Stepny of the Ass Ques Oscillation 2

8/8/2019 Stepny of the Ass Ques Oscillation 2

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1. Marks LOD MCA

Topic

Question: A particle of mass 5 x 10

-3

kg performing SHM of amplitude 150 mm takes 47 s tomake 50 oscillations. The kinetic energy in millijoules (x 10-3

J)of the oscillations at the restposition is:

(a) 0 mJ (b) 2.5 mJ

(c) 3.9 mJ (d) 5 mJ

Correct Option:

Hints:

Solution/Answer:

2. Marks LOD MCA

1

Topic Simple Harmonic Motion

Question: Consider that a particle as shown figure is executing Simple Harmonic Motion. The centre of

+A and A is equilibrium position of the particle.What is magnitude of force on particle executing Simple

Harmonic Motionat the equilibrium position where displacement is zero?

(a) (b)


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(c) (d)

Correct Option:

Hints:

Solution/Answer:

3. Marks LOD MCA

Topic

Question:

to know what is the nature of the graph of kinetic energy along y axis and displacement along x axis for a

particle performing simple harmonic motion ?

i think i am sure it is an inverted parabola symmetric about y axis

please i would like a confirmation or correction

(a) (b)

(c) (d)

Correct Option:

Hints:

Solution/Answer:

1. A particle oscillates with undamped SHM. Which one of the followingstatements is true about the acceleration of the oscillating particle?

A. ? It is least when the speed is greatest.

B. ? It is always in the opposite direction to the velocity.

C. ? It is proportional to the frequency.

D. ? It decreases as the potential energy decreases.

which statement out of following is true regard to two functions f (t) and f (t T ) out of following


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The time tis chosen to be zero when the particle is at +A and it returns to +A at t= T. For this motion the

time period is given by T. The time tis chosen to be zero when the particle is at +A and it returns to +A at t=

T.

a motion for which the displacementx(t) of the particle from a certain chosen origin is found to vary with time

may be termed as Simple Harmonic Motion A = amplitude

m = mass ofswinging body

y = actual position at time t

= angular velocity

The kinetic energy =

EK = m/2*A^2*^2*sin^2(t)

The potential energy =

EP = m/2*A^2*^2*cos^2(t)

EP = EK

m/2*A^2*^2*sin^2(t) = m/2*A^2*^2*cos^2(t)

sin^2(t) = cos^2(t)

t = pi/4

y = A *cos(t)

y = A*(pi/4)y = 0,707*A

1) Equation of simple harmonic motion: y= Asint if initial phase and

displacement are zero. Here y is the displacement, is the angular

frequency and A is the amplitude.

y= Acost also represents simple harmonic motion but it has a phase

lead of /2 compared to the above one.

If there is an initial phase of the equation is y= Asin(t + ).y= Asint + Bcost represents the general simple harmonic motion

of amplitude (A2 + B2) and initial phase tan-1(B/A).

(2) The differential equation of simple harmonic motion is d2y/dt2 = -

2y

Note that =(k/m) where k is the force constant (force per unit


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displacement) and m is the mass of the particle executing the SHM.

(3)Velocityof the particle in SHM, v= (A2 y2)

Maximum velocity, vmax = A

(4) Acceleration of the particle in SHM, a = - 2y

Maximum acceleration, amax = 2A(5)Kinetic Energyof the particle in SHM, K.E. = m 2( A2 y2)

Maximum Kinetic energy= m 2A2

Potential Energyof the particle in SHM, P.E. = m 2y2

Maximum Potential Energy= m 2A2

Total Energyin any position = m 2A2

Note that the kinetic energy and potential energy are maximum

respectively in the mean position and the extreme position. The sum of

the kinetic and potential energies which is the total energy is a constant

in all positions. Remember this:

Maximum K.E. = Maximum P.E. = Total Energy= m 2A2

m = mass ofswinging body

y = actual position at time t

= angular velocity

The kinetic energy =

EK = m/2*A^2*^2*sin^2(t)

The potential energy =

EP = m/2*A^2*^2*cos^2(t)

EP = EK

m/2*A^2*^2*sin^2(t) = m/2*A^2*^2*cos^2(t)

sin^2(t) = cos^2(t)

t = pi/4

y = A *cos(t)

y = A*(pi/4)y = 0,707*A

2)

3)

(6) Period of SHM = 2(Inertia factor/ Spring factor)

In cases oflinearmotion as in the case of a spring-mass system or a


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simple pendulum, period, T = 2 (m/k) where m is the mass and k

is the force per unit displacement.

