Physics Beyond 2000

Physics Beyond 2000

Chapter 5

Simple Harmonic Motion

http://library.thinkquest.org/28388/Mechanics/Motions/SHM.htm



Simple Harmonic Motion

• It is a particular kind of oscillation.

• Abbreviation is SHM.

• Some terms

Amplitude, period, frequency and angular frequency.

http://www.physics.uoguelph.ca/tutorials/shm/Q.shm.html








Definition of SHM

• The motion of the particle whose acceleration a is always directed towards a fixed point and is directly proportional to the distance x of the particle form that point.

xa .2where ωis a constant, the angular frequency.

Examples of SHM

0

x = 0

The particle is at the position x = 0.It has a velocity to the right.

V

Examples of SHM

a

A

A is the maximum distance, the amplitude.

0

x

The particle moves to the right with retardation a.Note that x and a are in opposite directions.

x =A

Examples of SHM

a

A


0

x

The particle moves back to the left with acceleration a.Note that x and a are still in opposite directions.

x = 0

Examples of SHM

a

A


0

x

The particle moves to the left with retardation a.Note that both x and a change directions. They are still in opposite directions.

x = -A

Examples of SHM

a

A


0

x

The particle moves to the right with acceleration a.Note that x and a are still in opposite directions.

x = 0

Examples of SHM

The position x = 0 is the equilibrium position.

0

x = 0

V

Examples of SHMThe position x = 0 is the equilibrium position at which the net force is zero.

xa In the oscillation,

0

x V

a

The negative sign indicates the direction of a is opposite to that of x.

Example 1

• Is it a SHM?

• a = -16.x

Differential equation of SHM• acceleration a of SHM

xa .2

xdt

xd.2

2

2

xx .2The left hand sides show the accelerations in different mathematical forms.

The kinematics of SHM

• Displacement x of a SHM.

• A solution of the differential equation

xdt

xd.2

2

2

is x = A.sin(ωt + ψ)

t is the time, ωis the angular frequency,A is the amplitude and ψis the initial phase.

The kinematics of SHM

Displacement x

If ψ=0 x = A.sin (ωt)

0

tA

-AT 2T 3T

time

• Displacement x = A.sin(ωt + ψ) of a SHM.• The displacement x of a particle performing SHM changes sinusoid

ally with time t.

2

T• The period of the SHM is

Phase and Initial Phase

• x = A.sin(ωt + ψ)ωt + ψis called the phase.

• At t = 0, phase reduces to ψ. x = A.sin(ψ)

ψis called the initial phase.

Phase

• x = A.sin(ωt + ψ)

• If ψ=0, x = A.sin(ωt )

• If ψ= π/2, x = A.cos(ωt )

• If ψ= π, x = -A.sin(ωt )

• etc.

Initial Phase ψ

The value of ψis determined by the initial position of x (at t = 0).

i.e. how the motion is started.

ψ= 0• x = A.sin(ωt)• At t = 0, x = 0.

– The motion starts at x = 0.

• In the first T/4, x increases with time t and approaches the amplitude A.

0 x = 0

V

At t = 0

a

ψ= 0• x = A.sin(ωt)• At t = 0, x = 0.

– The motion starts at x = 0.

• In the first T/4, x increases with time t and approaches the amplitude A.

Displacement x

0

time tA

-AT 2T 3T

ψ= π/2• x = A.cos(ωt)• At t = 0, x = A.

– The motion starts at x = A.• In the first T/4, x decreases with time t and approa

ches 0.

0

x = A

v = 0

At t = 0

a

ψ= π/2• x = A.cos(ωt)• At t = 0, x = A.

– The motion starts at x = A.• In the first T/4, x decreases with time t and approa

ches 0.

Displacement x

0

time tA

-AT 2T 3T

ψ= π• x = -A.sin(ωt)• At t = 0, x = 0.

– The motion starts at x = 0.• In the first T/4, x decreases with time t and approaches -A.

0 x = 0

V

At t = 0

a

ψ= π• x = -A.sin(ωt)• At t = 0, x = 0.• The motion starts at x = 0.• In the first T/4, x decreases with time t and approaches -A.

