia of a rotationally symmetric object

θgeneral =

∫V

dθ =

∫V

r2dm (4.5)

If the density of the object is constant then dm = ρdV . With this:

θgeneral =

∫V

r2dm = ρ

∫V

r2dV (4.6)

53

By means of the above formula the moment of inertia for several symmetrical objectscan be calculated:

• Tube: Here r = R =Const.

Figure 4.2: Tube

Since R is constant this can be factored out.

θtube = ρ

∫V

R2dV = ρR2

∫V

dV = R2ρV = mR2 (4.7)

• Cylinder or disc:

Let us put together the cylinder from several tubes with increasing radii. Here dV =2rπldris the case, where l is the length of the cylinder.

Figure 4.3: Cylinder

θcylinder = ρ

R∫0

r22rπldr = ρ2πl

R∫0

r3dr = ρ2πl

[r4

4

]r=Rr=0

=π

2ρlR4 (4.8)

54

On the other hand the mass of the cylinder is the product of its volume and density.

m = R2πlρ (4.9)

Divide them:

θcylinderm

=π

2

ρlR4

R2πlρ=R2

2(4.10)

From here the moment of inertia is found:

θcylinder =1

2mR2 (4.11)

• Spherical shell. (Hollow sphere)

The radius is denoted R, the thickness of the shell is S.

Figure 4.4: Spherical shell

According to the general expression:

θgeneral = ρ

∫V

r2dV (4.12)

The function to be integrated is: r = (R cosφ)2

θshell = ρ

φ=π2∫

φ=−π2

(R cosφ)2dV (4.13)

55

The dV volume here is as follows: dV = (Rdφ)S(2πR cosφ)

θshell = ρ

φ=π2∫

φ=−π2

(R cosφ)2(Rdφ)S(2πR cosφ) =ρR4S2π

φ=π2∫

φ=−π2

cos3 φdφ (4.14)

Let us deal with the integral for a while separately. The antiderivative is as follows:

−π2∫

−π2

cos3 φdφ =

[1

12(9 sinφ+ sin 3φ)

]φ=π2

φ=−π2

=4

3(4.15)

By means of the integral value the moment of inertia can be expressed:

θshell =4

3ρR4S2π (4.16)

On the other hand the total mass is the product of the density and the volume.

m = ρ4R2πS (4.17)

Divide them.

θshellm

=4

3

ρR4S2π

ρ4R2πS=

2

3R2 (4.18)

The result is clear.

θshell =2

3mR2 (4.19)

• The bulky sphere

This can be composed of spherical shells with increasing radii. The infinitesimal volumeis the product of the surface of the inflating sphere and the infinitesimal thickness (dr):dV = 4r2πdr

θsphere = ρ

R∫0

2

3r2dV = ρ

R∫0

2

3r2(4r2πdr) = ρ

8π

3

R∫0

r4dr = ρ8π

3

[r5

5

]r=Rr=0

= ρ8π

15R5

(4.20)

On the other hand the mass of the sphere is the product of its volume and the density:

m =4R3π

3ρ (4.21)

56

Divide them:

θspherem

= ρ8π

15R5 3

4R3πρ=

2

5R2 (4.22)

Thus the result is:

θsphere =2

5mR2 (4.23)

• Stick or rod.

The stick is put together from several particles with increasing radii. Early in this chapterit was shown that in case of particles the moment of inertia is as follows: θparticle = mr2.In addition the infinitesimal volume is equal todV = Adr.

Figure 4.5: Moment of inertia for a rod

θrod = ρ

∫V

r2dV = ρ

l∫0

r2Adr = ρA

l∫0

r2dr = ρA

[r3

3

]r=lr=0

= ρA

(l3

3

)=

1

3ρAl3 (4.24)

The mass of the rod can also be expressed:

m = Alρ (4.25)

Accordingly:

θrod =1

3ml2 (4.26)

Now, we should find out how much the moment of inertia is through the center of mass.A symmetrically rotating rod can be composed of two half size rods. Therefore halflength and half mass rod will be used and finally doubled:

θCenter = 2

(1

3

m

2(l

2)2

)=

1

12ml2 (4.27)

57

• Steiner’s theorem

If one calculates the difference between the moment of inertia of the rod with differentaxes, an interesting relation can be revealed:

θrod − θcenter = (1

3− 1

12)ml2 =

1

4ml2 = m(

l

2)2 (4.28)

The difference is the total mass multiplied with the square of the distance between theaxes. The calculated result is a special case of a far more general law which is called theSteiner’s theorem. This general law states that moment of inertia is smallest throughthe axis of center of mass. In addition the moment of inertia for any parallel axes canbe calculated as follows:

θparallel = θcenter +mD2 (4.29)

where m and D are the mass and the distance between the axes respectively.

4.2 Equation of motion of the rigid body:

(Merev test mozgasegyenlete)Let us take a look at the equation of motion of an extended rigid body. Two laws

will be used, such as the theorem of momentum∑i

Fiext = ac

∑i

mi (4.30)

and the theorem of angular momentum.∑i

Miext =dLtot

dt(4.31)

In case of rigid body the theorem of momentum is written with the relevant quantities:∑i

Fi = mtotac (4.32)

The theorem of angular momentum is written by means of the moment of inertia. Sub-script c means that the moment of inertia is calculated to a principal axis through thecenter of mass. ∑

i

Mi =d(θcω)

dt= θc

d(ω)

dt= θcβ (4.33)

58

So all together the equation of motion consists of two vector equations:∑i

Fi = mtotac (4.34)

∑i

Mi = θcβ (4.35)

They represent in principle six scalar equations, but in the most cases the system of equa-tions contains only three equations. This is due to the fact that mechanical problems aremostly 2D plane problems. This case the theorem of momentum contains two variablesx and y components in the plane of the sheet, and the theorem of angular momentumcontains z component normal to the sheet.

The time derivative of angular velocity is the angular acceleration β. In presentsimplified situation the angular acceleration is always parallel with the angular velocity.

4.2.1 Demonstration example 1.

The yoyo Find the acceleration and the tension of the rope (rope force). The mass is3 kg, g= 10m/s2 the radii are R = 6 cm and r = 3cm

Figure 4.6: The yoyo

The equations of motion:

mg −K = ma (4.36)

Kr = (1

2mR2)(

a

r) (4.37)

In the second equation the first factor is the moment of inertia for cylinder, the second

factor is the angular acceleration. From the second equation K = 12ma(Rr

)2comes out.

59

This and the first equation are added so K cancels out. The mathematical steps are asfollows:

mg =

[1 +

1

2

(R

r

)2]ma (4.38)

a =1

1 + 12

(Rr

)2 g (4.39)

a =1

1 + 12

(63

)2 g =1

3g =

10

3

m

s2(4.40)

The rope force results as the acceleration is substituted to the expression of K.

K =1

2ma

(R

r

)2

=1

2m

(R

r

)21

1 + 12

(Rr

)2 g =mg

1 + 2(rR

)2 =3 · 10

1 + 2(

36

)2N = 20N

(4.41)

4.2.2 Demonstration example 2.

Cylinder on the table Find the accelerations and the tension of the rope. Determinethe smallest possible static friction coefficient, which provides sliding free rolling for thecylinder. m = 3 kg, M = 8 kg.

Figure 4.7: Cylinder on the table

Equation of motion for the block:

I. mg −K = ma

Equations of motion for the cylinder:

60

II. Mg −N = 0

III. K + S = M a2

IV. KR− SR = (12MR2)( a

2R)

In addition there is a condition for the static friction force:

V. S ≤ µstN

IV. K − S = 14Ma

IV.+III. 2K = 34Ma

IV.+III. K = 38Ma

I.+IV.+III. mg = 38Ma+ma

I.+IV.+III. mg = (38M +m)a

I.+IV.+III. a = m38M+m

g = 8m3M+8m

g

a =8m

3M + 8mg =

8 · 33 · 8 + 8 · 3

g =1

2g = 5

m

s2(4.42)

Let us find the tension of the rope:

IV.+III. K = 38Ma = 3

8M 8m

3M+8mg = 3mM

3M+8mg = 3·3·8

3·8+8·310 = 15N

Now determine the friction force:

-IV. −K + S = −14Ma

IV.+III. K = 38Ma

(IV.+III).-IV. S = 38Ma− 1

4Ma = 1

8Ma = mM

3M+8mg = 5N

The static friction coefficient is the last to deal with:

V. S ≤ µstN

II. Mg −N = 0

61

II. Mg = N

After substitution:

V. S ≤ µstMg

V. mM3M+8m

g ≤ µstMg

V. m3M+8m

= 33·8+8·3 = 1

16=≤ µst

The summary of the solution is the following:The acceleration of the block is a = 5 m/s2, The acceleration of the cylinder is a/2

= 2,5 m/s2. The tension of the rope is K = 15 N, the static friction force is S = 5 N.The minimum required static friction coefficient is µst = 1/16 = 0.0625.

4.3 Kinetic energy of the rigid body

(Merev test mozgasi energiaja)The motion of the rigid body represents kinetic energy. In chapter 3 the kinetic

energy of system of particles was discussed. Let us extend the validity of this discussionto the rigid bodies. ∑

Ekin =1

2mtotv

2c +

n∑i=1

(1

2miυ

2i

)(4.43)

The first term on the right hand side represents the kinetic energy of the rigid bodydue to the speed of the center of mass. The second term is associated with the motionrelative to the center of mass. This is typically the rotation in case of rigid bodies,therefore υi = ω × ri. Since they are normal to each other the absolute value of thevelocity is υi = ωri. Let us substitute to the formula of the kinetic energy.∑

Ekin =1

2mtotv

2c +

n∑i=1

(1

2miω

2r2i

)=

1

2mtotv

2c +

1

2ω2

n∑i=1

(mir

2i

)(4.44)

The last summa term on the right hand side is the moment of inertia of the rigid body.The total kinetic energy of a rigid body can be written as follows:∑

Ekin =1

2mtotv

2c +

1

2θcω

2 (4.45)

So the total kinetic energy consists of two terms. One of them is in conjunction withthe translation of the center of mass, the other one is related to the rotation around thecenter of mass.

sectionCorrespondence between translation and rotation in the framework of the sim-plified model

62

Translation RotationLinear momentum p = mv Angular momentum L = θCω

Force F = dpdt

= ma Torque M = dLdt

= θCβKinetic energy Ekin = 1

2mv2 Kinetic energy Ekin = 1

2θCω

2

Power P = Fv Power P = MωThe corresponding quantitiesLinear momentum kg.m/s p L kg.m2/s Angular momentumVelocity m/sv ω 1/s Angular velocityForce kg.m/s2 = NF M kg.m2/s2 =Nm TorqueMass kg m θC kg.m2 Moment of inertiaAcceleration m/s2a β 1/s2 Angular acceleration

63

Chapter 5

Non-inertial (accelerating) referenceframes - Gyorgy Hars

(Gyorsulo vonatkoztatasi rendszerek)So far dynamics has been treated in inertial systems. The inertial systems are equiv-

alent, which means that the equations of physical laws show up in the same shape inall inertial systems. There are practical aspects however which make the applicationof the non-inertial reference frames necessary. In here the shape of equation of motionchanges relative to that in inertial system. New terms, the inertial forces will appear inthe equation of motion in addition to the real forces. The real forces are associated withreal interactions with other objects.

5.1 Coordinate system with translational accelera-

tion

Figure 5.1: Coordinate system with translational acceleration

64

There are two coordinate systems. One of them is an inertial system, the other one isa coordinate system with translational acceleration. The position vector of the origin ofthe accelerating system is r0. The position, the velocity and the acceleration in the twosystems are generated by consecutive derivations. The vectors in the accelerating systemare denoted with the subscript “rel” standing for relative.

r = r0 + rrel (5.1)

v = v0 + vrel (5.2)

a = a0 + arel (5.3)

Let us multiply the last equation with the mass m.

ma = ma0 +marel (5.4)

The left hand side of the equation is the sum of the real forces. Regroup the terms.

Fsum + (−ma0) = marel (5.5)

The second term is called inertial force.

Finertial = −ma0 (5.6)

So altogether the sum of the real forces should be completed with the inertial force. Thisway the equation of motion can formally be handled just like in the inertial system.

Example: A simple pendulum is hanging in a uniformly accelerating train. Find theangle of the rope in stationary state relative to the vertical direction. The rope force isdenoted K the angle is ϕ.

Inertial system approach: In practice the observer is standing on the ground next tothe train.

↑ K cosφ−mg = 0 (5.7)

→ K sinφ = ma (5.8)

65

Figure 5.2: Inertial system approach

The angle can be determined accordingly: tgφ = ag

Accelerating system approach: In practice the observer is on the train.

Figure 5.3: Accelerating system approach

66

↑ K cosφ−mg = 0 (5.9)

→ K sinφ+ (−ma) = 0 (5.10)

The result is obviously the same as earlier. The point is that by choosing acceleratingsystem, the equation of motion becomes equilibrium equation, which easier to handle inmost cases.

5.2 Coordinate system in uniform rotation

All planets rotate thus this phenomenon is very frequent in nature. Let us have twocoordinate systems with common origin, the inertial system (x, y, z with the unit vectorsi, j, k) and the uniformly rotating system (p, q, s with the unit vectors e1, e2, e3). Thei, j, k vectors are constant in time since they belong to an inertial system, but the e1,e2, e3 vectors rotate together with the spinning system.

Figure 5.4: Coordinate system in uniform rotation

Consider an r vector in both coordinate systems:

r = xi + yj + zk = pe1 + qe2 + se3 (5.11)

Let us derivate it:

dr

dt=dx

dti +

dy

dtj +

dz

dtk = (

dp

dte1 + p

de1

dt) + (

dq

dte2 + q

de2

dt) + (

ds

dte3 + s

de3

dt) (5.12)

Regroup the right hand side:

dr

dt= (

dp

dte1 +

dq

dte2 +

ds

dte3) + (p

de1

dt+ q

de2

dt+ s

de3

dt) (5.13)

67

The first three terms contain the derivative in the spinning system. Here it is worthintroducing a notation for the derivation in the spinning system. This will be similar tothe usual derivation symbol, but instead of d a Greek letter δ is used.

δr

δt=dp

dte1 +

dq

dte2 +

ds

dte3 (5.14)

In the kinematics chapter it has been shown that the velocity due to spinning can bewritten as follows:

de1

dt= ω × e1

de2

dt= ω × e2

de3

dt= ω × e3 (5.15)

So altogether:

dr

dt=δr

δt+ pω × e1 + qω × e2 + sω × e3 (5.16)

Let us factor out the angular velocity:

dr

dt=δr

δt+ ω × (pe1 + qe2 + se3) =

δr

δt+ ω × r (5.17)

So the rule of transformation between the derivatives in the inertial and in the spinningsystem is as follows:

dr

dt=δr

δt+ ω × r (5.18)

This has a direct consequence to the velocities: v = vrel + ω × rIf the above transformation rule is applied to the velocity vector then the relation

between the accelerations is recovered:

dv

dt=δv

δt+ ω × v (5.19)

a =δ

δt(vrel + ω × r) + ω × (vrel + ω × r) =

δvrel

δt+ ω × δrrel

δt+ ω × vrel + ω × (ω × r)

(5.20)

The following notations are used: arel = δvrel

δtvrel = δrrel

δt

By means of which the acceleration is expressed:

a = arel + 2ω × vrel + ω × (ω × r) (5.21)

Regroup the above formula and express the relative acceleration.

arel = a + 2vrel × ω + ω × (r× ω) (5.22)

68

Multiply the equation with mass.

marel = ma + 2mvrel × ω +mω × (r× ω) (5.23)

This equation has an important message. The equation of motion in a spinning system isvery much similar to that in the inertial system. The major difference is that additionalterms appear which are called the inertial forces. These forces are not exerted by someother object, but they are due to the fact that the coordinate system is spinning. Sowhen solving problem in a spinning system these inertial force should be added to thereal forces. Take a closer look at these forces.

marel = F + FCoriolis + FCentrifugal (5.24)

The second term is called Coriolis force. This force only appears if the particle is movingrelative to the spinning system. If we are on a spinning contraption in the amusementpark Coriolis force can be felt if we move our hand perpendicular direction to the axisof spinning.

FCoriolis = 2mvrel × ω (5.25)

Coriolis force does not make any work because its direction is normal to the relativevelocity. This only deflects the moving objects.

The third term is called the centrifugal force. This force affects also the stationaryobjects in the spinning system and shows up in a turning car. Its direction points awayfrom the axis of rotation.

FCentrifugal = mω × (r× ω) (5.26)

Therefore the absolute value of the centrifugal force depends on the distance from therotational axis and proportional with the square of the angular velocity.

FCentrifugal = mrω2 (5.27)

In technology it is a major limiting factor at manufacturing turbines and fast spinningmotors.