In the case ofangularmotion, as in the case of a torsion pendulum,

T = 2 (I/c) where I is the moment of inertia and c is the torque

(couple) per unit angular displacement.You may encounter questions requiring calculation of the period of

seemingly difficult simple harmonic oscillators. Understand that the

question will become simple once you are able to find out the force

constant in linear motion and torque constant in angular motion.

Angular cases will be rare in Medical and Engineering Entrance test

papers. Let us now discuss some typical questions.

The following simple question appeared in the AIIMS 1998 test paper:

If a simple pendulum oscillates with an amplitude 50 mm and

time period 2s, then its maximumvelocityis

(a) 0.1 m/s (b) 0.15 m/s (c) 0.8 m/s (d) 0.16 m/s

Maximum velocity vmax = A where is the angular frequency and A

is the amplitude. Therefore vmax = (2/T)A = (2/2)5010-3 = 0.157

m/s [Option (b)].

The following question appeared in Kerala Engineering Entrance 2005

test paper:

A particle executes linear simple harmonic motion with an

amplitude of 2 cm. When the particle is at 1 cm from the meanposition, the magnitude of its velocityis equal to that of its

acceleration. Then its time period in seconds is

(a) 1/ 23 (b) 23 (c) 2/3 (d) 3/2 (e) 3/

The magnitudes of the velocity and acceleration of the particle when its

displacement is y are (A2 y2) and 2y respectively. Equating them,

(A2 y2) = 2y, from which = [(A2 y2)]/y = (4 1) = 3. Period

T = 2/ = 2/3.

Suppose you place a sphere ofmass m and radius r inside a

smooth, heavyhemispherical bowl of radius of 37r placed on

a horizontal table. If the sphere is given a small displacement,

what is its period of oscillation?

(a) 2(m/37rg) (b) 2(m/rg) (c) 12(r/g) (d) 2(r/g) (e)

2(37r/g)

The arrangement depicted in this question is similar to that of a simple


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pendulum. Instead of the usual string, you have a concave surface to

confine the bob (sphere) to its path along the arc of a circle. The usual

expression for the period, T = 2(L/g) holds here also, where the

length L = 36r since the length of the pendulum is measured from the

centre of gravity of the bob. The point of suspension is evidently at thecentre of the hemispherical bowl. The correct option is 12(r/g) given

in (c).

What will be the period of oscillation of a simple pendulum of

length 100 cm in a spaceship in a geostationaryorbit?

Well, in any satellite orbiting the earth (in any orbit), the condition of

weightlessness exists (effective g = 0), the pendulum does not oscillate and the

period therefore is infinite.

Consider the following question:

A simple pendulum is arranged using a small metallic bob ofmass

mand a light rubber cord of length L (on suspending the bob),

area of cross section A and Youngs modulus Y. [One should use

inextensible cord onlyfor simple pendulum!]. When this

unconventional pendulum is at rest in its mean position, the bob is

pulled slightlydown and is released. Then, the period of the

vertical oscillation of the bob is (assuming that the size of the bob is

negligible compared to the length of the cord)

(a) 22L/g (b) 2(mL/YA) (c) 2 (m/YAL) (d) 2 (L/g) (e) 2(mY/AL)

The period as usual is given by T = 2(m/k). Here m is the same as the mass

of the bob. The force constant can be found by writing the expression for

Youngs modulus (since it arises from the elastic force in the cord): Y =

FL/A(L) where L is the increase in the length of the cord on pulling the bob

down with a force F. Therefore, the force constant, F/(L) = YA/L. On

substituting this value, the period is 2(mL/YA).

The following MCQ on simple harmonic motion may generate a little

confusion in some of you:

A sphere ofmass M is arranged on a smooth inclined plane of angle

, in between two springs of spring constants K1 and K2 . The

springs are joined to rigid supports on the inclined plane and to the

sphere (Fig). When the sphere is displaced slightly, it executes

simple harmonic motion. What is the period of this motion?


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(a) 2[Mgsin/(K1-K2)] (b) 2[M/{K1K2/(K1+K2)}] (c)

2[Mgsin/(K1+K2)] (d) 2[M/(K1+K2)] (e) 2[(K1+K2)/M]

You should note that gravity has no effect on the

period of oscillation of a spring-mass system since the restoring force is

supplied by the elastic force in the spring. (It can oscillate with the same

period in gravity free regions also). So, whether you place the system on an

inclined plane or a horizontal plane, the period is the same and is determined

by the effective spring constant and the attached mass only. The effectivespring constant is K1 + K2 since both the springs try to enhance the opposition

to the displacement of the mass. The period of oscillation, as usual is given by,

T = 2(Inertia factor/Spring factor) = 2[M/(K1 + K2)], given in option (d).

