HSC PYQ 05 Oscillations – Maharashtra Board Solutions

05 Oscillations

Multiple Choice Questions

1. The total work done by a restoring force in simple harmonic motion of amplitude

and angular speed

, in one oscillation is

(A)
(B) zero
(C)
(D)

Ans. (B) zero

2. An amplitude of a simple pendulum of a period and length in increased by . The new period of that pendulum will be
(A)
(B)
(C)

(D) T

Ans. (D) T

3. The motion of a simple pendulum is

(A) oscillatory but not periodic

(B) periodic but not oscillatory

(D) periodic as well as oscillatory

Ans. (D) periodic as well as oscillatory

4. The maximum velocity of a particle performing linear S.H.M. is and its maximum acceleration is , the amplitude of S.H.M. is
(A)
(B)
(C)
(D)

Ans. (C)

5. Two particles perform linear simple harmonic motion along the same path of length and period as shown in the graph below. The phase difference between them is

(A) zero rad
(B)
(C)
(D)

Ans. (B)

6. A particle executing linear S.H.M. has velocities and at distances and respectively from the mean position. The angular velocity of the particle is

(A)
(B)
(C)

Ans. (B)

7. The phase difference between displacement and acceleration of a particle performing S.H.M. is

(A)
(B)
(C)
(D)

Ans. (B)

8. The average displacement over a period of S.H.M. is

(A = amplitude of S.H.M.)
(A) 0
(B)
(C)
(D)

Ans. (A) 0

9. If the metal bob of a simple pendulum is replaced by a wooden bob of the same size, then its time period will

(A) increase

(B) remain same

(D) first increase and then decrease

Ans. (B) remain same

10. A seconds pendulum is suspended in an elevator moving with constant speed in downward direction. The periodic time (T) of that pendulum is

(A) less than two seconds

(B) equal to two seconds

(D) very much greater than two seconds

Ans. (B) equal to two seconds

11. Two springs of force constants and are stretched by same force. If and be the work done stretching the springs then
(A)
(B)
(C)
(D)

Ans. (B)

12. In a damped harmonic oscillator, periodic oscillations have amplitude.

(A) gradually increasing

(B) suddenly increasing

(D) gradually decreasing

Ans. (B) suddenly increasing

13. If the particle starts its motion from mean position, the phase difference between displacement and acceleration is

(A)
(B)
(C)
(D)

Ans. (C)

14. In simple harmonic motion, acceleration of the particle is zero, when its

(A) velocity is zero

(B) displacement is zero

(D) both velocity and displacement are maximum

Ans. (B) displacement is zero

15. The equation of a simple harmonic progressive wave travelling on a string is . The speed of the wave is
(A)
(B)

Ans. (D)

16. A simple harmonic oscillator has amplitude and period 4 seconds. The interval of time required by it to travel from to is

(A) second
(B) second
(C) second
(D) second

Ans. (B) second

Theory Questions

Differential Equation of S.H.M.

1. Obtain the differential equation of linear S.H.M.

Ans:

i. In a linear S.H.M., the force is directed towards the mean position and its magnitude is directly proportional to the displacement of the body from mean position.

where, is force constant and is displacement from the mean position.

ii. According to Newton’s second law of motion,

From equations (1) and (2),

iii. The velocity of the particle is given by, ,

Acceleration

iv. Substituting equation (4) in equation (3),

v. Substituting , where is the angular frequency,

This is the differential equation of linear S.H.M.

2. Define linear S.H.M. Obtain differential equation of linear S.H.M.

Ans: Definition: Linear S.H.M. is defined as the linear periodic motion of a body, in which force (or acceleration) is always directed towards the mean position and its magnitude is proportional to the displacement from the mean position.

Differential equation of linear S.H.M.:

Refer Subtopic 5.4: Q. No. I

Acceleration (a), Velocity (v) and Displacement (x) of S.H.M.

1. State the differential equation of linear simple harmonic motion. Hence obtain the expression for acceleration, velocity and displacement of a particle performing linear S. H. M.