5.2.1 Earth as a rotating coordinate system

It is well known that planet Earth rotates with one day period. Accordingly inertialforces appear. In most cases these are ignored, but in some special cases these forcesmake major qualitative changes in the physical processes.

69

The effects of the centrifugal force

Figure 5.5: The effects of the centrifugal force on planet Earth

The gravity force and the centrifugal force affect the object together. The resultingfree fall acceleration will increase as one goes closer to the poles. At higher latitude weare closer to the rotational axis therefore the centrifugal force is smaller. The vertical isdefined by the resulting direction of both effects. If planet Earth ceased to rotate strangeeffect would follow. The water of the oceans would move to the poles, and the oceanfloor close to the equator would be a huge desert.

Let us calculate the effect of the rotation on the equator numerically: The accelerationassociated with the centrifugal effect can be calculated as follows:

acentrifugal = Rω2 = 6, 37 · 106 4π2

(24 · 3600)2 = 3, 4 · 10−2m

s2(5.28)

The radius of Earth is denoted R which is 6370 km. The duration of a normal day showsup in the denominator in seconds. If the acceleration value is compared to the nominal

70

value of the free fall acceleration we find that it is around one third of a percent.

acentrifugalg

=3, 4 · 10−2

9, 81= 3, 4 · 10−3 (5.29)

The effects of the Coriolis force

Figure 5.6: Decomposition of the omega vector on the northern hemisphere

Since the angular velocity is a vector, this vector can be decomposed to two componentsas shown in the figure. The two resulting omega vector components are the horizontaland vertical component. On the surface of the Earth one makes rotation around two axessimultaneously. On the northern hemisphere there is an omega vector which is lying onthe ground horizontally and points to North, and another one which is standing out ofthe ground vertically and points upside direction. Each of these rotations has differentCoriolis effects.

ω = ωv + ωh (5.30)

FCoriolis = 2mvrel × ω = 2mvrel × (ωv + ωh) = 2mvrel × ωv + 2mvrel × ωh (5.31)

71

Let us study them separately.The first and second terms on the right hand side are denoted as Coriolis 1 and 2

forces.

FC1 = 2mvrel × ωv FC2 = 2mvrel × ωh (5.32)

The effects of Coriolis 1 force Coriolis 1 force is associated with the vertical omegavector.

• The hurricanes

The first effect to be treated is the whirling motion of the hurricanes, which is a counterclockwise (CCW) rotation on the northern hemisphere. Imagine that at certain spotthe air warms up more than in the surrounding places. Here the air lifts up and thesurrounding air will horizontally move to the place where the airlift occurred. This willcause an air flow vector field, in which the horizontal velocity vectors will all point tothe place of the airlift in

cylindrically symmetric arrangement. In order to describe the situation quantitativelya simplified mechanical model is introduced.

Consider a huge hypothetical frozen ocean on the northern hemisphere which is com-pletely flat and the ice is free of any friction. We put down a particle on the ice (imagineit weights some kilograms) and pull the particle with vrel speed by means of a weight-less ideal thread through several kilometers toward the center. The initial and the finalpositions of the particle are R1 and R2 respectively.

Figure 5.7: Elementary model for hurricane formation

The Coriolis 1 force is as follows:

FC1 = 2mvrel × ωv (5.33)

The associated torque to the center can be written:

M = r× FC1 = 2mr× (vrel × ωv) (5.34)

72

Now the following mathematical rule is used: a× (b× c) = b(ac)− c(ab)By applying this rule to the case we find the following outcome:

M = 2m (vrel(r · ωv)− ωv(r · vrel)) (5.35)

The first term in the parenthesis cancels out since the factors in the dot product arenormal to each other.

M = −2mωv(r · vrel) (5.36)

Now we use the integral form of the theorem of angular momentum. It declares that theangular impulse equals the variation of the angular momentum.

L2 − L1 =

t2∫t1

M(t)dt (5.37)

For present case:

L2 − L1 = −2mωv

t2∫t1

(r · vrel)dt = −2mωv

t2∫t1

r · (vreldt) (5.38)

The integration by time is switched to position integral by using the fact: vreldt = dr.

L2 − L1 = −2mωv

R2∫R1

r·dr = 2mωv

R1∫R2

r·dr (5.39)

After integration we find the result:

L2 − L1 = mωv(R21 −R2

2) (5.40)

On the other hand it is known that the angular momentum can be expressed as theproduct of the angular velocity and the moment of inertia. The angular velocity isdenoted as Ω.

L2 − L1 = Ω(mR22) (5.41)

The right hand sides are equal

Ω(mR22) = mωv(R2

1 −R22) (5.42)

Finally the angular velocity of the rotation can be expressed:

Ω = ωv(R2

1

R22

− 1) (5.43)

73

Back to the hypothetical experiment, it can be concluded that the mass on the threadwill rotate around the central position. The ratio of the initial and the final radii willdetermine the angular velocity of the rotation.

The result above for the angular velocity can also be interpreted in the inertial sys-tem. In this approach the law to be used is the conservation of the angular momentum.The result in the inertial system is very much similar to the actual formula except forminus one in the parenthesis, which is missing in inertial system approach. In rotat-ing coordinate system the angular velocity of the rotating coordinate system should besubtracted.

When the speed of the hurricane is of interest this can readily be expressed:

vhurricane = R2Ω = ωvR2

[(R1

R2

)2 − 1

](5.44)

The angular frequency of the Earth’s rotation at 45 degree latitude is as follows:

ωv =2π

24 · 3600sin 45o = 5, 13 · 10−5 1

s(5.45)

Let us take some realistic numerical values. Assume that R1 =400 km and R2 =100 km.Substitute these values to find out the wind speed in the hurricane.

vhurricane = 5, 13 · 10−5 · 105 [16− 1] = 77m

s= 277

km

h(5.46)

The result is realistic though the physical model was the most basic possible.

• The Foucault pendulum

The second effect to be treated is famous Foucault pendulum experiment. Leon FoucaultFrench physicist made the experiment in 1851. A 67 meter long pendulum was hung upin the Pantheon in Paris. As the pendulum was swinging for several hours the plain ofthe oscillation was gradually turning. The stick at the end of the pendulum was drawinga rosette shape figure into the sand on the floor. The angular velocity of the turningwas measured to be 11.29 degrees per hour. Which value gave an exact match with theactual vertical angular velocity ( ωv) in Paris. Since Paris is at the 48.83 degree latitude,the sine of this angle multiplies the total angular velocity of planet Earth.

ωv =15o

hoursin 48, 83o =

11, 29o

hour(5.47)

Let us discuss the motion of the pendulum in the rotating coordinate system of planetEarth. The problem is considered to be two dimensional. The pendulum is treated inlinear approximation. The equation of motion is as follows:

md2r

dt2= −Dr +mrω2

v + 2mdr

dt× ωv (5.48)

74

At little angular excursions the motion can be approximated with a horizontal planemotion. Therefore D = mg

land l is the length of the pendulum. On the right hand side

the first term is the returning force due to gravity, the second term is the centrifugalforce and the third one is the Coriolis force. After rearranging the equation the followingstate is reached:

d2r

dt2+ (ω2

0 − ω2v)r + 2ωv ×

dr

dt= 0 (5.49)

Here ω20 = g

l. This equation is difficult to handle due to the cross product. At this point

it is worth to consider the variable as a complex number rather than a two dimensionalvector. The complex calculation provides a straightforward means to rotate a complexnumber. Let us switch to z as a complex variable. By this way the following equation isrecovered: Here j is the imaginary unit.

d2z

dt2+ 2jωv

dz

dt+ (ω2

0 − ω2v)z = 0 (5.50)

This is an ordinary second order differential equation which is widely used. The solutionis looked for in the following form: z(t) = Aeλt. After substitution the equation is asfollows:

λ2 + 2jωvλ+ ω20 − ω2

v = 0 (5.51)

By solving this equation the roots emerge:

λ12 =−2jωv ±

√−4ω2

v − 4(ω20 − ω2

v)

2= j(−ωv ± ω0) (5.52)

Ultimately there are two linearly independent solutions:

z1(t) = Aej(−ωv+ω0)t z2(t) = Aej(−ωv−ω0)t (5.53)

Any linear combination of the above solutions is a valid solution. For the sake of sim-plicity wee choose the average of the above:

z(t) =1

2(z1 + z2) = Ae−jωv

ejω0t + ejω0t

2= Ae−jωv cos(ω0t) (5.54)

The real and the imaginary parts give the solutions:

z(t) = A cos(ωvt) cos(ω0t)− jA sin(ωvt) cos(ω0t) (5.55)

Finally the two coordinates of the motion can be written:

x(t) = A cos(ωvt) cos(ω0t) y(t) = −A sin(ωvt) cos(ω0t) (5.56)

75

In inertial system approach the situation is completely different. The plane of the oscil-lation is constant and only planet Earth rotates under the experiment with ωv angularvelocity. This makes possible that basically a kinematical solution is given to this prob-lem.

Figure 5.8: The Foucault pendulum in inertial system

The inertial coordinate system is characterized by the e1, e2 unit vectors. Therotating coordinate system, (which rotates together with Earth) is characterized by i, junit vectors.

The oscillation takes place along the e1 axis therefore it is written as follows:

r = Ae1 cos(ω0t) (5.57)

Looking from the rotating system the e1 unit vector seems turning to the negative ϕangle direction. Accordingly:

e1 = i cos(−φ) + j sin(−φ) = i cosφ− j sinφ (5.58)

Let us substitute and take it into consideration that φ = ωvt.

r = A(i cosφ− j sinφ) cos(ω0t) = A [i cos(ωvt) cos(ω0t)− j sin(ωvt) cos(ω0t)] (5.59)

The two coordinates of the motion can be written accordingly:

x(t) = A cos(ωvt) cos(ω0t) y(t) = −A sin(ωvt) cos(ω0t) (5.60)

This perfectly matches the result above, which was calculated in a more tedious way bysolving the differential equation of motion in the coordinate system of planet Earth.

The effects of Coriolis 2 force The Coriolis 2 force is associated with the horizontalomega vector component.

FC2 = 2mvrel × ωh (5.61)

76

Figure 5.9: The effects of horizontal omega vector

Dependent on the direction of the relative velocity two effects emerge.If the relative velocity points vertically down, the FC2 force will point to the East.

This force will deflect the freefall from the perfect vertical direction slightly to the East.Consider a tower with height denoted h. A particle is dropped with zero initial

velocity. The velocity of free fall is as follows: vrel = gt.The Coriolis acceleration inpresent case is: aC2 = 2vrelωh. After substitution the aC2 = 2gωht formula results.This acceleration should be integrated twice in order to find out the magnitude of thedeflection.

vC2(t) =

t∫0

2gωht,dt, = 2gωh

[t,2

2

]t,=tt,=0

= gωht2 (5.62)

The displacement is the integral of the above formula:

xC2(t) =

t∫0

gωht,2dt, = gωh

[t,3

3

]t,=tt,=0

=1

3gωht

3 (5.63)

On the other hand the total duration of the freefall (denoted τ ) can be determined as

follows: h = g2τ 2. From this formula τ =

√2hg

results. The amount of the deflection

emerges by substituting τ into xC2 formula above.

xC2(τ) =1

3gωhτ

3 =1

3gωh

(2h

g

) 32

= ∆C2 (5.64)

After arranging it to a decent form:

∆C2 =

(2

3ωh

√2

g

)h

32 (5.65)

77

Let us take the constant values numerically and find out the value in parenthesis above.The g equals 9,81 m/s2 and angular velocity at 45 degree latitude is the following:

ωh =2π

24 · 3600cos 45o = 5, 13 · 10−5 1

s(5.66)

After substitution the final formula shows up:

∆C2 =

(2

35, 13 · 10−5

√2

9, 81

)h

32 = 1, 54 · 10−5h

32 (5.67)

If the height is 30 meters the deflection is 2.53 millimeters, when the height goes upto 100 meters the deflection becomes 1,54 centimeters. These values are so small thatany air disturbance will causes much higher deflection, which makes this specific Corioliseffect practically ignorable.

If the relative velocity points horizontally in East or West direction the FC2 forcewill point to vertically up or down respectively. Consequently this effect will cause avirtual decrease or increase in the weight of any object. Let us calculate the magnitudeof the effect. Assume that an aircraft travels 1008 km/h (280m/s) speed in either di-rection mentioned. The absolute value of the FC2 force is easy to calculate due to theperpendicular position of the vectors in the cross product.

FC2 = 2mvrelωh (5.68)

The associated acceleration is aC2 = 2vrelωh. Substitute the numerical values:

aC2 = 2vrelωh = 2 · 280 · 5, 13 · 10−5 = 2, 87 · 10−2m

s2(5.69)

If one compares this to the nominal value of the freefall acceleration the following comesout:

aC2

g=

2, 87 · 10−2

9, 81= 2, 93 · 10−3 (5.70)

Roughly one third of a percent reduction or increase in the weight of any object seemsnegligible in most practical cases. And do not forget that the speed was high. At speedsusual in ground transportation the effect is far more insignificant.

78

Chapter 6

Oscillatory Motion - Gabor Dobos

A body is doing oscillatory motion when it is moving periodically around an equilibriumposition. In mathematical terms this can be expressed in the following form:

r(t) = r(t+ T ) (6.1)

where r is the position-vector of the body, t is the time and T is called the period of theoscillation. (6.1) means that if the body is found in position r at a given t time, it willreturn to the same position T time later. Any kind of motion that can be described byan r(t) function which satisfies (6.1) is called and oscillation.

Oscillations and vibrations are one of the most common types of motion in nature. Apendulum or a body attached to a spring is doing oscillatory motion. Musical instrumentsand our vocal cords also utilise vibrations to create sounds. Most modern clocks measuretime by counting oscillations. From switching mode power supplies to radio transmittersthere is a wide variety of electronic devices that rely on the periodic motion of chargedparticles. Atoms in solids are also vibrating around their equilibrium positions.

To understand all these phenomena, first we have to understand oscillations. We willstart our discussion of oscillatory motion with the simple harmonic oscillator, which isthe easiest to describe in mathematical terms. Then we will expand this simple modelto take into account other effects such as damping and external excitation. Finally wewill discuss how the superposition of oscillations can be used to describe any arbitraryperiodic motion.

6.1 The simple harmonic oscillator

A body is doing a simple harmonic motion if its displacement from the equilibriumposition can be described by a sinusoidal function:

x(t) = Acos(ωt+ φ0) (6.2)

79

Figure 6.1: Displacement of a body doing simple harmonic oscillation as a function oftime

where:

• x is the position (or displacement) of the body (Measured in meters)

• A is called the amplitude of the oscillation. This defines the range in which thebody is moving. If the equilibrium point is in the origin, the body is oscillatingbetween x = A and x = −A. (Measured in meters)

• The argument of the sinusoidal function (ωt+φ0) is called the phase of the oscilla-tion. (Measured in radians)

• At t = 0 the phase is equal to φ0, thus it is called the initial phase. (Also measuredin radians)

• The constant ω is called angular frequency (measured in radians/s, or simply 1/s),and it is determined by the period of the oscillation. cos() is a periodic function:it takes the same value when the phase is increased or decreased by 2π. Thus thetime required to change the phase by 2π is equal to the period of the oscillation.In mathematical terms:

(ω(t+ T ) + φ0)− (ωt+ φ0) = 2π (6.3)

ωT = 2π (6.4)

ω =2π

T(6.5)

80

• The number of oscillations in a unit of time is called the frequency of the oscillation.It is usually measured in Hertz (Hz), which is one oscillation per second. Like theangular frequency, it is determined by the period of the oscillation:

f =1

T=

ω

2π(6.6)

6.1.1 Complex representation of oscillatory motion

An interesting feature of harmonic oscillations is their connection to circular motion.Consider a body moving on a circular trajectory of radius A around the origin. Whenthe position of the body is projected to a straight line (such as the x or y axis) it canbe described by a sinusoidal function, thus the projection of the body is doing a simpleharmonic oscillation. The angular frequency of this oscillation is the same as the angularvelocity of the circular motion, and its amplitude is the radius of the circle.

Figure 6.2: The connection of circular and simple harmonic motion.

This connection can be used to describe harmonic oscillations in complex form. Acomplex number may be represented by a vector in the complex plane. The coordinatesof this vector can be given either in Cartesian or in polar coordinates:

C = x+ jy (6.7)

C = Aejφ (6.8)

where j =√−1 is the imaginary unit, x and y are called the real and imaginary parts

of the complex number C (usually marked by Re(C) and Im(C)), while A and φ are itsmagnitude and phase, respectively.