The following two questions (MCQ) appeared in Kerala Engineering

Entrance 2006 testpaper:

(1)The instantaneous displacement of a simple harmonic oscillator

is given byy= A cos(t + /4). Its speed will be maximum at the

time

(a) 2/ (b) /2 (c) / (d) /4 (e) /

This question was omitted by a fairly bright student who got selected with a

good rank. The question setter used the term speed (and not velocity) to make

things very specific and to avoid the possible confusion regarding the sign. So

what he meant is the maximum magnitude of velocity. The velocity is the time

derivative of displacement: v = dy/dt = -A(sin t + /4). Its maximum

magnitude equal to A is obtained when t = /4, from which t = /4.

(2) A particle ofmass 5 g is executing simple harmonic motion withan amplitude 0.3 m and time period /5 s. The maximumvalue of

the force acting on the particle is

(a) 5 N (b) 4 N (c) 0.5 N (d) 0.3 N (e) 0.15 N

If you remember the basic expression for period in the form, T = 2(m/k)

where k is the force constant, the solution becomes quite easy. From this, k =

42m/T2 = 42 510-3/(/5)2 = 0.5. Since k is the force for unit


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displacement, the maximum force is k times the maximum displacement

(amplitude). Therefore maximum force = kA = 0.50.3 = 0.15N.

[If you remember that = (k/m) you can arrive at the answer since T =

2/].

The frequency of vertical oscillations of a mass suspended at the end of a light spring is n. If

the system is taken to a location where the acceleration due to gravity is reduced by 0.1%, the

frequency of oscillation will be

(a) 1.01 n

(b) 0.99 n

(c) 1.001 n

(d) 0.999 n

(e) n

A spring-mass system (unlike the simple pendulum) does not require a gravitational force

for oscillations since the restoring force required for oscillations is supplied by the elastic

forces in the spring.

[Note that the period (T) of oscillations is given by T = 2(m/k) where m is the massattached to the spring and k is the spring constant].

Therefore the change in g does not affect the frequency and the correct option is (e).

(2) One end of a light spring is fixed to the ceiling and a mass Mis suspended at the other

end. When an additional mass m is attached to the massM, the additional extension in the

spring is e. The period of vertical oscillation of the spring-mass system now is

(a) 2[(M+m)e/mg]

(b) 2(me/mg)

(c) 2[(M+m) /mge]

(d) 2[me/(M+m)g]


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(e)2(M/mge)

The period (T)of oscillations is given byT = 2[(M+m)/k] where k is the spring constant.

Since an additional weight mg attached to the spring produces an additional extension e, the

spring constant k = mg/e.

Therefore, period of oscillations T = 2[(M+m)/(mg/e)] = 2[(M+m)e/mg], as given in

option (a).

(3) The period of vertical oscillations of a massMsuspended using a light spring of spring

constant k is T. The same spring is cut into three equalparts and they are used in parallel to

suspend the massMas shown in the adjoining figure. What is the new period of oscillations?

(a) T

(b) 3T

(c) 9T

(d) T/9

(e)T/3

The original period of oscillation Tis given by

T = 2(M/k)

When the spring is cut into three equal parts, each piece has spring constant 3 k.

[Since the length of each piece is reduced by a factor three, the extension for a given applied

force will be reduced by a factor three so that the spring constant (which is the ratio of force

to extension) will become three times].


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Since the three pieces are connected in parallel the effective spring constant of the

combination is 3k+3k+3k = 9k. The new period of oscillations T1is given by

T1= 2(M/9k) = T/3 since T = 2(M/k)

You will find some multiple choice questions with solution in this section here as well as

here.

POSTED BY M V AT 8:51 PM0 COM M ENTS LINKS TO THIS POST

LABELS: OSCILLATION , SIM PLE HARM ONIC M OTION , SPRING

Thursday, December 06, 2007

Two Questions on Oscillations

Today we will discuss two multiple choice questions on simple harmonic

motion:

(1) A simple pendulum of period 2 s has a small bob ofmass 50 g.