Ans: Differential equation of linear S.H.M.: The differential equation of linear S.H.M. is given as,

i. Expression for acceleration in linear S.H.M:

a. From differential equation,

b. But, linear acceleration is given by,

From equations (1) and (2),

Equation (3) gives acceleration in linear S.H.M.

ii. Expression for velocity in linear S.H.M:

a. From differential equation of linear S.H.M,

b. Integrating both sides of equation (4),

where, is the constant of integration. c. At extreme position, and

Substituting these values in equation (5),

d. Substituting equation (6) in equation (5),

This is the required expression for velocity in linear S.H.M.

iii. Expression for displacement in linear S.H.M:

a. From differential equation of linear S.H.M, velocity is given by,

But, in linear motion,

From equation (1) and (2),

b. Integrating both sides of equation (3),

where, is constant of integration which depends upon initial condition (phase angle)

This is required expression for displacement of a particle performing linear S.H.M. at time

2. From differential equation of linear S.H.M., obtain an expression for acceleration, velocity and displacement of a particle performing S.H.M.

Ans: Refer Subtopic 5.5: Q. No. 1

3. State the differential equation of linear S.H.M. Hence, obtain expression for :

i. acceleration

ii. velocity

Ans: Refer Subtopic 5.5: Q. No. 1 (i) and (ii)

5.7 Reference Circle Method

1. Define linear S.H.M.

Show that S.H.M. is a projection of U.C.M. on any diameter.

Ans: Definition: Refer Subtopic 5.4: Q. No. 2
i. Figure (a) shows the anticlockwise uniform circular motion of a particle , with centre at the origin . Its angular positions are decided with the reference .

ii. It means, if the particle is at , the angular position is zero, at it is , at it is , and so on. If it comes to E again, it will be .

iii. Let be the position vector of this particle. At , let the particle be at with reference angle . During time , it has angular displacement . Thus, the reference angle at time ‘ ‘ is .

iv. Let us choose the diameter along -axis as the reference diameter and label as the projection of OP on this.

v. Projection of displacement:

At time ,

the position vector

This is the equation of linear S.H.M. of amplitude r.

vi. Projection of velocity: Instantaneous velocity of the particle in the circular motion is the tangential velocity of magnitude ‘ ‘. Its projection on the reference diameter will be as shown in figure (b),

This is the expression for the velocity of a particle performing a linear S.H.M.

vii. Projection of acceleration: Instantaneous acceleration of the particle in circular motion is the radial or centripetal acceleration of magnitude , directed towards . Its projection on the reference diameter will be

This is the corresponding acceleration for the linear S.H.M.

viii. From this analogy it is clear that projection of any quantity for a uniform circular motion gives us the corresponding quantity of linear S.H.M. This analogy can be verified for any diameter as the reference diameter. Thus, the projection of a U.C.M. on any diameter is an S.H.M.

Graphical Representation of S.H.M.

1. Represent graphically the displacement, velocity and acceleration against time for a particle performing linear S.H.M., when it starts from the mean position. [Mar 08]

Ans: Graph:

2. Give graphical representation of S.H.M. when particle starts from the positive extreme position.

Ans: Graph:

3. Define phase of S.H.M.

Show variation of displacement, velocity and acceleration with phase for a particle performing linear S.H.M. graphically, when it start from extreme position.

Ans: Definition: Phase in S.H.M. (or for any motion) is the state of oscillation.

Particle performing S.H.M., starting from the positive extreme position:

Equations: As the particle starts from the positive extreme position,

Phase,

Displacement,

Velocity,

Acceleration,

Table:

0		T/2	3T/4	T

A	0		0		0
0		0		0
	0		0		0

Graph: Refer Subtopic 5.9: Q. No. 2

4. Define linear simple harmonic motion.

Assuming the expression for displacement of a particle starting from extreme position, explain graphically the variation of velocity and acceleration w.r.t. time. [July 17]

Ans: Definition: Refer Subtopic 5.4: Q. No. 2 (Definition only)

Variation of velocity and acceleration w.r.t. time: Refer Subtopic 5.9: Q. No. 3 (Velocity time graph and acceleration-time graph and explanation)