Consider a complex function C(t), whose magnitude is constant (A = const.), whileits phase is a linear function of time (φ = φ0+ωt). The endpoint of the vector representingC in the complex plane is moving on a circle around the origin. The real and imaginaryparts of C can be determined using Euler’s formula:

ejφ = cos(φ) + jsin(φ) (6.9)

81

Using this, the real and imaginary parts of C are:

x = Acos(φ) = Acos(φ0 + ωt) (6.10)

y = Asin(φ) = Asin(φ0 + ωt) (6.11)

Since the real and imaginary parts are basically projections of C to the real and imaginaryaxes, x and y are sinusoidal functions of time. This also means that a simple harmonicoscillation can be described as the real part of such a “rotating” complex function. Suchfunctions are usually referred to as phase vectors or phasors. The main advantage ofsuch a representation is that the dependencies on A, ω and φ0 can be separated intothree independent factors:

Aejφ = Aej(φ0+ωt) = Aejφ0︸ ︷︷ ︸A

ejωt (6.12)

As the first two terms on the right-hand side are time independent, they are usuallymerged to a single constant, called complex amplitude (A = Aejφ0), which describes boththe amplitude and the initial phase of the oscillation.

6.1.2 Velocity and acceleration in oscillatory motion

The velocity and the acceleration of the particle can be determined from (6.2):

v =dx

dt= −ωAsin(ωt+ φ0) (6.13)

a =d2x

dt2= −ω2Acos(ωt+ φ0) = −ω2x (6.14)

Figure 6.3 shows the position, velocity and acceleration as a function of time. All ofthese are sinusoidal functions, but in different phases. The velocity is late by a quarterperiod with respect to the displacement, and the acceleration is also shifted by anotherquarter period. This means, that in simple harmonic motion the acceleration is alwaysproportional and opposite to the displacement. This also gives us a hint about what kindof systems may exhibit simple harmonic oscillations. As the sum of the forces acting onthe body is proportional to its acceleration there must be a retracting force, pulling thebody back towards the equilibrium with a strength proportional to its displacement.

6.2 Motion of a body attached to a spring

One of the simplest mechanical systems that exhibit simple harmonic oscillations consistsof a body attached to a fix point by a spring. Since the spring force is proportional to

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Figure 6.3: Displacement, velocity and acceleration in simple harmonic motion

83

the elongation of the spring, this system satisfies the aforementioned criterion. Becauseof its simplicity it is ideal to demonstrate how harmonic motion can be described inmathematical terms. Using Newton’s law of motion:

F = ma (6.15)

−Dx = md2x

dx2(6.16)

md2x

dt2+Dx = 0 (6.17)

d2x

dt2+D

mx = 0 (6.18)

(6.18) is a homogenous second-order linear differential equation, which can be solvedby the ansatz (or trial function) x(t) = eλt. The values of λ can be determined bysubstituting the trial function into the differential equation:

d2

dx2eλt +

D

meλt = 0 (6.19)

λ2eλt +D

meλt = 0 (6.20)

λ2 +D

m= 0 (6.21)

λ = ±j√D

m(6.22)

(6.21) is usually referred to as the characteristic polynomial. Since (6.21) has two dif-ferent roots, (6.18) has two independent solutions. (These are the so called fundamentalsolutions of the differential equation.)

x1(t) = ej√

Dmt (6.23)

84

x2(t) = e−j√

Dmt (6.24)

According to the superposition principle any solution of a linear differential equationcan be built up as a linear combination of its fundamental solutions.

x(t) = C1ej√

Dmt + C2e

−j√

Dmt (6.25)

where C1 and C2 are complex constants, that can be determined from the initial condi-tions. (The state of the system – such as the position and velocity at the start of theexperiment – is usually referred to as initial conditions.) But there is a further importantpoint to consider: the displacement of the object is a measurable quantity, thus it mustalways be a real number. (There are no imaginary quantities in the real physical world.)This means that only those solutions have valid physical meaning, where the imaginaryparts of the fundamental solutions cancel out each other for all moments of time. Thiscan be easily achieved by making the two constant multipliers the complex conjugatesof each other. This way the imaginary parts of the two terms on the right-hand side of(6.25) are always the opposites of each other and their sum is a real number. The generalsolution of (6.18) is:

x(t) = Cej√

Dmt + C∗e−j

√Dmt (6.26a)

where C is determined by the initial conditions. Using Euler’s formula (6.26a) can betransformed to:

x(t) = A1sin

(√D

mt

)+ A2cos

(√D

mt

)(6.26b)

Or by using trigonometric identities:

x(t) = Acos

(√D

mt+ φ0

)(6.26c)

(6.26a), (6.26b) and (6.26c) are equivalent: they all give precisely the same result for allvalues of t, and they can be transformed into each other using mathematical identities.Comparing (6.26c) to (6.2) shows, that the body is doing a simple harmonic oscillation,

with angular frequency ω0 =√

Dm

. The amplitude (A) and initial phase (φ0) are deter-

mined by the initial conditions. (In fact A and φ0 can be expressed in terms of A1 andA2 or in terms of C, that are also determined by the initial conditions.)

An important feature of this system is that the period of oscillation (T =2π

ω0

=

2π

√m

D) is independent of the initial conditions. It is influenced only by the parameters of

85

the system itself (such as the mass of the body or the spring constant). If the amplitudeis increased, the body is going to move proportionally faster, and finish each cycle inprecisely the same amount of time. This means that such systems are ideal for buildingclocks: as each cycle takes the same time regardless of the initial conditions, time canbe measured by counting cycles.

6.3 Simple pendulum

Another mechanical system, that exhibits similar oscillations, is the simple pendulum,which is a bob suspended on a massless rope from a frictionless pivot. When it is displacedfrom the vertical (equilibrium) position the weight of the bob pulls it back towards theequilibrium. As the length of the rope is constant, the bob moves on a circular trajectory.We can apply Newton’s second law to describe its tangential acceleration:

−mg sin(θ) = mld2θ

dt2(6.27)

Figure 6.4: A simple pendulum

The left-hand side of (6.27) is the tangential component of the weight of the bob.(Ft = −mg sin(θ)) The minus sign indicates that it points towards the equilibriumposition. The right-hand side is the mass of the bob (m) multiplied by the tangential

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acceleration (which can be derived from the angular acceleration, by multiplying it bythe length of the rope). Rearranging (6.27) gives:

d2θ

dt2+g

lsin(θ) = 0 (6.28)

Although this is another second order homogenous differential equation, unlike (6.18) thisis not a linear differential equation, and it has no simple analytical solution. Howeverfor small angles sin(θ) is very close to θ, therefore when the amplitude of the oscillationis small we may approximate sin(θ) by θ, and (6.28) becomes:

d2θ

dt2+g

lθ = 0 (6.29)

(6.29) is identical to (6.18), with x replaced by θ andD

mreplaced by

g

l. Therefore the

solution is also identical:

θ(t) = θ0cos(ω0t+ φ0) (6.30)

where

ω0 =

√g

l(6.31)

and θ0 and φ0 are determined by the initial conditions. The period of the oscillation

(T =2π

ω0

= 2π

√l

g) depends only on the length of the pendulum (l), and the gravitational

acceleration (g). As both of these parameters can be kept reasonably constant, pendulumclocks were considered to be the most precise clocks from their invention in 1565 byChristiaan Huygens until the invention of quartz clocks in 1927.

Note, that the calculation above is not an exact solution of (6.27), merely an approx-imation. It is true only for small angles. For higher amplitudes the difference betweensin(θ) and θ increases, and the approximation becomes less and less precise. It can beshown, that at high amplitudes the behaviour of the pendulum becomes significantlydifferent from that of a harmonic oscillator, and the period of the oscillation is notcompletely independent of the amplitude.

In fact the calculation above is never completly accurate. Even at small angles sin(θ)and θ are not exactly the same, which means that the formulas above are never completlyprecise. But for small angles the deviations are so small, that they are indistinguishablefrom other errors.1 For example the length of the pendulum can be measured only

1It must be noted however, that approximations cause so called systematic errors, while some mea-surement errors are random. When the experiment is repeated many times sytematic errors remainthe same, while random errors are different each and every time. This means that random errors tendto average out when calculating the mean of a large number of measurement values, while systematicerrors cause a bias in the mean value, no matter how many measurements are made.

87

with a limited precision, and it may even change with temperature. The gravitationalacceleration may be different at different locations, the pivot is not completely frictionless,and the viscosity of air may also influence the movement of the pendulum.

These disturbing effects may be decreased by careful design, but they can never becompletely eliminated. Physical measurements always have some level of error. If theerror which is caused by our approximations is considerably smaller than measurementerrors, it becomes undetectable: it is hidden by other disturbing effects.

The important lesson of this sub-chapter is that in many cases mathematical descrip-tion of physical systems becomes too complicated, and no analytical solution can befound. In such cases one may attempt to find an approximate solution by simplifyingthe problem, like we did in case of the simple pendulum. But it is important to keepit in mind that these kinds of calculations are not completely precise. They are appli-cable only under certain circumstances, when the error of the approximation becomesnegligible relative to other errors.

6.4 Energy in simple harmonic motion

Equation (6.14) gives the acceleration of a body doing a simple harmonic oscillation.Using Newton’s law of motion we can calculate the force which is acting on the bodyduring harmonic oscillations:

F = ma = −mω2x (6.32)

This is a conservative force, which means there is a potential associated with it. Thepotential energy of the body can be obtained by integrating (6.32) .

F = −dEpdx

(6.33)

∫dEp =

∫−Fdx =

∫mω2xdx (6.34)

An indefinite integral is only defined up to an additive constant. We may choose thevalue of this constant, so that the potential energy is zero in the origin:

Ep =1

2mω2x2 =

1

2mω2A2cos2(ωt+ φ0) (6.35)

Besides its potential energy the body also has kinetic energy:

Ek =1

2mv2 =

1

2m(ωAsin(ωt+ φ0))2 =

1

2mω2A2sin2(ωt+ φ0) (6.36)

88

According to the Pythagoren theorem sin2(φ) + cos2(φ) = 1, thus:

Ek =1

2mω2A2(1− cos2(ωt+ φ0)) (6.37)

The total energy of the system is:

E = Ep + Ek =1

2mω2A2 (6.38)

which is a constant quantity. This is not surprising since there are no dissipative forcesin the system. As the body is moving, energy is transformed from one form to the other:at the equilibrium the potential energy is zero, while the kinetic energy is maximal. Atthe extrema the velocity and kinetic energy of the body becomes zero, and all of thesystem’s energy is stored in the spring as potential energy.

6.5 Damped oscillator

So far we have ignored friction and the drag force of the medium in which the oscillator ismoving. But in practice some level of dissipation is inevitable. Therefore it is necessaryto expand our model of the simple harmonic oscillator to take this into account. (6.17)can be modified by adding a new term representing a drag force, which is proportionalto the velocity of the body, but points to the opposite direction:

md2x

dt2+ k

dx

dt+Dx = 0 (6.39)

d2x

dt2+k

m

dx

dt+D

mx = 0 (6.40)

We may use the same trial function as before:

x(t) = eλt (6.41)

Substituting it into (6.40) gives:

λ2 +k

mλ+

D

m= 0 (6.42)

λ1,2 = − k

2m±

√(k

2m

)2

− D

m(6.43)

89

We already know that ω0 =

√D

mis the angular frequency of the undamped oscilla-

tor, which is usually referred to as the natural angular frequency. Introducing the new

constant β =k

2m, (6.43) becomes:

λ1,2 = −β ±√β2 − ω2

0 (6.44)

λ1,2 = −β ± j√ω2

0 − β2 (6.45)

Depending on wether β is less than, equal or greater than ω0 we have three distinct cases:

• ω0 > β (underdamped case):

x(t) = Ce

(−β+j√ω20−β2

)t+ C∗e

(−β−j√ω20−β2

)t

(6.46)

x(t) = e−βt(Cej√ω20−β2t + C∗e−j

√ω20−β2t

)(6.47)

introducing the new constant ω =√ω2

0 − β2, (6.47) becomes:

x(t) = e−βt(Cejωt + C∗e−jωt

)(6.48)

x(t) = e−βtAcos(ωt+ φ0) (6.49)

In the underdamped case oscillations occur due to the imaginary component ofλ. It must be noted however that the angular frequency of these oscillations (ω =√ω2

0 − β2) is lower than the natural angular frequency (ω0). The other maindifference between the undamped and the underdamped oscillator is the result ofthe real part of λ, which is responsible for the e−βt factor. Damped oscillators loseenergy due to the dissipative drag force, thus the amplitude of the oscillation isnot constant but decreases exponentially with time.

• ω0 = β (critical damping): Increasing the viscosity of the medium further decreasesthe angular frequency of the oscillations. The angular frequency reaches zero whenβ becomes equal to ω0. In this case the second term on the right-hand side of (6.45)is zero, and both roots are the same. (The solution is degenerated.) But a secondorder linear differential equation must have two independent fundamental solutions.The second fundamental solution can be found using the ansatz x(t) = teλt. Thegeneral solution is the linear combination of the two fundamental solutions:

x(t) = (A+Bt)eλt (6.50)

90

Figure 6.5: Underdamped oscillation

Figure 6.6: In the critically damped case even a single cycle takes infinitely long time

91

At this point the frequency of the oscillation decreases to zero, thus even a singlecycle would take an infinitely long time. This is the so called critically dampedoscillator.

• ω0 < β (overdamped case): In the criticaly damped case, and at higher viscositiesλ has no imaginary part, thus no oscillations can occur. The system simply crawlsback toward the equilibrium position (without ever reaching it):

x(t) = Aeλ1t +Beλ2t (6.51)

Figure 6.7: In the overdamped case there are no oscilations.

6.6 Forced oscillations

In most practical cases losses are inevitable, and vibrations continuously lose energy. Foran oscillator to be able to operate for an extended period of time these losses have to bereplenished. In other words a periodic external force has to act on the system to avoidthe decay of its vibration. This can be taken into account by expanding (6.39):

md2x

dt2+ k

dx

dt+Dx = F0Re

(ejωf t

)(6.52)

92

where the right hand side represent the external force. Dividing both sides by m, andusing the constants β and ω0 again:

d2x

dt2+ 2β

dx

dt+ ω2

0x =F0

mRe(ejωf t

)(6.53)

This is a second order inhomogeneous linear differential equation. The solution of suchequations is the sum of the solution of the homogeneous differential equation and aparticular solution of the inhomogeneous equation:

x(t) = xh(t) + xp(t) (6.54)

where xh(t) is one of (6.49), (6.50) or (6.51), depending on the values of k and D. Wemay look for xp(t)in the following form:

xp(t) = Aejωf t (6.55)

This means that x(t) is the superposition of two oscillatory motions. In the beginning(transient state) the system is oscillating like a damped oscillator with ω =

√ω2

0 − β2

angular frequency. But the amplitude of this oscillation decreases exponentially due todamping. On the other hand the system is forced to start vibrating with the frequency ofthe external driving force. While the damped oscillation quickly dies out, the amplitudeof the forced oscillation increases until the power lost to damping becomes equal to thepower supplied to the system by the external driving force. The amplitude and phase ofthe oscillation (both included in the complex amplitude) in this stationary state can bedetermined by substituting (6.55) into (6.53) :

−Aω2fejωf t + 2jβAωfe

jωf t + ω20Ae

jωf t =F0

mejωf t (6.56)

−Aω2f + 2jβAωf + ω2

0A =F0

m(6.57)

A =F0/m

ω20 − ω2

f + 2jβωf(6.58)

The magnitude of the complex amplitude is the amplitude of the oscillation:

A =F0/m√(

ω20 − ω2

f

)2+ 4β2ω2

f

(6.59)

93

Figure 6.8: The amplitude of the forced oscillation depends on the excitation frequency

The phase of the complex amplitude represents the phase difference between theoscillation and the periodic external driving force:

∆φ = arctan

(ω2

0 − ω2f

2βωf

)(6.60)

As the system is linear, the amplitude of the oscillation in stationary state is propor-tional to F0. A more interesting fact is that both the amplitude and phase-shift dependsnot only on the parameters of the oscillator (such as the mass off the body, the springconstant and the strength of damping), but also on the frequency of the external drivingforce. Figure 6.8 shows the amplitude of the oscillation as a function of the drivingfrequency for different dampings. When the damping is strong, the amplitude decreaseswith increasing frequency. But as the damping decreases a pronounced peak appears onthe graph: the amplitude is maximal, when the denominator on the right-hand side of(6.59) is minimal. This occurs when ωf reaches

ωA =√ω2

0 − 2β2 (6.61)

This is called the angular frequency of amplitude resonance. The smaller the damping,the more pronounced the resonance becomes and its frequency shifts towards the naturalfrequency of the oscillator.