Displacement x

0time t

A

-AT 2T 3T

ψ= 2π/3• x = -A.cos(ωt)• At t = 0, x = -A.• The motion starts at x = -A.• In the first T/4, x increases with time t and approaches 0.

0x = -A

v=0

At t = 0

a

ψ= 2π/3• x = -A.cos(ωt)• At t = 0, x = -A.• The motion starts at x = -A.• In the first T/4, x increases with time t and approaches 0.

Displacement x

0time t

A

-AT 2T 3T

Angular frequency ω

• Period T = time for one complete oscillation.

• Frequency f = number of oscillations in one second.

• Angular frequency = 2f

Tf

1

Angular frequency ω

• In a SHM, x = A.sin(ωt+ψ)

• After a time T, x must be the same again.

• A.sin(ωt+ψ) = A.sin(ωt+ ωT +ψ)ωT = 2π

fT 22 Unit of ω is rad s-1

Isochronous oscillation

• The period T in a SHM is independent of the amplitude A.

• A SHM is an isochronous oscillation.

Velocity in SHM

• x = A.sin(ωt+ψ)

dt

dxv

)cos(. tAv

Velocity in SHM

• x = A.sin(ωt+ψ)

)cos(. tAv

What is the maximum speed in this motion?

vo = Aω

The maximum speed occurs at the equilibrium position.i.e. when x = 0.

Velocity in SHM

• Example 2.

Acceleration in SHM

dt

dva

)cos(. tAv

Acceleration in SHM

)sin(.2 tAa

What is the maximum acceleration in this motion?

ao = -Aω2 or Aω2

The maximum acceleration occurs at the positionswith maximum displacement.i.e. when x = A or -A.

Displacement, Velocity and Acceleration in SHM

• x = A.sin(ωt + ψ)• v = Aω.cos(ωt + ψ)• a = -Aω2.sin(ωt + ψ)

Displacement, Velocity and Acceleration in SHM with ψ=0

0t

A

-AT 2T 3T

x

v

0t

Aω

-Aω T 2T 3T

a

x = A.sin(ωt)

v = Aω.cos(ωt)

0 tAω2

-Aω2 T 2T 3T a = -Aω2.sin(ωt)

Acceleration and Displacementin SHM

• x = A.sin(ωt + ψ)• a = -Aω2.sin(ωt + ψ)• a = - ω2.x• This is a characteristic of a

SHM.

Acceleration and Displacementin SHM

• a = - ω2.x• acceleration and displacement in SHM are i

n antiphase.(i.e. in opposite directions.)

0t

A

-AT 2T 3T

x = A.sin(ωt)

x

0 tAω2


a

Velocity and Displacement in SHM

0t

A

-AT 2T 3T

x = A.sin(ωt)

x

v

0t

Aω

-Aω T 2T 3T v = Aω.cos(ωt)

• v leads x by π/2 or 900 (or x lags behind x by π/2 or 900 ).

Velocity and Displacement in SHM

• v leads x by π/2 or 900.

• 22 xAv

Velocity and Acceleration in SHM

v

0t

Aω

-Aω T 2T 3T v = Aω.cos(ωt)

• a leads v by π/2 or 900 (or v lags behind a by π/2 or 900 ).

0 tAω2


a

The kinematics of SHM• We may look upon SHM as a projection of a uniform circular motion on its

diameter.• A green ball is performing a uniform circular motion with angular velocity

.

ω

The kinematics of SHM• A green ball is performing a uniform circular

motion with angular velocity . • Its projection on the diameter is the red ball

performing SHM.

ω

http://id.mind.net/~zona/mstm/physics/mechanics/simpleHarmonicMotion/vrmlshm.html





performing SHM.

ω



performing SHM.

ω



performing SHM.

ω



performing SHM.

ω

http://www.phy.ntnu.edu.tw/java/shm/shm.html



Period• The period of the circular motion is the same as

the period of the SHM.

Period T = time to completeone revolution

Period T = time to moveto and fro once.

fT

12

Displacement of SHM

Let A = radius of the circleIt is also the amplitude of the SHM

Let be the starting angle of the green ball.It is the initial phase of the red ball.