5.10 Composition of two S.H.M.s having same period and along the same path

1. Discuss the composition of two S.H.M’s along the same path having same period. Find the resultant amplitude and intial phase. [Oct 15]

Ans:

i. Consider a particle simultaneously subjected to two S.H.M.s having the same period and along same path (let it be along the -axis), but of different amplitudes and initial phases. The resultant displacement at any instant is equal to the vector sum of its displacements due to both the S.H.M.s at that instant.

ii. Let the two linear S.H.M’s be given by equations,

where are amplitudes; are initial phase angles and are the displacement of two S.H.M’s in time ‘ ‘. is same for both S.H.M’s.

iii. The resultant displacement of the two S.H.M’s is given by,

iv. Using equations (1) and (2), equation (3) can be written as,

v. As and are constants, we can combine them in terms of another convenient constants and as

and

vi. Using equations (5) and (6), equation (4) can be written as,

Equation (7) is the equation of an S.H.M. of the same angular frequency (hence, the same period) but of amplitude R and initial phase . It shows that the combination (superposition) of two linear S.H.M.s of the same period and occurring along the same path is also an S.H.M.

Resultant Amplitude:

Resultant amplitude is,

Squaring and adding equations (5) and (6) we get,

Equation (8) represents resultant amplitude of two S.H.M’s.

Initial Phase:

Dividing equation (6) by (5), we get,

Equation (9) represents resultant or initial phase of two S.H.M’s.

2. Discuss analytically the composition of two linear SHMs having same period and along the same path. Obtain the expression for resultant amplitude.

Find the resultant amplitude when the phase difference is
a. zero radians
b. radians.

Ans: Refer Subtopic 5.10: Q. No. 1 (except Initial Phase)

vii. a. When phase difference is zero radians:

Two S.H.M.s are in phase, i.e.,

From equation (8),

b. When phase difference is radians:

Two S.H.M.s are out of phase, i.e., ,

From equation (8),

5.11 Energy of a Particle Performing S.H.M.

1. Represent graphically the variations of K.E., P.E. and T.E. of a particle performing linear S.H.M. with respect to displacement. [Oct 08]

Ans:

2. State an expression for K. E. (kinetic energy) and P. E. (potential energy) at displacement ‘ ‘ for a particle performing linear S.H.M. Represent them graphically. Find the displacement at which K.E. is equal to . .

Ans: Refer Subtopic 5.11: Q. No. I for graphical representation.

i. Kinetic energy of particle executing S.H.M is given by,

ii. Potential energy of particle executing S.H.M. is given by,

iii. For a particle performing linear S.H.M.

If ,

Thus at ,

3. Obtain an expression for potential energy of a particle performing simple harmonic motion. Hence evaluate the potential energy

a. at mean position and

b. at extreme position.

Ans: Expression for potential energy:

i. The restoring force acting on the particle at point is given by,

where is the force constant.

ii. Suppose that the particle is displaced further by an infinitesimal displacement ‘ ‘ against the restoring force ‘ ‘.

iii. The external work done (dW) during this displacement is

iv. The total work done on the particle to displace it from to is given by,

v. This work done is stored as the potential energy (P.E.) of the particle at displacement .

vi. At time ,

Thus, with time, it varies as .

vii. At the mean position,

Potential energy,

viii. At the extreme positions,

Potential energy,

4. Prove the law of conservation of energy for a particle performing simple harmonic motion. Hence graphically show the variation of kinetic energy and potential energy . . t. instantaneous displacement.

Ans:

i. Expression for kinetic energy:

Consider a particle of mass m, performing a linear S.H.M. along the path MN about the mean position as shown in figure.

b. At a given instant, let the particle be at , at a distance from .

c. Velocity of the particle in S.H.M. is given as

where is the displacement of the particle performing S.H.M. and is the amplitude of S.H.M.

d Thus, the kinetic energy,

This is the kinetic energy at displacement .

c Also, at time , kinetic energy is,

Thus, with time, it varies as . ii. Expression for potential energy: Refer Subtopic 5.11: Q. No. 3 [(i) to (vi)]

iii. Expression for total energy:

a. The total energy of the particle is the sum of its kinetic energy and potential energy.

b. Using equation (1) and equation (2), we get

This expression gives the total energy of the particle at point .

iv. As and are constant, the total energy of the particle at any point is constant (independent of and ). In other words, the energy is conserved in S.H.M.