The velocity of the oscillator can be calculated by differentiating the displacementwith respect to time. In stationary state the velocity amplitude is:

94

v0 = ωfA =ωfF0/m√(

ω20 − ω2

f

)2+ 4β2ω2

f

(6.62)

v0 =F0/m√(

ω20

ωf− ωf

)2

+ 4β2

(6.63)

The velocity amplitude also depends on the frequency of the external driving force.vo is maximal, when the denominator on the right-hand side of (6.63) is minimal. Thisoccurs when the quantity within the parenthesis becomes zero.

ω20

ωf− ωf = 0 (6.64)

ωf = ω0 (6.65)

The velocity of the oscillator becomes maximal, when the frequency of the externaldriving force is equal to the natural frequency of the oscillator. As the kinetic energyalso reaches its maximum at this frequency, the phenomenon is called energy resonance.

In forced oscillators both the amplitude and the kinetic energy increases until lossesdue to damping become equal to the power supplied to the oscillator by the externaldriving force. When the damping is weak this may result extremely strong vibrations,giving the false impression that a well-tuned oscillator can generate energy. Many con-men try to utilise this false impression and claim that they are able to build perpetummobiles based on oscillators. Of course, this is not possible. It is important to under-stand, that oscillators cannot generate energy, only accumulate the power of the externaldriving. (In fact, they are continuously losing energy to damping.) Although the amountof accumulated energy can be very high, it is not generated by the oscillator. Withoutthe external driving force, vibrations decay exponentially.

6.7 Superposition of simple harmonic oscillations

As we have seen in in the previous section, different oscillations may get superimposedon each other. In the following we will discuss some special cases:

6.7.1 Same frequency, same direction

Let us consider two oscillations in the x direction with the same frequency, but withdifferent amplitudes and different initial phases:

95

x1(t) = Re(A1e

jωt)

(6.66)

x2(t) = Re(A2e

jωt)

(6.67)

The superposition of the two oscillations is:

x(t) = x1(t) + x2(t) = Re(

(A1 + A2)ejωt)

= Re(Aejωt

)(6.68)

where A = A1+A2. In other words the superposition of the two oscillations is anotherharmonic oscillation of the same frequency, with a complex amplitude, that is the sumof the complex amplitudes of the two oscillations. The amplitude of this oscillation(which is the magnitude of the complex amplitude) can be determined by using the lawof cosines:

Figure 6.9: Sum of the complex amplitudes

A2 = A21 + A2

2 + 2A1A2cos(φ1 − φ2) (6.69)

The initial phase is:

tg(φ) =sin(φ)

cos(φ)=A1sin(φ1) + A2sin(φ2)

A1cos(φ1) + A2cos(φ2)(6.70)

96

6.7.2 Different frequency, same direction

Let us consider two oscillations in the x direction with the same amplitude, but withdifferent frequencies:

x1(t) = ARe(ejω1t

)(6.71)

x2(t) = ARe(ejω2t

)(6.72)

The superposition of the two oscillations is:

x(t) = x1(t) + x2(t) = Re[A(ejω1t + ejω2t)

]= (6.73)

= Re[A(ej

ω1−ω22

t + e−jω1−ω2

2t)ej

ω1+ω22

t]

= (6.74)

= 2Acos

(ω1 − ω2

2t

)cos

(ω1 + ω2

2t

)(6.75)

Figure 6.10: The superposition of two oscillations with slightly different frequenciescreates a beat

The superposition of the two oscillations is an oscillation (represented by the cos

(ω1 + ω2

2t

)term in (6.75)) whose frequency is the average of the frequencies of the two oscillations,

with an amplitude modulated by a sinusoidal function (the cos

(ω1 − ω2

2t

)term in

(6.75)). For example when two musical instruments emit similar sounds, we cannot de-tect them individually. Instead we hear a single tone with a fluctuating intensity. Thephenomenon is called beat.

The amplitude, of the oscillation will be maximal when the cos() term equals 1 or -1.This means, that in each cycle of the modulation function we can detect two fluctuations.Even though the frequency of the modulation function is half of the difference of the twooriginal frequencies, the frequency of the beat is the difference of the two frequencies.

97

6.7.3 Lissajous figures

It is also possible to describe the superposition of oscillations in different directions. Thesimplest case is when two oscillations are perpendicular to each other. In this case wemay choose the x and y axes of our coordinate system to point in the direction of thetwo oscillations. The trajectory of the oscillating particle (the so called Lissajous figure)depends both on the frequencies of the two oscillations, and also on the difference of theirinitial phases. (We will assume, that both oscillations have the same amplitude. If thisis not true, the Lissajous figure will be scaled in the x and y directions proportionally tothe respective amplitudes.)

Figure 6.11 represents typical Lissajous figures. In the first four cases the frequenciesof the two oscillations are identical. When the initial phases are equal to each other, thex and y coordinate of the oscillating particle changes in unison and the particle moveson a straight line. Since cos(φ+π) = −cos(φ), the particle also moves on a straight line,

when the phase difference is π. When the phase difference isπ

2, the coordinates of the

particle are:

x(t) = Acos(ωt) (6.76)

y(t) = Acos(ωt+π

2) = Asin(ωt) (6.77)

These are the horizontal and vertical components of circular motion, and consequentlythe Lissajous figure becomes a circle. For any other arbitrary phase difference the trajec-tory of the oscillating particle is an ellipse, with an aspect ratio depending on the phasedifference.

Figure 6.11: Lissajous figures

When the ratio of the frequencies in the two directions is n:m, the particle finishesn cycles in one direction while it does m cycles in the other direction. This makes theLissajous figures considerably more complicated. When the frequencies are the same,the particle repeats the same trajectory in every cycle. When the horizontal and verticalfrequencies are different, the number of cycles it takes to close the Lissajous figure andstart repeating the same motion again is the least common multiple of n and m.

This also shows that Lissajous figures appear only when the ratio of frequencies is arational number. When the ratio is an irrational number there is no common multiple

98

other than infinity, which means that the trajectory is not a closed curve: although thehorizontal and vertical components of the motion are both periodic, their superpositionis not.

The ratio of frequencies can be determined by counting how many times the figuretouches the horizontal and vertical edges of the square that can be drawn around thefigure. Since the particle finishes n cycles in one direction while it does m cycles inthe other, it touches the edges perpendicular to the first direction 2n times, while ittouches the other edges 2m times. Consequently the ratio of the horizontal and verticalfrequencies is the inverse of the ratio of the number of cross-sections with the horizontaland vertical edges. This technique can be used to measure unknown frequencies with atwo channel oscilloscope (that can plot the two signals against each other in the so calledX-Y mode) and a reference oscillator of known frequency.

6.7.4 Fourier analysis

During this chapter we have focused on oscillatory motions that can be described bysinusoidal functions. But the definition of oscillatory motion is more general: it doesnot require the motion to be harmonic, only to be periodic. Therefore it is necessary toformulate a mathematical method to describe anharmonic oscillations. The problem isthat not all kinds of motion can be expressed in closed-form formulas. The best we cando is to create a method that enables us to approximate any periodic motion at arbitraryprecision.

Our task is slightly similar to what we had to do to describe the simple pendulum.Even though we cannot give a completely precise mathematical description, we may beable to give an approximation that is so close to the actual behaviour of the systemthat any deviations are unnoticable. Since we already have a basic understanding ofharmonic motion it is a sensible idea to approximate arbitrary periodic motions by thesuperposition of sinusoidal functions.

Consider a periodic function f(t), with a period of T . In the zero order approxi-mation we may try to calculate the time-average of f(t). Naturally, this is a very poorapproximation since a constant cannot describe how f(t) changes with time. We mayimprove the approximation by adding a sinusoidal function with the same T period asf(t) to the average value, and adjusting its phase and amplitude, to achieve the bestpossible fit. Since f(t) is not a sinusoidal function we may not achieve a perfect fit, butthe difference between the arbitrary function and this simple harmonic approximation ismuch smaller than the difference between f(t) and its average value.

If adding a single sinusoidal function has improved the approximation, we may try tofurther improve it, by adding even more sinusoidal functions. Adding a second sinusoidalcomponent of the same frequency is pointless, since the sum of two sinusoidal functionsof equal frequency is another sinusoidal function of the same frequency with differentphase and amplitude. As we have already optimised the phase and amplitude of the

99

first sinusoidal component, adding another one with the same frequency will not improvethe approximation. Therefore the second sinusoidal component needs to have a differentfrequency. But we also have to make sure, that the approximation has the same period asf(t). This can be achieved by choosing the frequencies of further sinusoidal componentsto be integer times the frequency of the first component. This way each sinusoidalcomponent will finish an integer number of cycles in T time, and their sum will have thesame period as f(t).

Figure 6.12: An arbitary periodic function (upper left) may be approximated by thesuperposition of a series of sinusoidal functions

Adding each new sinusoidal component will slightly increase the quality of the ap-proximation. By adding together a suitably high number of sinusoidal components wemay create an approximation of arbitrary precision.

f(t) = a0 + a1cos(ωt+ φ1) + a2cos(2ωt+ φ2) + a3cos(3ωt+ φ3) + ... (6.78)

Ususally it is more efficient to represent all sinusoidal components in complex form:

f(t) = Re[a0 + a1e

jωt + a2e2jωt + a3e

3jωt + ...]

(6.79)

To create an approximation of any arbitrary periodic function, all what we have to do, isto determine the optimal values of a0, a1, a2, . . . to achieve the best possible fit. These

100

values may be calculated by the formulas deduced by Joseph Fourier in the nineteenthcentury:

a0 =1

T

∫ T

0

f(t)dt (6.80)

an =1

T

∫ T

0

f(t)enjωtdt (6.81)

(6.82)

Using these formulas we may express any arbitrary periodic function as a sum of sinu-soidal functions. It must be noted however that this is merely an approximation. Themore components included in the sum, the more similar it becomes to the original func-tion, but to reach a perfect fit an infinite number of components would be required,which is not possible in practice. When choosing the parameters of the approximation,two criteria must be observed:

• The frequency of the first sinusoidal component is determined by the period of theoscillation. The longer the period, the lower the frequency needs to be.

• Low frequency sinusoidal functions change very slowly. Therefore if the f(t) func-tion changes rapidly, high frequency components are required to follow these rapidchanges. Since we have already chosen the frequency of the first sinusoidal com-ponent to set the period, this can be achieved only by increasing the number ofsinusoidal components. As a thumb rule the period of the highest frequency com-ponent should not be longer than the double of the required time resolution. Anychanges faster than that are going to be lost in the approximation.

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Chapter 7

Waves - Gabor Dobos

Most people imagine waves as some sort of periodic disturbance which is moving in amedium, such as the waves on the surface of a pond after a pebble is dropped into it.It must be noted however that in physics the term is used in a much broader sense.Electromagnetic waves can propagate in perfect vacuum without the presence of anymedium, shock waves are non-periodic, and standing waves are not moving in space.The most general definition of a wave that includes the above examples is a disturbancewhich can transport energy and momentum without any long range movement of matter.

Waves on the surface of water as well as sound waves, or waves in a string of amusical instrument are called mechanical waves as they mechanically distort an elasticmedium. Electromagnetic waves consist of oscillations of electrical and magnetic fieldsthat can travel in space. In a plasma these fields can influence the movement of chargedparticles forming different kinds of magnetohydrodynamic waves such as Alfven waves ormagnetosonic waves. In some cases elementary particles also behave as waves. Accord-ing to Einstein’s theory of general relativity even gravitational waves may be createdby distorting the fabric of the space-time continuum, although these have never beenexperimentally detected due to measurement difficulties.

In this chapter we will discuss mechanical waves. These waves may be created bysuddenly distorting an elastic medium. Its surroundings will exert a restoring force onthe displaced portion of the medium, pulling it back towards the equilibrium position.As we have seen in the previous chapter this may lead to vibrations. But in this casevibrations are not limited to the displaced section. As its surroundings exert a forceupon it, it will also exert an opposite force on its surroundings. Due to this force,the surrounding medium may also start to vibrate, and the disturbance may propagatetrough the medium.

An interesting feature of this type of motion is that although the disturbance maytravel large distances, the movement of a given piece of the medium is limited: it is oscil-lating around its own equilibrium position. The thunder of lightning can be heard from adistance of several kilometres, but this does not mean that individual air molecules travel

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this far. They collide with other particles, passing over their energy and momentum. Itis through these collisions that the disturbance reaches our ears, while the air as a wholedoes not move.

The direction of the oscillation can be either parallel or perpendicular to the directioninto which the wave is traveling. Longitudinal waves occur when the direction of thedistortion is parallel to the propagation, like in case of sound waves in air. In othercases the medium is oscillating at right angles to the direction into which the wave istraveling. These are called transverse waves. It must be noted however that transversemechanical waves are not possible in all media since the restoring force must be parallelto the displacement. In case of transverse waves this means that the restoring force hasto be perpendicular to the direction of propagation. In other words a shearing stressmust exist in the medium, and this is not possible in all materials. For example noshearing stress may exist in gases, thus sound waves in air are always longitudinal waves.

As we are living in a three dimensional world there are two directions that are perpen-dicular to the direction of propagation. This means that in case of transverse waves theoscillations at each point may be the superposition of two perpendicular oscillations. Thefrequency of these perpendicular oscillations must be the same, but their initial phasesmight be different. The situation is very similar to what we have discussed in section6.7.3. The Lissajous figures created by the superposition of perpendicular oscillationsof the same frequency may form a straight line (when the phase difference is 0 or π), a

circle (when the phase difference isπ

2or −π

2) or an ellipse (in case of any other arbitrary

phase difference). Accordingly transverse waves might have linear polarisation, circularpolarisation or elliptical polarisation.

7.1 Sine wave

We have started the discussion of oscillatory motion with the simple harmonic oscillator,as it was the easiest to describe in mathematical terms. For the same reason the firsttype of wave we will discuss in more detail is the one dimensional sine wave, which canbe described by:

y(x, t) = Asin(kx− ωt) (7.1)

where y is the displacement, A is the amplitude, ω is the angular frequency, and k iscalled the wave number. If x is fixed, the displacement is a sinusoidal function of time.In other words, each point of the medium is doing harmonic oscillations, with an initialphase depending on the position. If we fix the time (like taking a snapshot of the wave)one can see, that the displacement is also periodic in space. The spatial period of thewave (the distance between two consecutive corresponding points – like tow crests or

103

Figure 7.1: A sine wave

zero crossings) is called the wavelength and it is usually marked by λ. While the periodof the oscillation is determined by the angular frequency (ω), the wavelength dependson the wave number (k):

k =2π

λ(7.2)

The velocity of propagation can be determined by following the movement of a certainpoint of the wave, like a crest or a trough. In other words we shall determine how xshould change as a function of time, to keep the argument of the sin() function (thephase) constant:

kx− ωt = const. (7.3)

vph =dx

dt=ω

k(7.4)

As it is the rate at which the phase of the wave propagates through space, it is calledthe phase velocity.

In case of simple harmonic oscillations, the acceleration of the particle was propor-tional to the displacement. In case of a sine wave every point of the medium is doing asimple harmonic oscillation, thus:

∂y

∂t= −Aωcos(kx− ωt) (7.5)

∂2y

∂t2= −Aω2sin(kx− ωt) = −ω2y(x, t) (7.6)

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As y(x,t) depends on two parameters, we may also calculate its differentials withrespect to the space coordinate (x):

∂y

∂x= Akcos(kx− ωt) (7.7)

∂2y

∂x2= −Ak2sin(kx− ωt) = −k2y(x, t) (7.8)

Comparing (7.6) and (7.8) gives:

∂2y

∂t2=(ωk

)2 ∂2y

∂x2(7.9)

Since we already know, that vph =ω

kis the phase velocity of the wave:

∂2y

∂t2= v2

ph

∂2y

∂x2(7.10)

For sine waves the second differentials of the displacement with respect to time andto the space coordinate are proportional to each other, and their ratio is the square ofthe velocity of the wave. Systems that can be described by similar equations have thepotential to exhibit wave motion, thus (7.10) is called a wave equation

7.2 Transverse wave on a string

A string under tension is one of the simplest mechanical systems that can exhibit wavemotion. Imagine a thin wire with a linear mass density µ, fixed at one and, and pulledwith a constant F force at the other end. When a section of the string is displaced ina perpendicular direction, the tension in the string pulls it back towards the original,straight position. In other words it represents a restoring force, and we may expect thedisplaced section to start oscillating. But the sections of the string are not separatedfrom each other. If one section starts oscillating, it will exert a periodic force on theneighbouring sections, forcing them to start oscillating, too.