At time = 0, the position of the green ball and the red ball are as shown.

ω

A

O

Displacement of SHM

A

O

After time = t, the green travels and angular displacement t and the red ball moves a displacementx as shown.

ω

x

t

The displacement of SHM is x = A.sin(t + ) t + is the phase of the SHM.

Velocity of SHM

A+t

O

ω

v

The linear speed of the green ball is A.The velocity of the red ball is the horizontal component of A .

v = A .cos(t + )

Let A = vo

v = vo.cos(t + )

A

Acceleration of SHM

A2+t

O

ω The green has centripetal acceleration A.2

The acceleration of SHMis the horizontal componentof A.2

va

a = -A2.sin(t + )Let A2 = ao

a = - ao .sin(t + )

SHM and UniformCircular Motion

A2+t

O

ω

va

x = A.sin(t + )

and

a = -A2.sin(t + )

a = - 2.x

SHM and UniformCircular Motion

A2+t

O

ω

a

The projection of a uniform circular motion on its diameter is SHM.

SHM and Uniform Circular Motion

A2

O

ω

A

In a SHM, the red ball passes its equilibrium position:1. The speed of the red ball is a maximum.2. The acceleration of the red ball is zero.

SHM and Uniform Circular Motion

A2

O

ω

A

In a SHM, when the red ball is at its maximum displacement:

1. The speed of the red ball is a zero.

2. The acceleration of the red ball is a maximum.

Rotating Vector

• It is difficult to represent SHM using diagram because its quantities vary with time.

• Use a vector which is rotating in uniform circular motion.

• http://members.nbci.com/Surendranath/Shm/Shm01.html

O

http://members.nbci.com/Surendranath/Shm/Shm01.html



Rotating Vector



O

Rotating Vector



O

Rotating Vector



O

Rotating Vector



O

Rotating Vector


• Its projection on the diameter is SHM.

• Period T =

O

2

Rotating Vector• Use three rotating

vectors.

• The three projections on the diameter represent displacement, velocity and acceleration of the SHM respectively. The magnitude of the displacement vector is xo.

The magnitude of the velocity vector is vo. The magnitude of the acceleration vector is ao.

O

xo

voao

Rotating Vector

O

xo

voao

From the rotating vectors,the velocity v leads displacement x by π/2and acceleration a and displacement x are in antiphase.

http://electron6.phys.utk.edu/phys135/modules/m9/oscillations.htm



The Dynamics of SHM

a

x v

When the particle is at a displacement x, thereis an acceleration a. From Newton’s 2nd law, there must be a net force Fnet on it Fnet = m.aand in SHM a = -ω2.xThe above two equations Fnet = -mω2.x

Horizontal Block-spring system

Equilibriumposition

Is this a SHM?http://library.thinkquest.org/16600/intermediate/simpleharmonicmotion.shtml

The block isoscillating.

Mass of the block = m ; Mass of the spring can be ignored. There is not any friction.

http://library.thinkquest.org/16600/intermediate/simpleharmonicmotion.shtml

Horizontal Block-spring systemEquilibriumposition The block is

oscillating.

Suppose that the block has been displaced bya displacement x.Fnet = m.a andFnet = -k.x where k is the spring constant

x

m.a = -k.x a = - .x k

m


oscillating.

x

a = - .x k

m

Compare it to a standard SHM equationa = - .x 2We see that the motion of the block is a SHMwith ω =

m

k

a


oscillating.

x

a

Its period T = k

m

22

The period T is independent of the amplitudeof oscillation. This is an isochronous oscillation.


Equilibriumposition

The force is maximum when the displacement is a maximum.

The force is zeroat the equilibriumposition.

Fnet


Equilibriumposition

The speed is zero when the displacement is a maximum.