Graphical representation of K.E. and P.E. of S.H.M.: Refer Subtopic 5.11: Q. No. 1

5. Obtain an expression for potential energy of a particle performing S.H.M. What is the value of potential energy at (i) Mean position and (ii) Extreme position?

Ans: Refer Subtopic 5.11: Q. No. 3

6. Define epoch of S.H.M. State the factors on which the total energy of a particle performing S.H.M. depends. [July 19]

Ans: Definition The physical quantity which describes the state of oscillation of a particle performing S.H.M at the start of motion is called epoch of S.H.M.

Factors affecting total energy: Total energy of particle .. performing S.H.M. depends on the mass, amplitude and frequency of motion of the particle.

7. At which position, the total energy of a particle executing linear S.H.M. is purely potential?

Ans: At the extreme position, the total energy of a particle executing linear S.H.M. is purely potential.

5.12 Simple Pendulum

1. Define an ideal simple pendulum. Show that, under certain conditions, simple pendulum performs linear simple harmonic motion.

Ans: Definition: An ideal simple pendulum is a heavy particle suspended by a massless, inextensible, flexible string from a rigid support.

i. Let ‘ ‘ be the mass of the bob and ‘ be the tension in the string. The pendulum remains in equilibrium in the position , with the centre of gravity of the bob, vertically below the point of suspension .
ii. If now the pendulum is displaced through a small angle , called angular amplitude, and released, it begins to oscillate on either side of the mean (equilibrium) position in a single vertical plane.

iii. In the displaced position (extreme position), two forces are acting on the bob.

a. Force due to tension in the string, directed along the string, towards the support.

b. Weight , in the vertically downward direction.

iv. At the extreme positions, there should not be any net force along the string.

v. The component of can only balance the force due to tension.

Thus, weight is resolved into two components;

a. The component along the string, which is balanced by the tension .

b. The component perpendicular to the string is the restoring force acting on mass tending to return it to the equilibrium position.

Restoring force,

vi. As is very small ,

From the figure,

For small angle,

As and are constant,

vii. Thus, for small displacement, the restoring force is directly proportional to the displacement and is oppositely directed. Hence the bob of a simple pendulum performs linear S.H.M. for small amplitudes.

2. Derive an expression for the period of motion of a simple pendulum. On which factors does it depend?

Ans: Refer Subtopic 5.12: Q. No. 1 [(i) to (vii)]

viii. The period of oscillation of a pendulum is given by,

Using equation (1),

(in magnitude)

Substituting in the expression for ,

This gives the expression for the time period of a simple pendulum.

ix. Period of simple pendulum depends on the length of the pendulum and acceleration due to gravity.

3. Define practical simple pendulum.

Show that motion of bob of pendulum with small amplitude is linear S.H.M. Hence obtain an expression for its period. What are the factors on which its period depends?

Ans: Definition: A practical simple pendulum is a small heavy (dense) sphere (called bob) suspended by a light and inextensible string from a rigid support.

Motion of bob of pendulum is linear S.H.M.:

Refer Subtopic 5.12: Q. No. 1

Expression for time period of simple pendulum: Refer Subtopic 5.12: Q. No. 1 and Q. No. 2

Factors affecting time period of simple pendulum: Refer Subtopic 5.12: Q. No. 2 (ix)

4. Define an ideal simple pendulum. Show that the motion of a simple pendulum under certain conditions is simple harmonic. Obtain an expression for its period.

Ans: Refer Subtopic 5.12: Q. No. 1 and Q. No. 2

5. Define second’s pendulum. Derive a formula for the length of second’s pendulum.

Ans:

i. A simple pendulum whose period is two seconds is called second’s pendulum.

ii. Time period of simple pendulum,

For a second’s pendulum,

Where, is the length of second’s pendulum, having period .