To give a mathematical description of the phenomenon, consider an infinitesimallysmall section of the string. Both ends are pulled by the same F force, but in slightlydifferent directions. (If a transverse wave is traveling in the string, it is no longer straight,but slightly curved.) Applying Newton’s law of motion in the horizontal direction:

∆max = F cos(θ + ∆θ)− Fcos(θ) (7.11)

µ∆x ax = F (cos(θ + ∆θ)− cos(θ)) (7.12)

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Figure 7.2: Forces acting on a section of the string

where ∆m is the mass of the section, ∆x is its length, and ax is its acceleration. Atsmall angles the cosine function changes very slowly. If θ << 1, cos(θ + ∆θ) ≈ cos(θ).Thus the acceleration in the horizontal direction is zero. In the vertical direction:

µ∆x ay = F (sin(θ + ∆θ)− sin(θ)) (7.13)

If the distortion is small (θ << 1) the sine function may be approximated by the tangentof the same angle. As the tangent is basically the steepness of the curve, it may be

replaced by∂y

∂x:

sin(θ) ≈ tg(θ) =∂y

∂x(7.14)

Using this approximation (7.13) becomes:

µ∆x ay = F

(∂y

∂x

∣∣∣∣x+∆x

− ∂y

∂x

∣∣∣∣x

)(7.15)

ay =F

µ

∂y

∂x

∣∣∣∣x+∆x

− ∂y

∂x

∣∣∣∣x

∆x(7.16)

If the section is infinitesimally small (∆x→ 0):

∂2y

∂t2=F

µ

∂2y

∂x2(7.17)

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(7.17) is a wave equation, and shows, that if the distortion is small, transverse sine waves

are going to travel in the string with vph =

√F

µvelocity. This may be verified by

substituting (7.1) into (7.17):

−Aω2 sin(kx− ωt) =F

µAk2 sin(kx− ωt) (7.18)

As the equation must hold for all values of t and x:

ω2 =F

µk2 (7.19)

vph =ω

k=

√F

µ(7.20)

7.3 Energy transport by mechanical waves

When a sine wave is traveling in a medium, each point is doing a simple harmonicoscillation. As we have seen in the previous chapter the energy of a harmonic oscillatoris:

E =1

2mω2A2 (7.21)

where A is the amplitude of the oscillation, m is the mass of the oscillating body, and ωis the angular frequency. Using this, the energy of the wave in an infinitesimally smallsection of the medium is:

dE =1

2dmω2A2 (7.22)

where dm = µdl is the mass of the section, dl is its length, and µ is the linear mass

density of the string. A wave travelling by v velocity needs dt =dl

vtime to travel dl

distance. Therefore:

dE =1

2µ v dt ω2A2 (7.23)

The power transmitted by the wave is:

107

P =dE

dt=

1

2µvω2A2 (7.24)

The power transmitted by a wave traveling in a three dimensional medium may becalculated in a similar manner. Consider a small dx · dy · dz section of the medium:

dE =1

2dmω2A2 =

1

2ρ dx dy dz ω2A2 (7.25)

dE =1

2ρ v dt dy dz ω2A2 (7.26)

where ρ is the density of the medium. The power transmitted by the wave trough a unitarea of the medium is:

S =dE

dt dy dz=

1

2ρ v ω2A2 (7.27)

Human senses - such as hearing and vision - measure intensities on a logarithmicscale. This is the reason why the intensity of certain types of waves is usually measured

not inW

m2, but in other, special units, on a logarithmic scale. For example the intensity

level of sound is usually given in decibels (dB). By definition:

β = (10dB)lgS

S0

(7.28)

where S0 = 10−12 W

m2. A 10dB increase in the noise level actually means that the power

of sound waves reaching our ears is increased by one order of magnitude. Our ears have

a very impressive dynamic range: 0dB (10−12 W

m2) is usually the smallest sound level

that a human may detect. A whisper is around 20dB (10−10 W

m2), while normal talking is

approximately 60dB (10−6 W

m2). Our eras can tolerate noise levels up to 90dB (10−3 W

m2)

for an extended period of time, without any lasting damage, and only noises louder than

120dB (1W

m2) are immediately painful. Measuring a quantity on such a wide dynamic

range on a linear scale would be very difficult even with modern instruments. That iswhy most biological systems measure such quantities on a logarithmic scale.

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7.4 Group velocity

As we have seen in the previous chapter superposition of two oscillations of similarfrequencies creates a beat. A similar phenomenon can be observed with waves. Imaginetwo waves travelling in the same medium with slightly different angular frequencies andwave numbers. (This is actually a very realistic model. In practice it is impossible tocreate a perfectly monochromatic wave: usually we have to deal with the superpositionof several waves with slightly different parameters.) The wave that we can observe isgoing to be the superposition of the two waves:

y1(x, t) = Asin(k1x− ω1t) (7.29)

y2(x, t) = Asin(k2x− ω2t) (7.30)

y(x, t) = y1(x, t) + y2(x, t) (7.31)

y(x, t) = Asin(k1x− ω1t) + Asin(k2x− ω2t) (7.32)

using the trigonometric identity sinα + sinβ = 2cosα− β

2sin

α + β

2:

y(x, t) = 2Acos(k1 − k2)x− (ω1 − ω2)t

2sin

(k1 + k2)x− (ω1 + ω2)t

2(7.33)

y(x, t) = 2Acos(k1 − k2)x− (ω1 − ω2)t

2sin(kx− ωt) (7.34)

Figure 7.3: Superposition of two sine waves of similar frequencies

where k is the average of k1 and k2, and ω is the average of ω1 and ω2. The su-perposition of the two waves gives a wave whose wave number and angular frequencyis the average of the wave numbers and angular frequencies of the two original waves(this is called the carrier wave), but whose amplitude is not constant, but modulatedby a sinusoidal function. Since this modulation function also depends on both x and t,its nodes (the positions where the amplitude is minimal) and antinodes (the positionswhere the amplitude is maximal) are also going to move in the medium.

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Two waves traveling in the same medium with similar angular frequencies are goingto have similar wave numbers and phase velocities, too. Since the angular frequency andwave number of the carrier wave is the average of those of the two waves, its velocity isalso going to be similar.

vph =ω

k=

ω1 + ω2

2k1 + k2

2

(7.35)

The groups formed by the modulation are going to travel with a different velocity:

vg =ω1 − ω2

k1 − k2

(7.36)

If the frequencies and wave numbers of the waves are infinitesimally close to each other

vg =ω1 − ω2

k1 − k2

=∂ω

∂k(7.37)

This is called the group velocity and it is a very important parameter not only inphysics, but also in the field of telecommunication. Information is usually transmittedby the modulation of some sort of wave. The type of carrier waves may vary fromsound waves to electromagnetic waves, and different types of modulation techniques arein use, but in most cases the velocity with which information can be transmitted fromone point to another is the same as the velocity of the groups formed by the modulation.Even though electromagnetic waves travel at the speed of light, this does not mean thatinformation can be transmitted at the same velocity. For most media the group velocityis lower than the phase velocity, therefore the speed of information transmission is alsolower.

It must be noted however that there are exceptions from this rule: the group velocityis not necessarily the same as the “signal velocity”. The ω(k) function is called thedispersion relationship. For most materials the slope of this function decreases for higherfrequencies, thus the group velocity is lower than the phase velocity. But there are someexotic materials with a so called anomal dispersion relationship. In these materials thegroup velocity may be higher than the phase velocity (or the speed of light). This doesnot mean however that these materials enable faster than light information transmission.As a thumb rule the signal velocity cannot be higher than the lesser of the phase andgroup velocities.

110

7.5 Wave packets

Another interesting parameter that can be understood by considering the superpositionof waves is the amount of information that can be transmitted in a unit of time. Althoughactual communication protocols are very complicated the rate of information transmis-sion depends on the duration of pulses that can be formed by modulating the carrierwave. (The shorter the pulses the more data can be transmitted in a unit of time.)

In section 6.7.4 we have used Fourier analysis to describe anharmonic oscillations asthe superposition of a large number of sinusoidal components. In a similar manner wemay attempt to construct short pulses from the superposition of sine waves. (This kindof construct is usually referred to as a wave packet.) The problem is that Fourier analysisis applicable only to periodic functions, and a pulse is not periodic. This problem maybe remedied by treating the non-periodic function describing the pulse as a periodicfunction with an infinitely long period. (If the period is infinitely long, the functionnever repeats itself...) As the frequency difference between Fourier components is theinverse of the period, the Fourier components of an aperiodic function are infinitely closeto each other. In other words the components do not belong to discrete frequencies: theyform a continuous distribution. This distribution is called the Fourier transform of thefunction.

f(ν) = F (f(t)) =

∫ ∞−∞

f(t)e−2π j t νdt (7.38)

where t is the time, ν is the frequency, and f(t) is the function whose Fourier transformis f(ν).

Let us consider the Fourier transform of a Gaussian-shaped pulse (In nature quantitiestend to have a so called normal distribution which is described by the Gaussian function.Therefore it is reasonable to discuss Gaussian-shaped pulses. Of course, other pulseshapes may also be described in a similar fashion.):

f(t) =1√

2πσte− t2

2(σt)2 (7.39)

The duration of the pulse is determined by the constant σt, which is the full with at halfmaximum of the Gaussian function. The Fourier transform of this function is:

f(ν) =

∫ ∞−∞

1√2πσt

e− t2

2(σt)2 e−2π i t νdt (7.40)

f(ν) = e−2π2σ2t ν

2

(7.41)

This function gives the amplitudes of the harmonic components whose superposition isthe σt wide Gaussian-shaped pulse. This is another Gaussian function whose full width

111

at half maximum is σν :

1

2σ2ν

= 2π2σ2t (7.42)

σν =1

2πσt(7.43)

It must be noted, that information is transmitted by the modulation of a carrier wave.This means, that the components, that make up the pulse should centre around thefrequency of the carrier wave (ν0). In other words we shall shift the f(ν) function by ν0.The pulse formed by these components is:

f ′(t) = F−1(f(ν + ν0)

)=

1√2πσt

e− t2

2(σt)2 e−2πjν0t (7.44)

Of course this is not a pure Gaussian-shaped pulse, since it is modulated be the e−2πjν0t

factor, which is basically a sinusoidal modulation with the frequency of the carrier wave.Nonetheless the envelope of the pulse is still a Gaussian function whose full width at halfmaximum satisfies:

σtσν =1

2π(7.45)

This means that in order to construct short pulses (and reach high data transmissionrates) we have to include Fourier components from a wide frequency range. The widerthe available frequency window, the shorter the pulses may become.

This explains why telecommunication corporations are willing to pay billions to ob-tain the right to use certain frequency windows. Although we usually say that - forexample - cell phones transmit data at 900 MHz frequency, in reality they use not just asingle frequency but a range of frequencies around 900 MHz. Each provider has its ownfrequency window, for which it pays a concession fee to the government. The wider thiswindow is, the faster your mobile internet connection may become.

Based on this, we may also understand why the number of radio and television stationsis limited. Although in layman terms we usually say that a station transmits at a givenfrequency, this refers to the frequency of the carrier wave. In reality the station uses arange of frequencies around this frequency, to modulate the carrier wave. This is why thefrequencies of radio stations cannot be arbitrarily close to each other lest they interfere.

It must also be noted that (7.45) describes a much more basic principle, which has awide range of implications beyond telecommunication. Equations like (7.45) are usuallyreferred to as uncertainty relations. Although we may be able to form wave packets bycombining a large number of sinusoidal components, the length of these wave packetscannot be infinitely short, and they always include a large number of different frequencies.The dimensions of the wave packet in the time domain and in the frequency domain are

112

related to each other. The shorter the pulse is the wider range of frequencies it mustinclude. When we use only a narrow frequency range, the pulse is going to get longer.

It is easy to see, how a similar relation can be deuced between the length of a pulse(its x dimension), and the range of wavenumber (k) components that must be combinedto form the wave packet:

σxσk =1

2π(7.46)

If we try to measure the position of the wave packet, the precision of our measurementswill be limited by σx. In a similar fashion σk limits the precision of wavenumber mea-surements. Equation (7.46) shows that the position and wave number of the wave packetcannot be measured at arbitrarily high precision at the same time. The large number ofsinusoidal components that are required to form a short pulse makes wavenumber mea-surements uncertain. On the other hand, if we use only a narrow range of wavenumbercomponents σx increases and the pulse becomes blurred, which decreases the precisionof position measurements.

In general, where waves are concerned there are certain parameters that are connectedto each other. These parameters cannot be measured at the same time at arbitrarily highprecision. As we will see during the discussion of quantum mechanics, the quantum statesof physical systems are represented by wave functions, and quantum uncertainties areclosely related to the principles that we have discussed above.

7.6 Standing waves

Imagine two waves of the same frequency traveling in the same medium in oppositedirections. (Such situations may arise when the wave is reflected back from the end ofthe medium.)

y1(x, t) = Asin(kx− ωt) (7.47)

y2(x, t) = Asin(kx+ ωt) (7.48)

The superposition of the two waves is:

y(x, t) = y1(x, t) + y2(x, t) (7.49)

y(x, t) = Asin(kx− ωt) + Asin(kx+ ωt) (7.50)

Using the same trigonometric identity as in the previous section:

y(x, t) = 2Asin(kx)cos(ωt) (7.51)

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The superposition of the two waves is a so called standing wave. Each point ofthe medium is oscillating with the same frequency, but unlike other waves, in case ofstanding waves, the initial phase is the same in every point of the medium. This meansthat the wave is not moving. This is because the position and time dependencies of thedisturbance are separated into two terms. All points are oscillating in unison, with anamplitude determined by the position. (The situation is slightly similar to that whatwe have seen in section 7.4, but in this case the group velocity is zero: the groups arenot moving.) The positions where the amplitude of the oscillation is minimal are callednodes, the ones, where the amplitude is maximal, are called anti-nodes.

Most musical instruments are designed to efficiently form certain types of standingwaves and reject others. For example in case of string instruments - such as guitarsand violins - the ends of the strings are held tight. This means that the standing wavemust have nodes at both ends, otherwise its energy dissipates away very quickly. Whenthe player uses the instrument many waves with different frequencies are created, butonly those may form stable standing waves, that have nodes at both ends. The distance

between two neighbouring nodes isλ

2, therefore only those waves are stable, whose

half wavelength fits integer times into the length of the string. The wavelength (λ) isdetermined by the wave number (k), which in turn depends on the velocity of the wave(v) and its frequency (f):

λ =2π

k= 2π

v

ω=v

f(7.52)

The so called principal mode is the standing wave with the lowest frequency that isstable in the string. It has nodes at both ends, but there are no further nodes in between.Therefore the wavelength is twice the length of the string:

L =λ0

2(7.53)

λ0 = 2L (7.54)

f0 =v

λ0

=v

2L(7.55)

This is called the fundamental frequency of the instrument. The first harmonic hasone extra node between the ends:

L = λ1 (7.56)

f1 =v

L(7.57)

114

The second harmonic has two nodes between the ends

L = 3λ2

2(7.58)

λ2 =2L

3(7.59)

f2 =v

λ2

= 3v

2L(7.60)

And so on... In general:

λn =2L

n(7.61)

fn = nv

2L(7.62)

Figure 7.4: Standing waves in different types of resonators

Only waves with these frequencies form stable oscillations in the string, all other fre-quencies are rejected. The phenomenon is a type of resonance, and such instruments arecommonly referred to as resonators. The ratio of the intensities of the stable harmonics(or overtones) depends on the design of the instrument, and it is characteristic to eachresonator. It is these intensity ratios that give each musical instrument its own uniquetone.

Other types of musical instruments have different resonators. For example a pan fluteconsists of small tubes closed at one end, and open at the other end, so that the playercan blow air into it. The standing waves forming in such a resonator needs to have anode at the closed end an anti-node at the open end. The distance between a node andan anti-node is a quarter wavelength, thus:

115

L =λ0

4(7.63)

λ0 = 4L (7.64)

f0 =v

λ0

=v

4L(7.65)

The first overtone has an additional node between the two ends:

L = 3λ1

4(7.66)

λ1 =4L

3(7.67)

f1 =v

λ1

= 3v

4L(7.68)

The second overtone has two nodes between the ends:

L = 5λ2

4(7.69)

λ2 =4L

5(7.70)

f2 =v

λ2

= 5v

4L(7.71)

In general:

λn =4L

2n+ 1(7.72)

fn = (2n+ 1)v

4L(7.73)

Resonators open at both ends, have anti-nodes at both ends. The principal mode hasa single node in between:

L =λ0

2(7.74)

λ0 = 2L (7.75)

f0 =v

λ0

=v

2L(7.76)

116

The first harmonic has two nodes between the anti-nodes at the ends:

L = λ1 (7.77)

f1 =v

L(7.78)

The second harmonic:

L = 3λ2

2(7.79)

λ2 =2L

3(7.80)

f2 =v

λ2

= 3v

2L(7.81)

In general:

λn =2L

n(7.82)

fn = nv

2L(7.83)

7.7 The Doppler Effect

When a racing car drives through the finish line the spectators on the stands will heara shift in the frequency in its sound: it seems to be higher when the car races towardsthe spectators and drops quickly as it drives past. The phenomenon is called DopplerEffect, and it is detected when the source or the observer is moving with respect to themedium in which the wave is traveling.