The amplitude of oscillation of the bob is 10 cm and it is at a height

of 45 cm from the ground in its mean position. While oscillating,the string breaks just when the bob is in its mean position. The

horizontal distance R `from the mean position where the bob will

strike the ground is nearly

(a) 35.2 cm (b) 23 cm (c) 15.3 cm

(d)12.4 cm (e) 9.4 cm

The angular frequency () ofoscillationofthe pendulum is given by


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LABELS: OSCILLATION , SIM PLE HARM ONIC M OTION

Sunday, June 10, 2007

Force, Momentum and Circular Motion

You might have noted that the angle between the force acting on a body and the momentum ofthe

body can be anything between zero and 2. (But the changeofmomentum is always in the direction

of force). Here is a simple question which you should be able to answer in a minute. Ifyou find it

difficult you should work harder to understand basic points inmechanics thoroughly.

A particle moves in a plane such that the rectangular components of its momentum vary simple

harmonically with the same periodandamplitude, but with a constant phase difference of /2.

The angle (in radian)between the momentum of the particle and the force acting on it is

(a) (b) (c) zero (d) /2 (e) varying between zero and 2

The particle is forced to move simple harmonically along two mutually perpendicular directions. In

other words, this is a case ofthe superpositionoftwo simple harmonic motions ofthe same perod

and amplitude at right angles to each other. Since the phase difference is /2, the resultant motion

is uniform circular motion. The angle between the resultant momentum ofthe particle and the force

acting on the particle is therefore /2.

Mow consider the following MCQ which appeared inIIT-JEE 2007 question paper:

A particle moves in the X-Y plane under the influence ofa force such that its linear momentum is p(t)

= A[ cos(kt) sin(kt)], where A and k are constants. The angle between the force and the

momentum is

(a) 0 (b) 30 (c) 45 (d) 90

Simply by noting that the momentum vector p has simple harmonically varying components A

cos(kt) and A sin(kt) in the X andYdirections respectively, you can conclude that this is a case ofthe

superposition two simple harmonic motions of the same frequency and amplitude at right angles,

with a constant phase difference of/2. [The phase difference is /2 since one is a sine function

while the other

If a particle moves in simple harmonic motion with a frequency of 3.00 Hz and an amplitude of 5.00

cm through what total distance does the particle move during one cycle of its motion, what is its


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maximum speed, where does this maximum speed occur and what and where does the maximum

acceleration of the particle happen?

Answer

Total distance = 10 cm

let w be the symbol for omega, so that w = 2(pi)f

y = Asin(wt) = 5sin[6(pi)t]

v = dy/dt = Awcos(wt)

so the max. velocity = Aw = 30(pi) cm/s which occurs as the particle passes through the midpoint

a = dv/dt = -Aw^2sin(wt) so the max. acceleration is

Aw^2 = 180(pi^2) cm/s^2 which occur at the extremes of the range of the particle.

a. The maximum displacement from the equilibrium position A = 10.0 cm.

b. The time for one complete oscillation T = T/2 s. Notice the maximum positive

displacement x = +10.0 cm occurs at t = 0 and the next time at t = T/2 s.

It occurs again at t = Ts.

2.

a. v(t) = - (2T/T)(A sin 2Tt/T). The maximum value of the sine is 1. The

maximum absolute value of v = 2TA/T. The signs account only for the

direction of the velocity.

b. From Fig. 1b above, |vmax| = 40.0 cm/s = 2TA/T = 2T(10cm)/Ts/2.

3.


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a.

i. a(t) =- (2T/T)2(A cos 2Tt/T). Maximum value of the cosine is

1. |amax| = (2T/T)2A. The signs account only for the direction of

the acceleration.

ii. a(t) = - (2T/T)2 (A cos 2Tt/T) = - (2T/T)2 x(t)

since x(t) = A cos 2Tt/T.

-a(t)/x(t) = (2T/T)2.

b. From Fig. 1c above,

|amax| = 160.0 cm/s2 = (2T/T)2A = (2T/Ts/2)2 10 cm.

4. a. For a mass attached to a spring, F = - kx or F/x = - k, where k is aconstant. The applied force F is directly proportional to the displacement xand the minus sign says it is in the opposite direction to x.

b. When the spring is extended and released, Fnet = ma or

- kx = ma or - a/x = k/m.

c. From 3a(ii), - a(t)/x(t) = (2T/T)2

By comparison, k/m = (2T/T)2 or T = 2T(m/k)1/2.

5. a. Given x(t) = A cos (2Tt/T + H), where A is the maximum displacement from

the equilibrium position. The maximum value of cos (2Tt/T + H) is 1, so the

equation accurately describe the definition of A.

b. x(t + T) = A cos [2T(t + T)/T + H]

= A cos [2Tt/T + 2TT/TH]

= A cos [2Tt/T + 2TH]

= A cos (2Tt/T + H)

= x(t).