The speed is fastestat the equilibriumposition.

v


• A simulation program

• You may download the program from– http://www.programfiles.com/index.asp?ID=74

91

http://www.programfiles.com/index.asp?ID=7491

http://www.programfiles.com/index.asp?ID=7491

Vertical Block-spring system

• http://webphysics.ph.msstate.edu/javamirror/explrsci/dswmedia/harmonic.htm

• http://www.exploratorium.edu/xref/phenomena/simple_harmonic_motion.html

• http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html

http://webphysics.ph.msstate.edu/javamirror/explrsci/dswmedia/harmonic.htm



http://www.exploratorium.edu/xref/phenomena/simple_harmonic_motion.html

http://www.exploratorium.edu/xref/phenomena/simple_harmonic_motion.html

http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html

http://hyperphysics.phy-astr.gsu.edu/hbase/shm.html


naturallength

Equilibriumposition

e = extension

mg

ke

m = mass of the blockk = spring constantmg = ke Mass of the spring can be ignored.

oscillation

Is this a SHM?


oscillation

natural length

Equilibriumposition

Suppose that the blockhas moved a displacementx below the equilibrium position.At the instant, what arethe forces acting on theblock?x

e


oscillation

natural length

Equilibriumposition

Fnet = -k(x+e)+mgandmg = keFnet = -k.x

x k(x+e)

e

mg


oscillation

natural length

Equilibriumposition

Fnet = -k.xand Fnet = m.am.a = -k.x

x k(x+e)

e

mga = - .x

k

m


oscillation

natural length

Equilibriumposition

x k(x+e)

e

mg

a = - .x k

m

So it is a SHM with

m

k

and

Its period T = k

m

22


• The motion is similar to that of a horizontal block spring system.

• At the equilibrium position, the net force on the block is zero though the tension of the spring is ke.

• At the equilibrium position, the speed of the block is fastest.

• At the maximum displacement, the net force is a maximum and the speed is zero.

Examples

• Example 4• Horizontal block-spring system.

• Example 5• Vertical block-spring system.

• Example 6• Combination of springs.

Simple pendulum

The simple pendulumis set into oscillation.Is it a SHM?http://webphysics.ph.msstate.edu/javamirror/explrsci/dswmedia/harmonic.htm



Simple pendulum

Equibriumposition

θ

is the length of the simple pendulum.

m is the mass of the bob.Suppose that the bob hasan angular displacementθ from the equilibriumposition.

Simple pendulum

Equibriumposition

θ

What are the forcesacting at the bob atthis instant?

Simple pendulum

What are the forcesacting at the bob atthis instant?T = tension from the stringmg = weight of the bob

Equilibriumposition

θ

T

mg

Simple pendulum

As the bob is in a circularmotion, there is a net force(the centripetal force) on the bob.

Equilibriumposition

θ

T

mg

Simple pendulum

The tangential component Ft = -mg.sinFor small angle, sin . So Ft -mg. ……… (1) Only the tangential componentFt is involved in the change ofspeed of the bob. So Ft = m.a …………. (2)Equilibrium

position

θ

T

FtFr

Simple pendulum

Equilibriumposition

θ

T

FtFr

x

From equations (1) and (2), a = -g.………….. (3)The displacement of the bobis x …………. (4)

Simple pendulum

Equilibriumposition

θ

T

FtFr

x

From equations (3) and (4), a = - .x

So it is a SHM with

g

g

Note that this is true forsmall angle of oscillation

Simple pendulum

Equilibriumposition

θ

T

FtFr

x

g

Note that this is true forsmall angle of oscillation

gT

22

The period of oscillation is

Example 7

• To find the period of a simple pendulum

A floating object

• It is a SHM.

water

mg

Floating forceIn equilibrium

A floating object

• It is a SHM.

water

A floating object

• It is a SHM.

water

http://www.deepscience.com/shm.html

http://www.deepscience.com/shm.html

Liquid in a U-tube

• It is SHM.

Energy in SHM

• There is a continuous interchange of kinetic energy and potential energy.

• Horizontal Block-spring system– kinetic energy of the block <-> elastic potential

energy of the block

• Simple pendulum– kinetic energy of the bob <-> gravitational

potential energy

Energy in SHM

• Kinetic energy is maximum when the particle passes the equilibrium position.

• The maximum kinetic energy is

22 )(2

1

2

1 AmmvU oko

Energy in SHM

• Kinetic energy is maximum when the particle passes the equilibrium position.

22 )(2

1

2

1 AmmvU oko

• If we write m

k2

2

2

1kAU ko

Energy in SHM

• Kinetic energy is zero when the particle reaches its maximum displacement A.