To better understand the phenomenon imagine a stationary observer and a soundsource moving towards the observer at a given vs velocity. If the sound source would bestationary, the wavelength (λ) of the sound waves could be calculated from the ratio of

the speed of sound (v) and the frequency (f) emitted by the source (λ =v

f). But the

movement of the source compresses the waves and changes their wavelengths. This canbe understood by calculating the distance between two crests of the wave emitted by the

source. During one period (T =1

f) the crest of the wave emitted by the source travels

λ distance towards the observer. But in the same time the source is also moving, and it

gets closer to the observer by vsT =vsf

distance. Because of this, when the next crest of

117

the wave leaves the source, its distance from the previous one is not λ, but:

λ′ = λ− vsT =v − vsf

(7.84)

Due to this shift in the wavelength of the emitted sound, the observer is detecting a shiftin the frequency of the sound:

f ′ =v

λ′= f

v

v − vs(7.85)

The Doppler Effect is also observed when the source is stationary, and the observer ismoving towards it. To determine the frequency detected by the observer we have tocalculate how many cycles of the wave reaches him or her in a unit of time. (You mayimagine this by following the crests of the sound waves. The detected frequency is thenumber of crests reaching the observer in a unit of time.) Assume that the velocity of

the wave is v and its wavelength is λ. The wave requires T =λ

vtime to travel λ distance.

Therefore the frequency detected by a stationary observer is f =1

T=

v

λ. (If a crest

reaches the observer every T time, the frequency is f =1

T.) But when the observer is

moving towards the source with a vo velocity it will encounter a furthervoλ

oscillations

in the same unit of time. Therefore the frequency detected by the moving observer is

f ′′ =v

λ+voλ

= fv + vov

(7.86)

(7.85) and (7.86) can be combined into a single formula:

fd = fv + vov − vs

(7.87)

where f is the frequency emitted by the source, fd is the frequency detected by theobserver, v is the speed of sound, vo is the velocity of the observer and vs is the velocityof the source, respectively.

The phenomenon has widespread practical applications. Speed traps are importanttools in the hands of the police to enforce speed limits. Doppler radars may measurewind speed. Modern ultrasonic imaging devices may utilise this principle to measureblood flow in our veins. Even the expansion of the universe was detected by the red shiftin the spectrum of distant stars due to the Doppler Effect. (It must be noted however,that the formula for electromagnetic waves is different form (7.87) due to relativisticeffects.)

118

Chapter 8

First law of thermodynamics andrelated subjects - Gyorgy Hars

The subject of study is the thermodynamic system, which is separated from the envi-ronment by the boundary. The boundary does not allow material transport between theenvironment and the system so the mass contained is constant. In general the bound-ary allows transfer of heat and mechanical work. The heat transfer may be inhibitedby insulation and so the system is called “thermally isolated” or in other words “adia-batic” system. The mechanical interaction may also be excluded by using rigid boundary,which system is called “mechanically isolated”. If both thermal and mechanical isolationis active then the system is considered “isolated” system.

The thermodynamic system contains gas in this chapter with definite primary stateparameters such as pressure (p), temperature (T ) and volume (V ). These parametersare considered primary concepts with no further definition in the phenomenological dis-cussion of the subject. The pressure and the temperature are intensive, driving force-likeparameters. The volume is extensive, quantity-like parameter. Microphysical substanti-ation of the state parameters is provided by the kinetic theory of gases.

8.1 Ideal gas equation

The ideal gas equation is an empirical law with units to be substituted as follows:

pV = NkT p [Pa] V[m3]

N [1] T [K] k = 1.36 · 10−23J/K (8.1)

p = nkT p [Pa] n[1/m3]

T [K] k = 1.36 · 10−23J/K (8.2)

pV =m

MRT p [Pa] V

[m3]

m [kg] M [kg/mol] T [K] R = 8.3J/molK (8.3)

119

pV =N

ART p [Pa] V

[m3]

N [1] A = 6 · 10231/mol T [K] R = 8.3J/molK

(8.4)

One mol of material consists of 6 1023 pieces (Avogadro number A) of particles. This iscalled the amount of substance. The molar mass (M) is the mass of one mol substance,which is equal with the same number of grams as the atomic mass number of the particle.In System International the molar mass should be substituted in kilograms per mol. Thefirst and second version of the ideal gas law contain the Boltzmann constant (k), thelast two use the universal gas constant (R). The temperature is measured in (K) Kelvindegrees which is the absolute temperature scale with the starting value at -273 degreesCelsius. The following relations can be found between the constants: R = kA andNk = m

MR

8.2 The internal energy of the gas (U)

The particles of the gas make chaotic motion in the container. This motion representskinetic energy which is considered the internal energy of the gas. The calculation iscarried out in the framework of the kinetic theory of gases with the following basicassumptions.

• The particles interact by elastic collisions with the chamber walls and with eachother.

• No external potential is applied, so the total energy is kinetic.

• Uniform spatial distribution is in the chamber.

• There is no dedicated direction (isotropic structure).

Figure 8.1: Particles collide to the wall of the chamber

120

Particles bounce back from the wall of the chamber. Let us calculate the amount oflinear momentum transferred to the wall in a short dt time. The x direction is normalto the wall, vx is the x component of the velocity. Half of those particles reach the wallin dt time which are in vx.dt proximity of the wall, since the other half moves oppositedirection from the wall. Multiplying this thickness with the half of the density and thearea, the total colliding particles result. The mass of the particle is denoted µ. Thevariation of momentum is 2µvx due to bouncing back with negative vx. So ultimatelythe total momentum (dI ) transferred to the wall is as follows:

dI =1

2(vxdt · A · n) 2µvx (8.5)

dI = A · nµv2x · dt (8.6)

The time derivative of the momentum is the force:

dI

dt= F = A · nµv2

x (8.7)

The pressure is the ratio of the force over the area:

F

A= p = nµ · v2

x (8.8)

p = nµ · v2x (8.9)

This is true in all three directions.

p = nµ · v2y (8.10)

p = nµ · v2z (8.11)

Let us summarize. The root means square velocity is the Pythagorean sum of the com-ponent velocities:

3p = nµ · v2rms (8.12)

The pressure can be expressed:

p =1

3nµ · v2

rms (8.13)

The ideal gas equation can also be taken into consideration:

p = nkT (8.14)

121

The right hand sides of the equations are equal:

1

3nµ · v2

rms = nkT (8.15)

After some ordering the kinetic energy of the particle is revealed:

1

2µ · v2

rms =3

2kT (8.16)

In the argument above the kinetic energy of the particle was associated with three trans-lational coordinates, which represent three thermodynamic freedom degrees. The con-clusion demonstrates the principle of equipartition. According to this principle, all kindsof energy storage capabilities contain equal amount of energy which is one half kT. Inreality the particles may also rotate in three normal directions, so maximum six thermo-dynamic freedom degrees are at disposal. If in general case the thermodynamic freedomdegree is denoted f the total kinetic energy stored in one particle is as follows:

εkin =f

2kT (8.17)

If one considers N pieces of particles the value of the internal energy (U) results.

U = Nεkin =f

2kT ·N (8.18)

Let us use the relation expressed earlier: Nk = mMR

With this the expression of the internal energy can be transformed:

U =m

M

f

2RT (8.19)

Let us introduce the molar heat capacitance of constant volume with the following defi-nition:

CV =f

2R (8.20)

So ultimately the internal energy of the gas is revealed:

U =m

MCV T (8.21)

The internal energy is a secondary state parameter of the gas, which means that it isunambiguously generated from a primary state parameter (concretely from the temper-ature). The final parameter is T2 the initial is T1. The variation of the internal energyat finite and infinitesimal temperature variations is as follows:

∆U =m

MCV · (T2 − T1) dU =

m

MCV · dT (8.22)

122

8.3 The p-V diagram

The p-V diagram is a phase-plane on which each point represents a state of thermody-namic equilibrium. The horizontal axis is the volume the vertical is the pressure. Dueto the ideal gas law points of identical temperature are located on hyperbolas whichare called isotherms. The higher the temperature the further is the hyperbola from theorigin. The thermodynamic process which progresses reasonably slowly to reach equilib-rium state at all times is called reversible process, provided no heat dissipation happeneddue to friction. Therefore the process can be displayed by a solid line on the p-V dia-gram. Rapidly occurring irreversible processes (irreversible jumps) with finite variationsdo not reach equilibrium during the process they reach equilibrium in the final state only.State parameters such as p and T are defined for equilibrium, so these parameters arenon-existing during irreversible jump. A doted line from the starting point to the finalpoint shows the jump on the p-V plane.

8.4 Expansion work of the gas

If the volume of a thermodynamic system increases then the system carries out work onthe environment. This is called the work of the gas which is a positive number uponexpansion. The infinitesimal amount of work can be expressed as follows: dW = pdV If afinite expansion process is concerned the work of the gas is the integral of the infinitesimalcontributions.

Figure 8.2: p-V diagram on work of expansion

Wgas =

V2∫V1

pdV (8.23)

123

The work of gas (Wgas) can be viewed as the area under the curve in the p-V diagram.The advantage is that the integral is positive or negative simultaneously with the workof the gas.

8.5 First law of thermodynamics

The variation of internal energy can be caused by two effects. Either heat transfer (∆Q)or external mechanical work (∆W ) can change the internal energy of the gas. Thetransferred heat is obviously positive if the system accepted heat. The mechanical workcan either be external work, which is positive upon compression and work of the gas,which is positive upon expansion. One has to be careful not to make confusion. Frompractical point of view it is advised to insist on using the work of the gas.

The first law of thermodynamics is essentially the conservation of energy. Mostpractical forms for infinitesimal and finite cases are as follows:

δQ =m

MCV · dT + pdV (8.24)

∆Q = ∆U + ∆W =m

MCV · (T2 − T1) +

V2∫V1

pdV (8.25)

The above described form of the first law can be interpreted as follows: The transferredheat (∆Q) is utilized for two purposes. Partly the internal energy (∆U) of the system isincreased and partly the gas carried out mechanical work (∆W ) on the environment.

sectionThermodynamic processesIn this section reversible processes are discussed. Let us transform the ideal gas

equation in the following manner:

pV

T=m

MR (8.26)

If the amount of the gas does not change during a process the right hand side of theabove equation is constant.

pV

T= Const (8.27)

Accordingly can be written:

p1V1

T1

=p2V2

T2

(8.28)

This is called the united law of gases. Historically it is worth to mention the more specificlaws which date back to those scientists who discovered the actual laws.

124

8.5.1 Isochoric process

Here the volume is constant during the process. The volume cancels out from the unitedlaw of gases:

p1

T1

=p2

T2

(8.29)

This is the Gay-Lussac II. law.The work of the gas is zero since there is no change in the volume (∆W =0 ). The

internal energy of the gas is increased by the total amount of the heat transferred.

Figure 8.3: p-V diagram of the isochoric process

∆Q = ∆U =m

MCV (T2 − T1) (8.30)

This equation explains the why the notation CV is used. In isochoric process this quantityshows up as molar heat capacitance of the gas.

In any processes however the variation of the internal energy can be expressed bymeans of CV with the above formula.

8.5.2 Isobaric process

Here the pressure is constant during the process. The pressure cancels out from theunited law of gases:

V1

T1

=V2

T2

(8.31)

This is the Gay-Lussac I. law. (The name with hyphen corresponds to one person.)

125

The work of expansion is the mere product of the pressure and the change in thevolume.

∆W = p(V2 − V1) (8.32)

Let us use the ideal gas equation.

pV2 =m

MRT2 pV1 =

m

MRT1 (8.33)

The difference of the above equations is as follows:

p(V2 − V1) =m

MR(T2 − T1) (8.34)

Now the first law can be used:

Figure 8.4: p-V diagram of the isobaric process

∆Q = ∆U + ∆W =m

MCV · (T2 − T1) +

m

MR(T2 − T1) =

m

M(CV +R) · (T2 − T1) =

m

MCp · (T2 − T1)

(8.35)

Accordingly the molar heat capacitance in case of isobaric process is denoted Cp.

Cp = CV +R (8.36)

8.5.3 Isothermal process

Here the temperature is constant during the process. The temperature cancels out fromthe united law of gases:

p1V1 = p2V2 (8.37)

126

This is the Boyle-Mariotte law. (The name with hyphen corresponds to two persons whoindependently discovered this law.)

In here the internal energy is constant (∆U =0 ). The pressure is expressed from theideal gas equation.

p =1

V

m

MRT (8.38)

The work of the gas can be calculated by integrating the pressure between the bordervolumes.

Figure 8.5: p-V diagram of the isothermal process

∆W =

V2∫V1

pdV =

V2∫V1

1

V

m

MRT · dV =

m

MRT

V2∫V1

dV

V=m

MRT ln(

V2

V1

) (8.39)

∆W =m

MRT ln(

V2

V1

) (8.40)

The first law looks like as it follows: ∆Q = ∆WAccording to this result the total heat transferred has been turned to mechanical

work.

8.5.4 Adiabatic process

It needs to be emphasized that reversible adiabatic process what we are dealing with. Theword adiabatic on it own means a process without heat exchange with the environmenthowever the mechanical work can be transferred to the system. In contrast to this neitherheat nor mechanical work can be transferred to the system which is said to be isolated.

127

Since there is no heat exchange with the environment (δQ =0 ), the infinitesimal formof the first law can be written as follows:

m

MCV · dT + pdV = 0 (8.41)

Earlier the constant volume molar heat capacitance has already been introduced: CV =f2R

After substitution one can find: mM

f2R · dT + pdV = 0(m

MR)· f

2dT + pdV = 0 (8.42)

On the other hand consider the temperature derivative of the ideal gas equation.

pV =m

MRT (8.43)

d(pV )

dT=(mMR)

(8.44)

The right hand side of the last equation can be recognized in the equation above inparenthesis. Let us substitute it.

d(pV )

dT

f

2· dT + pdV = 0 (8.45)

Here dT cancels out.

f

2d(pV ) + pdV = 0 (8.46)

The variation of the product can be separated.

f

2V dp+

f

2pdV + pdV = 0 (8.47)

−f2V dp =

f

2pdV + pdV (8.48)

−fV dp = (f + 2)pdV (8.49)

Here we introduce the adiabatic exponent denoted by the Greek letter (kappa κ).

−V dp =f + 2

fpdV

f + 2

f= κ (8.50)

128

−V dp = κ · pdV (8.51)

This is a separable differential equation:

−dpp

= κ · dVV

(8.52)

Let us integrate from the initial value to the final value:

−p2∫p1

dp

p= κ ·

V2∫V1

dV

V(8.53)

− ln(p2

p1

) = κ ln(V2

V1

) (8.54)

ln(p1

p2

) = ln

(V2

V1

)κ(8.55)

Since the natural logarithm function is unambiguous so the arguments are equal.

p1

p2

=

(V2

V1

)κ(8.56)

p1Vκ

1 = p2Vκ

2 (8.57)

In general can be written the exponential equation of the adiabatic process:

pV κ = Const (8.58)

Figure 8.6: p-V diagram of the adiabatic process

129

The equation of the adiabatic process can also be expressed by means of other primarystate parameters as well. Let us combine it with the universal gas law.

pV κ = ConstpV

T= Const (8.59)

After dividing the equations p cancels out:

T · V κ−1 = Const (8.60)

Express the V from the second equation and substitute back to the first one:

T κ

pκ−1= Const (8.61)

8.6 Summary of the molar heat capacitances

The molar heat capacitances have been used extensively at the discussion of the thermo-dynamic processes. The following parameters have been discussed: (R =8.3 J/molK )

The thermodynamic freedom degree is the number of the energy storing capabilitieseach of which can contain kT kinetic energy, according to the equipartition principle. Thenoble gases with mono-atomic molecule have three freedom degrees, which are the x, y, ztranslational directions. The diatomic molecules such as the air and lot of others havethe three x, y, z translations and two additional rotational axes except for one directionwhich connects the centers of the atoms. The exception is explained by the fact thatlooking at the molecule in the central line it looks like a point without extension, thusit lacks of moment of inertia. Higher number of atoms in the molecule generates sixthermodynamic freedom degrees since three translational and three rotational motionsare all available.

Number ofatoms in themolecule

Thermodynamicfreedom degreef

Constant vol-ume molar heatcapacitanceCV = f

2R

Constant pres-sure molar heatcapacitanceCp =

(f2

+ 1)R

Adiabatic expo-nentf+2f

= κ

1 3 32R 5

2R 5

3

2 5 52R 7

2R 7

5

3 or more 6 3R 4R 43

130

8.7 The Carnot cycle

The Carnot cycle is a reversible cycle which consists of two isothermal and two adiabaticprocesses. This is the classical example of the heat engine which is capable of generatingmechanical work from a certain part of the heat.