The definition ofT is accurately described by the equation of motion for

simple harmonic motion, A cos (2Tt/T + H), because it allows the value of

x at t to equal the value of x at t + T or t + nT where n = 1, 2, 3 . . .

c. For x(t) = A cos (2Tt/T -T/2), x(0) = xo = A cos (- T/2) = 0.


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v(t) = - (2T/T)A sin (2Tt/TT/2).

v(0) = vo = - (2T/T)A sin (T/2) = (2T/T)A = +vmax.

At t = 0, the object is at the equilibrium position and travelling with themaximum velocity in the +X-direction.

6. Given that d2x/dt2 + (k/m)x = 0 (Equation 1")

Show that x(t) = A cos (Tt/T + H) is a solution. (Equation 1')

v = dx/dt = - (2T/T)A sin (2Tt/T + H).

dv/dt = d2x/dt2 = - (2T/T)2[A cos (2T\t/T + H)] = - (2T/T)2x (Equation 2)

Substituting Eq. 2 into Eq. 1", - (2T/T)2x + (k/m)x = 0.

This equation is true if (2T/T)2= (k/m) or T = 2T(m/k)1/2.

7.

In general, x(t) = A cos (2Tt/T + H) and v(t) = dx/dt = -A(2T/T) sin (2Tt/T + H).

(a) and (d)

In Fig. for #7a above, x(t) is plotted for H = 0.

In Fig. for #7d above, v(t) is plotted for H = 0.

For H = 0 and t = 0 , the initial conditions are:

the initial displacement = xo = A and

the initial velocity = vo = 0.

Immediately after t = 0, the object moves to the left (with a negative velocity).


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(b) For H = - T/2, x(t) = A cos (2Tt/T - T/2).

By trigonometric identity,

cos (C - D) = cos C cos D + sin C sin D (with C = 2Tt/T and D = T/2)

x(t) = A[cos 2Tt/T cos T/2 + sin 2Tt/T sin T/2]

x(t) = A[cos 2Tt/T(0) + sin 2Tt/T(1)] = A sin 2Tt/T

Notice the first maximum in Fig. b (immediately above) lags that in

Fig. a (above) by T/2 radians.

(e) v(t) = - 2TA/T sin (2Tt/T - T/2)

By trigonometric identity,

sin (C - D) = sin C cos D - cos C sin D

v(t) = -2TA/T[(sin 2Tt/T)(0) - (cos2Tt/T)(1)]

v(t) = (2TA/T) cos 2Tt/T


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For H = - T/2 and t = 0, xo = 0 and vo = +2TA/T

(c) For H = - T, x(t) = A cos (2Tt/T - T)

= A[cos 2Tt/T cos T + sin 2Tt/T sin T]

= A[(cos 2Tt/T)(-1) + (sin 2Tt/T)(0)]

x(t) = - A cos 2Tt/T

Notice that Fig. c (immediately above) lags Fig. a (above) by T radians.

(f) v(t) = - A(2T/T) sin (2Tt/T - T)

= - (2TA/T)[sin 2Tt/T cos T-cos 2Tt/T sin T]

=(- 2TA/T)[(sin2Tt/T)(-1) - (cos2Tt/T)(0)].

v(t) = (2TA/T) sin 2Tt/T

For H = - T and t = 0, xo = - A and vo = 0.

Immediately after t = 0, the object moves to the right with a positive velocity.

(g) The function ofH is to state the initial conditions.

Note: H!T or - T gives the same result.

8. a. x(t) = A cos ([t + H)

x(0) = xo = A cos H (Equation 1)

b. v(t) = dx/dt = - [A sin ([t + H)

v(0) = vo = -[A sin H (Equation 2)

Dividing both sides of Eq. 2 by - [:

- vo/[ = A sin H (Equation 3)

c. Dividing Eq. 3 by Eq. 1, tan H = - vo/x[o.

d. Squaring Eq. 1 and Eq. 3 and adding:

xo2 + (- vo/[)

2 = A2(cos2H + A sin2H) or [xo2 + (- vo/[)

2]1/2= A.

9.


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In Fig. for #9(a) above, the spring is not stretched. In Fig. for #9(b) above, the

mass is attached and the spring is stretched a distance xo.

a. The mass comes to rest and Fnet = ma = m(0).Taking down to be positive,

- kxo + mg = 0 (Equation 1)

b. In Fig. for #9(c) above, the spring has been displaced an additionaldistance x. Now Fnet = ma, where a 0 once the spring is released.Taking the direction of x, which is down, as positive,

- kx - kxo + mg = ma (Equation 2)

From Eq. 1, we see that - kxo + mg = 0. Eq. 2 becomes:

- kx = maor

- a/x = k

/m.