• Kinetic energy is completely transformed into potential energy.

• The maximum potential energy Upo is equal to the maximum kinetic energy Uko

Energy in SHM

• If the energy is conserved (the motion is undamped), the total energy is

Uo = Uko = Upo

22

2

1

2

1kAmvo

•At any position, the total energy is 222

2

1

2

1

2

1kAkxmvUo

Example 10

• Angular speed of a simple pendulum

Energy versus displacement

• When the displacement = x,

– the potential energy Up = k.x2

– the kinetic energy Uk = m.v2

– the total energy Uo = k.A2

– Up + Uk = Uo

2

1

2

1

2

1

Energy versus displacement

22

2

1

2

1kxkAUUU pok

2

2

1kxU p

0 A-Ax

Energy

Uk

Up

Uo

2

2

1kAUo

Example 11

• Find the total energy and the maximum speed.

Energy versus time

• For simplicity, choose ψ= 0.

• Use x = A.sin(ωt) and v = A ω.cos(ωt)

))2cos(1(2

1

)(cos

)(cos2

1

2

222

tU

tU

tmAU

o

o

k

Energy versus time

• For simplicity, choose ψ= 0.

• Use x = A.sin(ωt) and v = A ω.cos(ωt)

))2cos(1(2

1

)(sin.

)(sin2

1

2

22

tU

tU

tkAU

o

o

p

Energy versus time

tUU

tUU

tAx

op

ok

2

2

sin.

cos.

sin.

time0 T 2T

Displacement UpUk

Damped Oscillation

• There is loss of energy in this kind of oscillation due to friction or resistance.

• As a result, the amplitude decreases with time.

• There are– Slightly damping– Critical damping– Heavy damping

http://www3.adnc.com/~topquark/fun/JAVA/dho/dho.html

http://www.math.ualberta.ca/~ewoolgar/java/Hooke/Hooke.html





Slightly damping• The amplitude decreases exponentially with

time. An = A1.n

where is a constant

Critical damping

• The system does not oscillate but comes to rest at the shortest possible time.– e.g. galvanometer

Displacement x

time t

0 T 2T

Heavy damping

• The resistive force is very large. The system returns very slowly to the equilibrium position.

Displacement x

time t

0 T 2T

Example 12

• Slightly damping

Forced oscillations • The vibrating system is acted upon by

an external periodic driving force. The system is vibrating at the frequency of the external force.

• Without the external driving force, the frequency of free oscillation is called natural frequency.

• With the external periodic driving force, the amplitude of vibration varies

• http://physics.uwstout.edu/physapplets/a-city/physengl/resonance.htm

hand vibratingup and down

http://physics.uwstout.edu/physapplets/a-city/physengl/resonance.htm

http://physics.uwstout.edu/physapplets/a-city/physengl/resonance.htm

Forced oscillations

• The amplitude of a forced oscillation depends on– damping – frequency of the driving force


Forced oscillations • The amplitude of a forced oscill

ation is a maximum when the frequency of the driving force is equal to the natural frequency fo of the system


fo

Amplitude

Driving frequency

no damping

slight damping

heavy damping

Forced oscillations • Resonance occurs when the driving

frequency equals the natural frequency.

• The amplitude is a maximum at resonance.


fo

Amplitude

Driving frequency

no damping

slight damping

heavy damping

Phase relationship

• When the driving frequency is less than the natural frequency fo of the vibrating system, the two motions are in phase.

• When the driving frequency is greater than or equal to the natural frequency fo of the vibrating system, the driven motion always lags behind the driving motion.

Phase relationship

Driving frequencyDriven motion

Phase lag Less than fo 0

equal to fo /2

greater than fo

slight dampingheavy damping

/2

0 fodriving frequency

Tacoma Narrows Bridge

The bridge was driven by the wind.


The bridge was in resonance with the wind.


The bridge was in resonance with the windand finally collapsed.

Why is a simple harmonic motion ‘simple’?

• http://www.physlink.com/ae135.cfm

http://www.physlink.com/ae135.cfm

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Physics Beyond 2000