Figure 8.7: The Carnot cycle in p-V diagram

The corner points of the cycle are denoted A,B,C and D. The points A and B arelocated on the T1 isotherm while the C and D points are located on T2 isotherm. TheT1 temperature is higher than T2 therefore T2 isotherm is closer to the origin of the p-Vplane. The process starts in point A.

• AB section: (work output is positive in this section)

This is an isothermal expansion on T1 temperature. The heat intake from the heat bathis ∆Q1. The variation of the internal energy is zero (∆U =0 ) therefore the transferredheat is fully converted to mechanical work ∆W1.

∆W1 = ∆Q1 =m

MRT1 ln(

VBVA

) (8.62)

• BC section: (work output is positive in this section)

Adiabatic expansion (∆Q =0 ) as long as the temperature cools down to T2 temperature.The gas carries out positive mechanical work on the expense of its own internal energy.

∆W2 = −∆U2 =m

MCV (T1 − T2) (8.63)

131

• CD section: (work output is negative in this section)

Here an isothermal compression takes place on T2 temperature. The variation of theinternal energy is zero (∆U =0 ) therefore mechanical work ∆W3 is fully converted toheat ∆Q2 which in turn has been transferred to the sink.

∆W3 = ∆Q2 =m

MRT2 ln(

VDVC

) (8.64)

• DA section: (work output is negative in this section)

Adiabatic compression (∆Q =0 ) as long as the temperature warms up to T1 temperature.The mechanical work increased the internal energy of the gas.

∆W4 = −∆U2 =m

MCV (T2 − T1) (8.65)

The total amount of work carried out by the cycle ∆W is the sum of the mechanicalworks of the processes above: Apparently ∆W2 and ∆W4 cancel out.

∆WCycle = ∆W1 + ∆W2 + ∆W3 + ∆W4 = ∆W1 + ∆W3 (8.66)

∆WCycle =m

MR

(T1 ln(

VBVA

) + T2 ln(VDVC

)

)(8.67)

The above formula can be simplified further by considering the fact that the BC and DAsection are adiabatic processes. According to the laws of adiabatic process the followingequations can be written: (See at the end of the section “adiabatic process”.)

T1 · V κ−1A = T2 · V κ−1

D (8.68)

T1 · V κ−1B = T2 · V κ−1

C (8.69)

After dividing the two equations the temperatures cancel out on both sides:

VAVB

=VDVC

(8.70)

This can be substituted to the expression of the work of the cycle.

∆WCycle =m

MR

(T1 ln(

VBVA

)− T2 ln(VBVA

)

)=m

MR(T1 − T2) ln(

VBVA

) (8.71)

The work of the cycle can be viewed as the unusual rectangular area on the p−V plane.If the direction of the circumference is clockwise then the cycle carries out a positivework on the environment. In other words this is a heat engine.

132

The thermal efficiency of the heat engine (η) is defined as the ratio of the work ofthe cycle (Wcycle) over the heat intake from the heat bath (∆Q1). In other words this isthat percentage of the heat intake which has been converted to mechanical work duringthe cycle. The rest of the heat intake necessarily goes to the sink.

η =Wcycle

∆Q1

=mMR(T1 − T2) ln(VB

VA)

mMRT1 ln(VB

VA)

=T1 − T2

T1

(8.72)

Many terms cancel out.

η = 1− T2

T1

(8.73)

The work of the cycle can be expressed in an alternative way as well:

∆WCycle = ∆W1 + ∆W3 = ∆Q1 + ∆Q2 (8.74)

The efficiency is as follows:

η =∆Q1 + ∆Q2

∆Q1

= 1 +∆Q2

∆Q1

(8.75)

The two expressions of the efficiency are equal:

1 +∆Q2

∆Q1

= 1− T2

T1

(8.76)

After some ordering an important formula is the result.

∆Q1

T1

+∆Q2

T2

= 0 (8.77)

The two terms of the above equation are called “reduced heat”, which are the ratio of theheat and the actual temperature at which the heat transfer took place. This equationdeclares the fact that the sum of the reduced heats for the Carnot cycle is zero. This hasfar reaching consequences in connection with the concept of entropy.

133

Chapter 9

The entropy and the second law ofthermodynamics - Gyorgy Hars

At the end of the previous chapter an important relation has been revealed.

∆Q1

T1

+∆Q2

T2

= 0 (9.1)

The two terms of the above equation are called “reduced heat”, which are the ratio of theheat and the actual temperature at which the heat transfer took place. The “reducedheat condition” declares the fact that the sum of the reduced heats for the Carnot cycleis zero. This has far reaching consequences in connection with the concept of entropy.

9.1 The entropy

Consider the p-V phase plane. Let us draw in solid line numerous isotherms in equidistantsmall temperature steps. Similarly draw in doted line numerous adiabatic curves withsmall equidistant increments. Now an arbitrary reversible cycle is plotted on the top ofthe isotherm and adiabatic grid.

The cycle on the figure can be approximated with little motions partly on the isothermand partly on the adiabatic curves. This way the arbitrary cycle is composed of severaltiny Carnot cycles. The reduced heat condition is valid for each of them. The reducedheat intake at the high temperature side for a tiny Carnot cycle is equal to the reducedheat egress at the low temperature side. Therefore the reduced heat contributions cancelout by pairs, thus the total sum of the reduced heats is equal to zero for the wholearbitrary cycle. If one makes the grid infinitely fine the sum of reduced heats is converted

134

to the closed loop integral. Accordingly this can be written:∮Arb.cycle

dQ

T= 0 (9.2)

Figure 9.1: Arbitrary cycle with grid

If the arbitrary closed loop integral is zero this is equivalent with the statement thatthe integral between two points does not depend on the path of the integration.

Figure 9.2: Integration on two alternative paths

Let us break the closed loop with two points A and B. Point A will be considered

135

the starting point while B is the final point.∮Arb.cycle

dQ

T=

B∫A

dQ

T

Path1

+

A∫B

dQ

T

Path2

= 0 (9.3)

Integrating from B to A is the negative of the opposite direction integration. B∫A

dQ

T

Path1

=

B∫A

dQ

T

Path2

(9.4)

The integral depends on the initial and final points only.Since the integral is independent of the path therefore the concept of entropy (S)

can be introduced. Variation of the entropy can be calculated on the most convenientpath, since the actual path is unimportant. First an isochoric process takes us to theintermediate point (X) then an isothermal process reaches the final point (B).

Figure 9.3: Calculation of entropy variation

In isochoric process dQ = dU = mMCV dT therefore the following integral is evaluated

∆S1 =

X∫A

dQ

T=m

MCV

X∫A

dT

T=m

MCV ln

TBTA

(9.5)

After this an isothermal process follows: dQ = pdV .

∆S2 =

B∫X

dQ

T=

B∫X

pdV

T=

B∫X

p

TdV (9.6)

136

Here we use the ideal gas equation in the following form:

p

T=m

MR

1

V(9.7)

∆S2 =

B∫X

m

MR

1

VdV =

m

MR

B∫X

dV

V=m

MR ln

VBVA

(9.8)

So the total entropy variation is the sum of the two sub-processes:

SB − SA =m

MCV ln

TBTA

+m

MR ln

VBVA

(9.9)

The entropy variation can be expressed by any two of the three (p, V, T ) primary stateparameters. We have made the (T, V ) version. The (p, V ) and the (p, T ) versions follow.

The ideal gas equation for the final and the initial states are as follows:

pBVB =m

MRTB (9.10)

pAVA =m

MRTA (9.11)

Let us divide them.

pBpA· VBVA

=TBTA

(9.12)

Take the natural logarithm of both sides:

lnpBpA

+ lnVBVA

= lnTBTA

(9.13)

The above equation can substituted into the formula of the entropy variation.

SB − SA =m

MCV

(lnpBpA

+ lnVBVA

)+m

MR ln

VBVA

=m

M(CV +R) ln

VBVA

+m

MCV ln

pBpA

(9.14)

Accordingly: SB − SA = mMCp ln VB

VA+ m

MCV ln pB

pA

Similarly ln VBVA

= ln TBTA− ln pB

pANow the above equation is substituted to the entropy variation:

SB − SA =m

MCV ln

TBTA

+m

MR(ln

TBTA− ln

pBpA

) =m

M(CV +R) ln

TBTA− m

MR ln

pBpA(9.15)

137

Accordingly: SB − SA = mMCp ln TB

TA− m

MR ln pB

pADirect physical meaning can be associated to the variation of entropy between two

points of the p − V plane. By choosing a universal initial point as reference point, theentropy function can be converted to a secondary state parameter. This time any pointon the p − V plane is characterized by a single entropy value. Practical choice can bethe following reference point. (T0 =273K, p0 =105Pa, V0 = (m/M) 2.266 10−2 m3/mol).

9.2 The isentropic process

Let us find out where those points are located on the p− V plane, which have identicalentropy value. In other words we are looking for the isentropic curves. Here SA = SB isthe condition. The entropy variation is expressed as follows:

SB − SA =m

MCp ln

VBVA

+m

MCV ln

pBpA

= 0 (9.16)

Cp lnVBVA

+ CV lnpBpA

= 0 (9.17)

CpCV

lnVBVA

= lnpApB

(9.18)

Here we recall the adiabatic exponent (κ) which has already been introduced in chapter8 as the ratio of the molar heat capacitances.(

VBVA

)κ=pApB

(9.19)

pAVκA = pBV

κB (9.20)

This is a known formula, which was derived first at the discussion of the reversibleadiabatic process in the previous chapter. Now it has been proven that the isentropicprocess is identical with the reversible adiabatic process.

9.3 The microphysical meaning of entropy

Let us make some elementary combinatorial calculations to reveal the fundamental rootsof entropy. We have two compartments (1 and 2) and four particles (N =4 ) with thenames a, b, c and d. From the point of view of micro states the particles are distin-guishable, which means that a new micro state is generated if two particles replace each

138

other. However these micro states represent the same macro state since in macro stateonly the number of the particles count. Accordingly, from point of view of macro statesthe particles are indistinguishable. The concept of thermodynamic weight is associatedwith the macro state and it is defined as the number of those micro states which generatethe given macro state. In the table below all the possible micro states are listed as wellas the corresponding thermodynamic weights.

Observablemacro state

Macro state charac-teristics(N1, N2)

Names of particlesin compartment 1.(Micro states)

Number of micro states(Thermodynamic weight)W = N !

N1!N2!

A 4, 0 abcd 1B 3, 1 bcd, acd, abd, abc 4C 2, 2 ab, cd, ac, bd, ad, bc 6D 1, 3 a,b,c,d 4E 0, 4 - 1

In present case the thermodynamic weight (W ) can be expressed as follows:

W =N !

N1!N2!N = N1 +N2 (9.21)

In general case there are several elementary cells. The number of cells is denoted with k.A series of numbers provide the characteristics of the macro stateN1, N2, N3, N4, N5.........Nk

where the numbers represent the numbers of particles in the elementary cells from 1 tok. The corresponding thermodynamic weight is as follows:

W =N !

N1!N2!N3!.............Nk!N = N1 +N2 + ......Nk (9.22)

In terms of combinatorics the above formula is a permutation with repetition. In thenumerator the total number of permutation shows up provided all the particles aredistinguishable. In the denominator there is the divider that contains those cases whichdiffer in the order of particles within the cell. These cases do not constitute new macrostate since only the amount of particles in the cell counts.

9.4 Gay-Lussac experiment

The gas subjected to an isolated irreversible jump in this classical experiment. Thephenomena will be studied by both phenomenological and statistical thermodynamics.

139

9.4.1 Phenomenological approach

Consider a gas container which consists of two internal compartments with volumes V1

and V2. Between the volumes there is a valve that can be operated remotely. The wholecontainer is thermally isolated from the environment. One of the volumes (V1) containsgas with known parameters. Vacuum is in the other compartment (V2). Suddenly thevalve is opened the gas flows partly to the empty volume. During the gas flow a whistlenoise can be heard. Thermometers attached to both compartments show that V1 cooleddown and V2 warmed up by some extent. When the noise died out and some minutespassed both thermometers show the initial temperature, in accordance with the first lawof thermodynamics. No work has been done by the gas no heat has been transferred tothe system so the total internal energy must have been conserved. This means that thetemperature is necessarily unchanged.

Though the temperature is identical with that in the initial state, this is far of beingan isothermal process. This is an isolated irreversible jump without work done. Theinitial and the final states are in thermal equilibrium therefore drawing these two pointson the p − V plane is justified. However due to the irreversible jump it is not is notallowed to draw any solid line much a rather a doted line from the initial to the finalpoint.

Figure 9.4: Gay-Lussac experiment on the p− V plane

Since the entropy is a state parameter the variation of entropy can be expressedwithout any respect to the events which took the system from the initial to the finalstate (here I did not use the word “process” intentionally). The known formula from the

140

previous page can readily be used with the condition that VA = V1 and VB = V1 + V2.

SB − SA =m

MCV ln

TBTA

+m

MR ln

VBVA

(9.23)

The first term cancels out due to the identical temperatures values.

SB − SA =m

MR ln

VBVA

(9.24)

The final volume is obviously bigger than the initial therefore the entropy variation ispositive. The irreversible jump takes place necessarily to the right hand side directionto a higher entropy value adiabatic curve. In addition the initial and final states are onthe same isotherm.

9.4.2 Statistical approach

In the figure below the gas container of the experiment is shown. Let us subdivide thetotal VB volume to very small identical cells with the volume denoted with δ. The numberof cells (ωA and ωB) comes out as follows:

ωA =VAδ

ωB =VBδ

(9.25)

Figure 9.5: Gay-Lussac experiment (statistical approach)

The actual emerging macro states carry the highest thermodynamic weight. In thesemacro states all the cells contain the same amount of particles. Here N is the total

141

number of particles. The initial and final macro states are the following:

N

ωA

∣∣∣∣1

,N

ωA

∣∣∣∣2

,N

ωA

∣∣∣∣3

...................N

ωA

∣∣∣∣ωA

, 0|ωA+1 , 0|ωA+2 ................... 0|ωB−1 , 0|ωB (9.26)

N

ωB

∣∣∣∣1

,N

ωB

∣∣∣∣2

,N

ωB

∣∣∣∣3

...........................N

ωB

∣∣∣∣ωB

(9.27)

The corresponding thermodynamic weights can be readily written:

WA =N ![(

NωA

)!]ωA· [0!](ωB−ωA)

WB =N ![(NωB

)!]ωB (9.28)

It is worth to be are aware of the fact that 0!=1. Let us calculate the ratio of thethermodynamic weights.

WB

WA

=

[(NωA

)!]ωA[(

NωB

)!]ωB (9.29)

An approximate formula for the natural logarithm of the factorial will be used here:

lnn! ≈ n lnn− n (9.30)

The relative error of this formula is tending to zero when n goes to the infinity. In presentcase when n is extremely high the approximation is especially accurate. Sketch of prooffollows at the end of this section.

lnWB

WA

= ωA

(N

ωAln(

N

ωA)− N

ωA

)− ωB

(N

ωBln(

N

ωB)− N

ωB

)(9.31)

lnWB

WA

=

(N ln(

N

ωA)−N

)−(N ln(

N

ωB)−N

)= N

(ln(

N

ωA)− ln(

N

ωB)

)(9.32)

lnWB

WA

= N lnωBωA

= N lnVBVA

(9.33)

Multiply the equation with the Boltzmann constant.

k lnWB

WA

= Nk lnVBVA

Nk =m

MR (9.34)

142

k lnWB

WA

=m

MR ln

VBVA

(9.35)

Compare it with the variation of entropy stated earlier:

SB − SA =m

MR ln

VBVA

(9.36)

The right hand sides match perfectly so the left hand sides are equal:

SB − SA = k lnWB

WA

(9.37)

The above formula is called the Boltzmann equation. Now this has been proven for thespecial case of the Gay-Lussac experiment QED.

9.5 The Boltzmann equation

The Boltzmann equation is a fundamentally significant equation which establishes thebridge between phenomenological and the statistical approach of the thermodynamics.Its importance is far greater than that it has just been proven for.

SB − SA = k lnWB

WA

(9.38)

Here k is the Boltzmann constant (k = 1.36 · 10−23J/K), WA and WB are the thermo-dynamic weights of the A and B macro states. The message of Boltzmann equation inconnection with the Gay-Lussac experiment is clear: The entropy increases according tothe phenomenological formula. Therefore the variation of entropy is positive so the loga-rithm value in the Boltzmann equation should also be positive. Accordingly the ratio ofthermodynamic weights is bigger than unit which means that far greater thermodynamicweight belongs to the final macro state than to the initial macro state. The fact can begenerally stated that in the isolated system at irreversible jump the final equilibriumstate is located on a higher value isentropic curve on the p-V plane. The thermodynamicweight of the final state is bigger than that of the initial. At irreversible jumps the sys-tem is tending to the state of highest possible thermodynamic weight which is the finalequilibrium macro state.