The ratio of a to x is the same whether the spring is mounted

horizontally or vertically.

c. As before, - a/x = (2T/T)2 = (2Tf)2 = k/m and f = (1/2T)(k/m)1/2.

10.

a. The forces acting on the pendulum bob are its weight mg and the tension Tin the string.

b. The only force tangent to the path is a restoring force - mg sin 5. From

the triangle with lengths, we find that

sin 5 = x/L and - mg sin 5 = - (mg/L)x.

For small displacements, x s, we can think of the displacement and the

restoring force acting horizontally.

Fnet = ma

- (mg/L)x = ma


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c. Since m, g, and L are constants, the restoring force, - (mg/L)x, is directlyproportional to the displacement and in the opposite direction. The

pendulum is an example of simple harmonic motion.

-a/x = (g/L) = (2T/T)2= (2Tf)2. f = (1/2T)(g/L)1/2.

For another approach, write5 = s/L. For small angles 5 is approximately

equal to sin 5. The restoring force - mg sin 5 - mg5 = m d2s/dt2

= m d2(L5)/dt2, or - (g/L)5 = d25/dt2and d25/dt2 + (g/L)5 = 0.

This is the same form as d2x/dt2 + (k/m)x = 0, for whichT = 2T (m/k)1/2

and x = A cos (2T/T). By comparison with the spring, for the pendulum T

= 2T (L/g)1/2 and f = 1/T = (1/2T)(g/L)1/2 and 5!5max cos (2T/T).

11. Given x(t) = 0.01 m cos (0.02T s-1 t - T/2) compare with x(t) = A cos (Tt/T + H)

and find:

a. the amplitude A = 0.01 m

b. 2T/T = 0.02T s-1; period T = 100 s,

c. the frequency f = 1/T = 0.01 s-1, and

d. the initial phase H = - T/2.

12.

From Fig. 3 above, we see that:

a. The cosine curve repeats itself every 4.0s so the period T = 4.0 s.

b. The amplitude of the motion A =10.0 cm.

c. If we write the equation of motion as a function of the cosine, we let H = 0.

x(t) = A cos 2Tt/T = 10.0 cm cos 2Tt/4.0 s = 10.0 cm cos Tt s-1/2.

d. v(t) = dx/dt = -10.0T/2 cm/s sin Tt s-1/2

|vmax| = 5.0T cm/s

e. a(t) = dv/dt = -10.0 (T/2)2cm/s2cos Tt s-1/2

|amax| = 2.5T2 cm/s2


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13.

For the X-component:

(a) x = r cos 5 = r cos [t

(b) vx = dx/dt = - [r sin [t

ax = dvx/dt = - [2(r cos [t) = - [2x

(c) ax/x = - [2 = - 4T2f2

(d) (e) Since the acceleration is directly proportional to the displacement and in the

opposite direction, the motion is simple harmonic. Remember by Newton's second

law of motion the acceleration is directly proportional to, and in the same direction

as, the force.

(f) ax/x = - [2 = - 4T2f2= - k/m or [!2Tf = (k/m)1/2

For the Y-component:

(a) y = r sin 5 = r sin [t

(b) vy = dy/dt = [r cos [t

ay = dvy/dt = - [2(r sin [t) = - [2y

(c) ay/y = - [2 = - 4T2f2

(d) (e) The motion is again simple harmonic, and

(f) ay/y = - [2 = - 4T2f2 = - k/m or [ = 2Tf = (k/m)1/2

14. For a mass-spring system,

- kx = ma = md2x/dt2 or

d2x/dt2+ (k/m)x = 0, where the period T = 2T (m/k)1/2.

By comparison with


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b2d2x/dt2+ c2x = 0 or

d2x/dt2+ (c/b)2x = 0,

we see that in this case T = 2T (b/c).

15. For a total swing back and forth of 4.0 cm, the amplitude A is 2.0 cm.

For x(t) = A cos 2Tt/T, v(t) = dx/dt = - A2T/T sin 2Tt/T.

The maximum velocity of the pendulum occurs at the center of the swing equal to

10 cm/s.

|vmax| = 2TA/T. T = 2TA/vmax= 2T (2.0 cm)/10 cm/s = 0.4T s.

16. x(t) = 4.0 cm cos (Tt s-1 - T/6).

2.0 cm = 4.0 cm cos (Tt s-1 - T/6).

cos (Tt s-1 - T/6) = 0.5 and (Tt s-1 - T/6) = T/3 (60o).

v(t) = dx/dt = - 4.0T cm/s sin (Tt s-1 - T/6).