5.13 Angular S.H.M. and its Differential Equation

1. Define angular S.H.M. and prove that a bar magnet vibrating in uniform magnetic induction perform uniform angular S.H.M. Obtain an expression for its frequency.

Ans: Angular S.H.M. is defined as the oscillatory motion of a body in which the torque for angular acceleration is directly proportional to the angular displacement and its direction is opposite to that of angular displacement.

Expression for frequency:

i. If a bar magnet is freely suspended in the plane of a uniform magnetic field, it remains in equilibrium with its axis parallel to the direction of the field.

ii. If it is given a small angular displacement about an axis passing through its centre, perpendicular to itself and to the field and released, it performs angular oscillations.

iii. Let be the magnetic dipole moment and the magnetic field.

iv. In the deflected position, a restoring torque acts on the magnet that tends to bring it back to its equilibrium position.

Magnet vibrating in a uniform magnatic field

v. The magnitude of this torque is

If is small,

vi. For clockwise angular displacement , the restoring torque is in the anticlockwise direction.

Where, is the moment of inertia of the bar magnet and is its angular acceleration.

vii. Since and are constants, equation (1) shows that angular acceleration is directly proportional to the angular displacement and directed opposite to the angular displacement. Hence the magnet performs angular S.H.M. viii. The period of vibrations of the magnet is given by,

Thus, by considering magnitude of angular acceleration ,

ix. Frequency

2. Define angular SHM. State its differential equation.

Ans: Refer Subtopic 5.13: Q. No. 1 (Definition only)

Differential equation for angular S.H.M.:

3. Write the differential equation for angular S.H.M.

Ans: Refer Subtopic 5.13: Q. No. 2 (Equation only)

amped Oscillations

1. Define the term ‘damped oscillations’.

Ans: Periodic oscillations of gradually decreasing amplitude are called damped harmonic oscillations.

Numericals

Linear Simple Harmonic Motion (S.H.M.)

1. A body of mass is made to oscillate on a spring of force constant dyne/cm. Calculate the magnitude of angular velocity and frequency of vibrations of the body. [July 19]

Solution:

Given:

To find: Angular velocity ( ), Frequency (n)

Formulae:

Calculation: From formula (i),

From formula (ii),

Ans: The angular velocity of the body is and the frequency of vibration is .

5.4 Differential Equation of S.H.M.

1. The maximum velocity of a particle performing linear S.H.M. is . If its maximum acceleration is , calculate its period.

Solution:

Given:

To find:

Formulae:

Calculation: From formulae,

Ans: Period of particle is .

2. A particle in S.H.M. has a period of 2 seconds and amplitude of . Calculate the acceleration when it is at from its

Solution:

positive extreme position. [Mar 15]

Given:

To find: Acceleration (a)

Formula: (in magnitude)

Calculation: Particle is from its positive extreme position, hence

From formula,

Ans: The acceleration of the particle when at from its extreme is .

3. The periodic time of a linear harmonic oscillator is second, with maximum displacement of . If the particle starts from extreme position, find the displacement of the particle after seconds.

Solution:

Given:

To find: Displacement (x)
Formula:

Calculation: From formula,

Ans: The displacement of the particle after seconds is .

4. A particle performing linear S.H.M. has a period of 6.28 seconds and a pathlength of . What is the velocity when its displacement is from mean position?

Solution:

Given:

To find: Velocity (v)

Formula:

Calculation: Since,

Ans: The velocity of the particle at is .

5. A particle performing linear S.H.M. has maximum velocity of and maximum acceleration of . Find the amplitude and period of oscillation. [Mar 18]

Solution:

Given: .

To find: Amplitude (A), Period (T)

Formulae: i. ii.

Calculation: From formulae (i) and (ii),

From formula (iii),

From formula (i),

Ans: Amplitude is and time period is

6. A body of mass is made to oscillate on a spring of force constant . Calculate

i. angular frequency

ii. frequency of vibrations.

Solution:

Given:

To find: Angular frequency , frequency of vibration (n)

Formulae: i. ii.

Calculation: From formula (i),

We have, for S.H.M.