9.6 Approximate formula (lnn! ≈ n lnn − n) a sketch

of proof:

lnn! = ln 1 + ln 2 + ln 3 + ......+ lnn (9.39)

143

The sum of logarithms of integers can be approximated with the following integral:

n∫1

lnx · dx = [x lnx− x]x=nx=1 = n lnn− n+ 1 ≈ n lnn− n (9.40)

Figure 9.6: Approximation of lnn! function. The crosshatched area is the absolute error

9.7 Equalization process

So far the subject of study was a single system consisting of gaseous particles. In equal-ization processes more thermodynamic systems take part, typically two systems willbe considered here. Initially these systems are in thermal equilibrium on their own.Suddenly they are united and the equilibrium state of both systems vanishes. Violentirreversible actions can proceed thus no equilibrium state parameters can be defined.Strictly speaking temperature, pressure, internal energy and entropy are not existingparameters in course of the events. After some time these actions are gradually relaxingand the united system occupies its newly born equilibrium state parameters.

144

9.7.1 Equalization between gaseous components

Let us consider two separated gas systems. They contain the same type of gas withthe following equilibrium state parameters V1, p1, T1 and V2, p2, T2.The gas systems areunited by removing the separating wall between them so the volumes are added. Wewonder the final state parameters and the variation of entropy.

The ideal gas equation will be used in the following form: pV = nRT Here n denotesthe number of moles.

System1 System2Parameters n1, p1, V1, T1 n2, p2, V2, T2

Volume V1 = n1RT1p1

V2 = n2RT2p2

Internal energy U1 = n1CVT1 U2 = n2CVT2

The internal energy, the volume, the number of moles and the entropy are addedtogether upon uniting the systems. The parameters without subscripts characterize theunited system.

U = U1 + U2 = CV (n1T1 + n2T2) = CV (n1 + n2)T (9.41)

The final temperature can be written accordingly:

n1T1 + n2T2

n1 + n2

= T (9.42)

This is a weighted arithmetic mean in terms of mathematics where the weights are thecorresponding mole numbers.

Now the pressure of the united system needs to be found. This is expressed from theideal gas equation:

p =nRT

V=

(n1 + n2)Rn1T1+n2T2n1+n2

R(n1T1p1

+ n2T2p2

)(9.43)

Two terms cancel out. The pressure finally can be expressed.

p =n1T1 + n2T2

n1T1p1

+ n2T2p2

(9.44)

This is a weighted harmonic mean in terms of mathematics where the weights are thecorresponding nT products. For later use it is worth to express the T over p ratio in

145

the final state. This is a weighted arithmetic mean of the T/p ratios in the initial stateswhere the weights are the corresponding moles.

T

p=n1

T1p1

+ n2T2p2

n1 + n2

(9.45)

Now the entropy variations need to be dealt with. The term variation means the differ-ence when the initial state parameters are subtracted from the final state parameters.

∆S1 = n1Cp lnT

T1

− n1R lnp

p1

∆S2 = n2Cp lnT

T2

− n2R lnp

p2

(9.46)

The total entropy variation of the equalization is the sum of the two variations above:

∆Seq = ∆S1 + ∆S2 = Cp(n1 lnT

T1

+ n2 lnT

T2

)−R(n1 lnp

p1

+ n2 lnp

p2

) (9.47)

∆Seq = Cp lnT (n1+n2)

T n11 T n2

2

−R lnp(n1+n2)

pn11 p

n22

= (n1 + n2)

(Cp ln

T(n1+n2)

√T n1

1 T n22

−R lnp

(n1+n2)√pn1

1 pn22

)(9.48)

∆Seq = (n1 + n2)

(Cp ln

T(n1+n2)

√T n1

1 T n22

+R ln(n1+n2)

√pn1

1 pn22

p

)(9.49)

The universal gas constant is factored out.

∆Seq = (n1 + n2)R

(CpR

lnT

(n1+n2)√T n1

1 T n22

+ ln(n1+n2)

√pn1

1 pn22

p

)(9.50)

For convenience reasons a new parameter is introduced here:

CpR

=f

2+ 1 = β (9.51)

With this parameter the formula can be transformed:

∆Seq = (n1 + n2)R

(β ln

T(n1+n2)

√T n1

1 T n22

+ ln(n1+n2)

√pn1

1 pn22

p

)(9.52)

∆Seq = (n1 + n2)R

lnT β

(n1+n2)

√T βn1

1 T βn2

2

+ ln(n1+n2)

√pn1

1 pn22

p

(9.53)

146

Let us organize the fraction under a single logarithm.

∆Seq = (n1 + n2)R ln

(T β

p(n1+n2)

√(p1

T β1)n1 · ( p2

T β2)n2

)(9.54)

The total variation of entropy due to the unification is finally revealed.

∆Seq = (n1 + n2)R ln

Tβ

p

(n1+n2)

√(Tβ1p1

)n1 · (Tβ2

p2)n2

(9.55)

The newborn equilibrium parameters show up in the numerator while the initial param-eters are in the denominator. The purpose is to prove the second law of thermodynamicsin mathematical precision. The second law states that the variation of entropy is alwayspositive at irreversible jumps of an isolated system.

Let us consider the numerator which is associated with the final state parameters. Thetemperature over pressure has already been expressed above from the original formulae.

According to the well-known mathematical theorem weighted geometric mean is al-ways not bigger than the weighted arithmetic mean.

T β

p= T (β−1)T

p= T (β−1)

n1T1p1

+ n2T2p2

n1 + n2

≥ T (β−1) (n1+n2)

√(T1

p1

)n1(T2

p2

)n2 (9.56)

So replace the numerator with a smaller or equal quantity. This way the calculatedformula is inevitably diminished or unchanged.

∆Seq ≥ (n1 + n2)R ln

T (β−1) (n1+n2)

√(T1p1

)n1(T2p2

)n2

(n1+n2)

√(Tβ1p1

)n1 · (Tβ2

p2)n2

(9.57)

Organize the fraction under a single root sign.

∆Seq ≥ (n1 + n2)R ln

(T (β−1) (n1+n2)

√(T1

p1

p1

T β1)n1(

T2

p2

p2

T β2)n2

)(9.58)

Here the pressures cancel out.

∆Seq ≥ (n1 + n2)R ln

(T (β−1) (n1+n2)

√(

1

T(β−1)1

)n1(1

T(β−1)2

)n2

)(9.59)

∆Seq ≥ (n1 + n2)R ln

T (β−1)((n1+n2)

√T n1

1 · T n11

)(β−1)

= (n1 + n2)R(β − 1) ln

(T

(n1+n2)√T n1

1 · T n11

)(9.60)

147

Here f denotes the thermodynamic freedom degree which can be 3, 5 and 6.

β − 1 =f

2∆Seq ≥ (n1 + n2)R(

f

2) ln

(T

(n1+n2)√T n1

1 · T n11

)(9.61)

Let us substitute the value of the final temperature to the numerator.

∆Seq ≥ (n1 + n2)R(f

2) ln

(1

n1+n2(n1T1 + n2T2)

(n1+n2)√T n1

1 · T n11

)≥ 0 (9.62)

In the numerator the weighted arithmetic mean of the initial temperatures is present.In the denominator the weighted geometric mean of the same temperatures shows upwith the identical weights. Here again one has to refer to the mathematical theoremof the inequality between the arithmetic geometric means. The numerator is surely notsmaller than the denominator so the fraction is not smaller than unit, which meansthat the logarithm value is a non-negative number. So ultimately the second law ofthermodynamics has been proven for the concrete case of the isolated irreversible jumpat gas equalization. Q.E.D.

Figure 9.7: Isentropic curves are displayedThe curve of S1 + S2 entropy is jumped over by the isolated equalization

Fig. 9.7

148

Some special cases of the general entropy variation formula will be discussed here.First the mole numbers of the initial components are equal. (n1 = n2 = n)

∆Seq = 2nR ln

Tβ

p√Tβ1p1· T

β2

p2

(9.63)

The initial temperatures can be equal too: (T1 = T2)

∆Seq = 2nR ln

1p√

1p1· 1p2

= 2nR ln

(√p1p2

p

)(9.64)

The final pressure can be written for the actual conditions: This is the harmonic meanof the initial pressures.

p =n1T1 + n2T2

n1T1p1

+ n2T2p2

=2

(1/p1) + (1/p2)(9.65)

This is substituted to the entropy variation above.

∆Seq = 2nR ln

√p1p2(2

(1/p1)+(1/p2)

) (9.66)

Here we should refer to the mathematical theorem that the harmonic mean is alwayssmaller or equal than the geometric mean. So the entropy variation is non-negative.

Finally the initial pressures are considered equal.

∆Seq = 2nR ln

T β√T β1 · T

β2

= 2nRβ ln

(T√T1 · T2

)(9.67)

The final temperature can be written for the actual conditions:

T =n1T1 + n2T2

n1 + n2

=1

2(T1 + T2) (9.68)

∆Seq = 2nR(f

2+ 1) ln

( 12(T1 + T2)√T1 · T2

)(9.69)

Here we should refer to the mathematical theorem that the arithmetic mean is alwaysgreater or equal than the geometric mean. So the entropy variation is non-negative again.

149

9.7.2 Equalization of non-gaseous materials without phase tran-sition

Heat equalization between liquid-solid and solid-solid components is studied mostly inthis point. In general the liquid-liquid equalization process is not part of this discussionexcept for the case when both components are water for instance, when the mixing isnot associated with growth of entropy.

The product of mass and specific heat is the heat capacitance denoted capital C.

U1 = m1c1T1 = C1T1 U2 = m2c2T2 = C2T2 (9.70)

The internal energies are added together.

U = C1T1 + C2T2 = (C1 + C2)T (9.71)

The temperature after equalization is denoted T without subscript.

T =C1T1 + C2T2

C1 + C2

=C1

C1 + C2

T1 +C2

C1 + C2

T2 (9.72)

Now the entropy variation is studied. The infinitesimal heat transfer can be used tocalculate the entropy variation. The actual temperature of the heat transfer is denotedT∗.

dQ = dU = C · dT∗ (9.73)

∆S1 =

T∫T1

dQ

T∗=

T∫T1

C1 · dT∗T∗

= C1

T∫T1

dT∗T∗

= C1 lnT

T1

(9.74)

∆S2 =

T∫T2

dQ

T∗=

T∫T2

C2 · dT∗T∗

= C2

T∫T2

dT∗T∗

= C2 lnT

T2

(9.75)

The total entropy variation of the equalization is the sum of the two components.

∆Seq = ∆S1 + ∆S2 = C1 lnT

T1

+ C2 lnT

T2

= (C1 + C2)

(C1

C1 + C2

lnT

T1

+C2

C1 + C2

lnT

T2

)(9.76)

After mathematical transformations this can be written:

∆Seq = (C1 + C2)

(ln

(T

T1

)(C1

C1+C2)

+ ln

(T

T2

)(C2

C1+C2))

(9.77)

150

The final result is as follows:

∆Seq = (C1 + C2) ln

T

T(

C1C1+C2

)

1 · T(

C2C1+C2

)

2

(9.78)

Let us substitute the final temperature calculated earlier:

∆Seq = (C1 + C2) ln

C1

C1+C2T1 + C2

C1+C2T2

T(

C1C1+C2

)

1 · T(

C2C1+C2

)

2

≥ 0 (9.79)

In the numerator of the fraction the weighted arithmetic mean of the initial temperaturescan be found, in the denominator the weighted geometric mean of the same temperaturesshow up with the identical weights. Now the mathematical theorem of the inequalitybetween the arithmetic and geometric mean needs to be referred to. According thistheorem the arithmetic mean is never smaller than the geometric mean. Ultimatelythe fraction is never smaller than unit which means that the entropy variation of theequalization process is never negative. The second law of thermodynamics has beenproven for this case. Q.E.D.

The non-negative result of the entropy variation can also be understood by verifyingan infinitesimal entropy transfer from the warmer component to the colder one. Thewarmer component carried out a dQ heat egress at T1 temperature while the coldercomponent took up the same dQ heat at a lower T2 temperature. The infinitesimalentropy egress dS 1 and intake dS 2 can be written as follows:

T1 ≥ T2 dS1 =−dQT1

dS2 =dQ

T2

(9.80)

The total infinitesimal entropy production is the sum of the formulae above:

dSeq = dS1 + dS2 = dQ(1

T2

− 1

T1

) = dQT1 − T2

T1 · T2

≥ 0 (9.81)

The difference in the numerator is obviously non-negative thus the original statement isproven.

9.7.3 Ice cubes in the water

The next problem contains phase transition which is associated with latent heat of melt-ing. This is a known parameter of any material. The actual latent heat (Qlat) is calculatedas a product: (Qlat = mLmelt). The entropy variation is the ratio of the latent heat overthe melting point temperature in Kelvin degrees. ∆Smelt = mLmelt

TmeltFrequent practice is to cool down the drinks by means of ice cubes.

151

We have a glass of water (m1 = 0.3 kg) at room temperature (T1 = 27 oC = 300K).Three ice cubes (m2 = 0.03 kg, T 2 =-13 oC = 260K) are dropped into the water. Afterthe cubes melted we want to find the final temperature and to determine the variationof the entropy of the system.

The chosen reference state is water at zero Celsius (T0 = 270K). The internal energies(U1, U2)are calculated relative to the reference state.

U1 = m1cw(T1 − T0) (9.82)

U2 = −m2L−m2cice(T0 − T2) (9.83)

The total internal energies of the initial components are equal with that of the final state.

U1 + U2 = m1cw(T1 − T0)−m2 [L+ cice(T0 − T2)] = (m1 +m2)cw(T − T0) (9.84)

After the system got into equilibrium the new temperature (T ) is formed.

T = T0 +m1

m1 +m2

(T1 − T0)− m2

m1 +m2

[L

cw+cicecw

(T0 − T2)

](9.85)

The numerical constants are as follows:

cw = 4.187 · 103 J

kgKcice = 2.108 · 103 J

kgKL = 3.34 · 105 J

kg(9.86)

cicecw

= 0.503L

cw= 79.8K (9.87)

After substitution the final temperature is revealed:

T = 273 +300

33027− 30

330[79.8 + 0.503 · 13] = 289.8K (9.88)

The final temperature is 289.8 K which is 16.8 Celsius. So the drink is cooler by 10.2 oC.The entropy variation of the water in the glass (∆S1) is negative.

∆S1 = m1cw lnT

T1

(9.89)

The entropy variation of the ice cubes (∆S2) is positive.

∆S2 = m2cice lnT0

T2

+m2L

T0

+m2cw lnT

T0

. (9.90)

152

The entropy production of the equalization is the sum of the variations:

∆Seq = ∆S1 + ∆S2 = m2cice lnT0

T2

+m2L

T0

+m1cw lnT

T1

m2cw lnT

T0

(9.91)

After some transformations can be written as follows:

∆Seq = m2

(cice ln

T0

T2

+L

T0

)+ cw(m1 +m2)

(m1

m1 +m2

lnT

T1

+m2

m1 +m2

lnT

T0

)(9.92)

∆Seq = m2

(cice ln

T0

T2

+L

T0

)+ cw(m1 +m2) ln

T

T(

m1m1+m2

)

1 · T(

m2m1+m2

)

2

(9.93)

The numerical constants can be substituted:

∆Seq = 0.03(2.108 · 103 ln273

260+

3.34 · 105

273) + 4.187 · 103 · 0.330 ln

289.8

300( 1011

) · 260( 111

)= 7.13

J

K(9.94)

The entropy variation of the equalization is positive as expected according to the secondlaw.

9.8 The second law of thermodynamics

In previous points the increase of entropy in isolated systems during irreversible eventshas been demonstrated for several concrete cases. The second law of thermodynamics isan empirically validated postulate. Increase of entropy in the isolated system is statisti-cally substantiated by the Boltzmann equation by declaring that highest thermodynamicweight belongs to the thermal equilibrium state.

The most accepted statement of the second law is as follows:In isolated system (no heat and no work exchange with the environment) at nat-

urally occurring irreversible events the total entropy production is positive.In limit case the entropy production can be zero in reversible process.

To be isolated is a fundamentally important condition in the statement above. Ifthe system is not isolated, the entropy variation can be anything. This can be highlynegative too. By cooling and compressing the gas the entropy is dropping fast all theway when it gets liquefied.

There are some related statements of the second law with the common feature ofdenying the possibility of the followings: Perpetual motion machine of the second kind,Heat engine operating cyclically with a single heat reservoir, Heat engine operating cycli-cally to convert heat to work in full extent, Heat that moves to the warmer object fromthe colder one on its own. These statements are merely the consequences of the “mostaccepted” version of the second law above.

153