When (Tt s-1 - T/6) = T/3, sin T/3 = 0.866 and v = - 4.0T (0.866) cm/s

= - 10.9 cm/s.

17.

[= 2Tf = 2T (3/2 s-1) = 3T s-1. xo = 0.25 m and vo = - 1.5 m/s.

We know that 0 < H


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18.

Fnet = ma = m d2x/dt2

For Fig. 5a,

- k1x - k2x = - (k1 + k2)x = m d2x/dt2 or

d2x/dt2+ [(k1 + k2)/m]x = 0.

Compare with

d2x/dt2+ [k/m]x = 0 when f = (1/2T)(k/m)1/2

and find for this case,

f = (1/2T)[(k1 + k2)/m]1/2.

The "effective" spring constant for springs in parallel is keff = k1 + k2. . + kn.

For Fig. 5b, the spring with constant k2 is in contact with mass m that has a

displacement x = x1 + x2, where x1 is the extension of the spring with constant k1

and x2 is the extension of the spring with constant k2. The force on the object is:

- k2x2 and - k2x2 = m d2x/dt2 (Equation 1)

Also,

x = x1 + x2 (Equation 2)

and the magnitude of the force on the second spring due to the first spring equals

the magnitude of force on the first spring due to the second spring, or

k1x1 = k2x2 or x1 = k2x2/k1 (Equation 3)

Substituting Eq. 3 into Eq. 2,

x = (k2x2/k1) + x2 = (k1 + k2)x2/k1 , orx2 = k1x/(k1 + k2) (Equation 4)

Substituting Eq. 4 into Eq. 1,

- [k2k1/(k1 + k2)]x = m d2x/dt2 or

d2x/dt2+ [{k1k2/(k1 + k2}/m)]x = 0.


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Comparing with

d2x/dt2+ [k/m]x = 0

For this case,

f = (1/2T)[{k1k2/(k1 + k2)}/m]1/2

.

The "effective" spring constant for series is:

keff = {k1k2/(k1 + k2} or

1/keff= 1/k1 + 1/k2 + . .+ 1/kn.

19. Imagine the spring cut into thirds with each part having a spring constant k. When

the three springs are connected in series, the spring constant is

k = 10.0 N/m.

For series,

1/k = 1/10 N/m = 1/k + 1/k + 1/k= 3/k

k = 3k = 30.0 N/m.

When two of these springs with k are connected in series,

1/k = 1/k + 1/k = 2/30 N/m,

or the spring constant with2/3 of the spring left (1/3 cut off) is

k = 15 N/m.

T= 2T(m/k)12 = 2T(0.30/15)1/2s = 2T (0.02)1/2 s.

20.

a. The differential mass dm = msdy/L

(Fig. 6 above)

b. Assuming the velocity vy at y increases linearly with y from 0 at y = 0to v at y = L, vy = vy/L.

c. dK = 1/2 dm v 2 = 1/2(msdy/L)(vy/L)2.


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The effective mass that takes place in the oscillation is ms/3.The fraction of the mass is 1/3.

21.

a. The slope of force F as a function of the extension x is

k = (5.0 - 0)N/(0.50 - 0)m = 10 N/m.

b. For a total mass M and force constant k, the period of the motion is


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T = 2T(M/k)1/2.

For a spring of effective mass ms, M = m + ms, where m is the variable

mass added to the spring load. For this case

T = 2T (m + ms)/k]1/2

orT2 = (4T2/k)m +(4T2/k)ms (Equation 1)

If we compare Eq. 1 to the equation of a line, y = (slope)x + (y-intercept),

we see that T2 versus m should yield a straight line and the intercept on the

T2axis is (4T2/k)ms. From Fig. 7a above, we find k = 10 N/m. From Fig.

7b above, we find the intercept on the T2 axis = 0.154 s2 = (4T2/k)ms =

(4T2/10 N/m)ms. ms = (1.54/4T2)kg = 0.390 kg and ms = 3 x (0.390 kg)

= 0.117 kg = 117 g.

c. When T2 = 0 in Eq. 1, 0 = (4T2/k)m + (4T2/k)ms or

ms = - m = - (-0.0390 kg).

ms = 0.0390 kg, agreeing with the value found in part (b).

d. P.S. It helps to have my graphing program that reads off coordinates.

22.

For simple harmonic motion, a/x = - [2.

For maximum acceleration, |amax| = A[2.

For motion in the vertical direction, (Fnet)y = may = m(0) orN - mg = 0 and N = mg.