From formula (ii),

Ans: The angular frequency of the body is and the frequency of vibration is .

7. A particle performing linear S.H.M. has maximum velocity and maximum acceleration . Find period of oscillations.

Solution:

Given:

To find: Period (T)

Formulae: i. ii.

Calculation: From formula (i), (ii) and (iii),

Ans: Time period of oscillation is .

Acceleration (a), Velocity (v) and Displacement (x) of S.H.M.

1. Calculate the velocity of a particle performing S.H.M. after 1 second, if its displacement is given by .

Ans: Given that,

Velocity of a particle performing S.H.M. is given by,

In second,

Amplitude(A), Period(T) and Frequency (n) of S.H.M.

1. A mass attached to a spring oscillates with a period of 2 second. If the mass is increased by , the period increases by 1 second. Find the initial mass, assuming that Hooke’s law is obeyed.

Solution:

Given:

To find: Mass (M)
Formula:

Calculation: From formula,

From (1) and (2),

Ans: The initial mass attached to the spring is .

2. A body of mass is made to oscillate on a spring of force constant . Calculate:

a. Angular frequency,

b. Frequency of vibration

Solution:

Given:

To find:

Angular frequency , frequency of vibration (n)

Formulae:

Calculation: From formula (i),

Ans: The angular frequency of the body is and the frequency of vibration is .

5.10 Composition of two S.H.M.s having same period and along the same path

1. The displacement of a particle performing S.H.M. is given by

Determine the amplitude, period and initial phase of the motion.

Solution:

Given: .

This expression can be written as superposition of two parallel SHMs of same period, as where and .

and

Resultant amplitude,

Now, period,

Initial phase,

5.11 Energy of a Particle Performing S.H.M.

1 A particle executes S.H.M. with a period of 10 seconds. Find the time in which its potential energy will be half of its total energy. [July 16]

Solution:

Given:

To find: Time

Formulae: i. P.E

ii. T.E

Calculation: From formula (i) and (ii),

… (From given condition)

Ans: The time in which the potential energy will be half of total energy is .

5.12 Simple Pendulum

1. When the length of a simple pendulum is decreased by , the period changes by . Find the original length of the pendulum.

Solution:

To find: Length

Formula:

Calculation: From formula,

Squaring both sides,

Ans: The original length of the pendulum is .

2. A clock regulated by seconds pendulum, keeps correct time. During summer, length of pendulum increases to . How much will the clock gain or lose in one day?

and

Solution:

Given:

To find: Loss in period

Formula:

Calculation: From formula,

The period of a seconds pendulum is 2 second.

Hence, the given pendulum clock will lose in (during summer).

Ans: The clock will lose in one day.

2. A simple pendulum of length has a mass of and oscillates freely with amplitude . Find its potential energy at extreme

Solution:

position.

Given:

To find: Potential energy (P. ) .

Formula: P.

Calculation: From formula,

Ans: The potential energy of the simple pendulum at extreme positions is .

3. The length of the second’s pendulum in a clock is increased to 4 times its initial length. Calculate the number of oscillations completed by the new pendulum in one minute.

Solution:

Given:

To find: Number of oscillations completed in one minute ( )

Formula:

Calculation: From formula,

No. of oscillations completed in one minute

Ans: Number of oscillations completed by the pendulum in one minute is .

5. A simple pendulum of length has mass and oscillates freely with amplitude of . Calculate its potential energy at extreme position.

Solution:

Given: ,

To find: Potential energy at extreme position (P.E

Formulae:

Calculation: From formula (i) and (ii),

Ans: Potential energy at extreme position is .

5.13 Angular S.H.M. and its Differential Equation

1. A bar magnet of mass in the form of a rectangular parallelepiped, has dimensions and , with its dimension ‘ ‘ vertical, the magnet performs angular oscillations in the plane of the magnetic field with period seconds. If the magnetic moment is , determine the influencing magnetic field.

Solution:

Given:

To find: Magnetic field (B)

Formulae: i.

Calculation: From formula (i),

From formula (ii),

Squaring both sides,

Ans: The influencing magnetic field is of strength or T.