The frictional force that keeps the block from slipping on the plate f = N = mg =

ma = mA[2. A = g/[2 = 0.60(10 m/s2)/(1.5 s-1)2= 2.7 m.

23. a. For the equilibrium position, dU/dr = 0 = (-5a/r6) + (3b/r4).Calling r = ro at the equilibrium position, ro = (5a/3b)

1/2.

b.

The reduced mass = m/2 and the frequency f = (1/2T)(k/)1/2=


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(1/2T)[(12b/m)(3b/5a)5/2]1/2.

24.

a. U = mgy

b. dU/dx = d(mgy)/dx = mg.d[R - (R2 - x2)1/2]/dx = mgx/(R2 - x2)1/2.For x < < R, dU/dx = mgx/R. Fx = - dU/dx = - (mg/R)x.The force Fx is directly proportional to the displacement x and in theopposite direction. The motion is simple harmonic.

c. Fx = ma or

- (mg/R)x = m d2x/dt2 and d2x/dt2 + (g/R)x = 0.

Compare d2x/dt2 + (k/m)x = 0 (for whichT = 2T(m/k)1/2)

and see for this case T= 2T(R/g)1/2.

25. a. When x = A, v = 0 and K = 0. In general, E = U + K = 1/2 kx2 + 1/2 mv2.

When x = A, E = 1/2 kA2 + 1/2 m(0) = 1/2 kA2. Since E is a constant, Ealways equals:

1/2 kA2= 1/2 kx2 + 1/2 mv2 (Equation 1)

b. Multiplying Eq. 1 by 2/m gives:

(k/m)A2= (k/m)x2 +1/2 mv2 or

(k/m)(A2 - x2) =v2 and

v =dx/dt =(k/m)1/2(A2 - x2)1/2.

c. Separating variables dx/(A2 - x2)1/2 = (k/m)1/2 dt.

Let x = A sin 5, dx = A cos 5 d5.

(A2 - x2)1/2 = A(1 - sin25)1/2 = A cos 5.

For limits on 5, xo = A sin 5o and sin 5o = xo/A.

The lower limit is sin -1 xo/A and the upper is sin-1 x/A.

Integrating the above equation gives:

sin -1 x/A + sin -1 xo/A = (k/m)1/2t or


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sin -1 x/A = (k/m)1/2t + sin -1 xo/A.

Letting sin -1 xo/A = H,

sin -1 x/A = (k/m)1/2t + H or

x(t) = A sin [(k/m)1/2t + H].

26.

a. X = rx F. X= rF sin r, F. About the pivot point, the torque for the rod of

mass m1 is -(L/2)m1g sin 5 and for the point mass it is -Lm2g sin 5. The

negative signs occur because they are restoring torques. When thependulum is swinging counterclockwise, the torque tends to make it swing

clockwise. The moment of inertia of the rod about an end is 1/3 m1L2. The

moment of inertia of a point particle of mass m2 a distance L from the axisis m2L

2.

b. X = -E

-Lg(m1/2 + m2)sin 5= (m1/3 + m2)L2 d25/dt2

For small 5sin 5 is approximately equal to 5 and

d25/dt2+ [g(m1/2 + m2)/(m1/3 + m2)L]5 = 0.

Comparing with

d2x/dt2+ [k/m]x = 0 (for which Period = 2T (m/k)1/2),

we find for this pendulum,

Period = 2T [2(m1 + 3m2)L/(m1 + 2m2)3g]1/2

27.


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For oscillations about pin axis (Fig. 24 above),

a. X= rx F.

X = rF sin r,F = -rmg sin 5 -rmg 5 for small 5.

The moment of inertia about the pin a distance r from the center of mass,

I = (1/2 mR2 + mr2).

b. X= IE = (I)d

2

5/dt

2

-rmg5 = (1/2 mR2 + mr2)d25/dt2

c. or

d25/dt2+ [rg/(1/2 R2 + r2)]5 = 0.

Compare with

d2x/dt + [k/m]x = 0 (for which Period = 2T[m/k]1/2)

and see for the disk,

Period = 2T[(R2 + 2r2)/2rg]1/2.

d. For a minimum period,

d(Period)/dr = 0 =

2T(1/2)[(2rg)(4r) - (R2+ 2r2)2g]/4r2g2][(R2 + 2r2)/2rg]1/2 or

(2rg)(4r) = (R2+ 2r2)2g or

4r2= R2+ 2r2 and 2r2 = R2 or

r = R/(2)1/2.


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Stepny of the Ass Ques Oscillation 2