A point particle of mass 0.1 kg is executing SHM of amplitude 0.1$m$. When the particle passes through the mean position, its KE is $8 \times {10^{ - 3}}{\rm{J}}$. The equation of motion of this particle, if its initial phase of oscillation is ${45^0}$ is

$y = 0.1\sin \left( {\frac{r}{4} + \frac{\pi }{4}} \right)$

$y = 0.1\sin \left( {\frac{t}{2} + \frac{\pi }{4}} \right)$

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JEE • PHYSICS • OSCILLATIONS

Oscillations practice · Mixed difficulty

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 483 questions

1 Medium

1 of 483

Analyzing +4 | -1

Assertion: A particle executing simple harmonic motion comes to rest at the extreme positions.

Reason: The resultant force on the particle is zero at these positions.

View answer and explanation

Correct option: C

If Assertion is correct but Reason in incorrect.

Option C is correct. In SHM, the particle momentarily comes to rest at extreme positions, but the resultant force is maximum, not zero, at these points.

2 Medium

2 of 483

Understanding +4 | -1

A particle is subjected simultaneously to two SHM’s one along the

x

-axis and the other along the

y

-axis. The two vibrations are in phase and have unequal amplitudes. The particle will execute

View answer and explanation

Correct option: A

Straight line motion

Option A (Straight line motion) is correct because when two SHMs are in phase, the resulting motion is a straight line with a slope determined by the ratio of their amplitudes.

3 Medium

3 of 483

Application +4 | -1

A particle is executing SHM with amplitude

a

. When the PE of a particle is one-fourth of its maximum value during the oscillation, its displacement from the equilibrium position will be

View answer and explanation

Correct option: C

a/2

Option C (

a/2

) is correct because the displacement is calculated when potential energy is one-fourth of its maximum value.

4 Hard

4 of 483

Understanding +4 | -1

The displacement, velocity and acceleration in a simple harmonic motion are related as the

View answer and explanation

Correct option: A

displacement, velocity and acceleration all act in the same direction

Option A (displacement, velocity and acceleration all act in the same direction) is the correct answer because in SHM, at extreme positions, all three quantities point towards the mean position.

5 Medium

5 of 483

Understanding +4 | -1

Assertion: Angular velocity of earth around sun is lesser than the angular velocity of moon about earth.
Reason: Time taken by moon revolve around earth is less than time taken by earth to revolve around sun.

View answer and explanation

Correct option: A

Both Assertion (A) and Reason (R) are true & correct explanation of Assertion 'A

Option A (Both Assertion (A) and Reason (R) are true & correct explanation of Assertion 'A') is correct because the moon's orbital period around the Earth is shorter than the Earth's orbital period around the Sun, resulting in a higher angular velocity for the moon.

6 Medium

6 of 483

Application +4 | -1

A particle is executing

SHM

. At a displacement

{y_1}

its potential energy is

{U_1}

and at a displacement

{y_2}

its potential energy is

{U_2}

. The potential energy of the particle at displacement

\left( {{y_1} + {y_2}} \right)

View answer and explanation

Correct option: D

{U_1} + {U_2} + 2\sqrt {{U_1}{U_2}}

Option D (

{U_1} + {U_2} + 2\sqrt {{U_1}{U_2}}

) is correct because the potential energy at the combined displacement is the square of the sum of the individual displacements, related to their respective potential energies.

7 Medium

7 of 483

Application +4 | -1

T

is the time period of simple pendulum on the earth's surface. Its time period becomes

x T

when taken to a height

R

(equal to earth's radius) above the earth's surface. Then, the value of

x

will be:

View answer and explanation

Correct option: C

Option C (2) is correct because the acceleration due to gravity decreases to g/4 at a height R, causing the time period to double.

8 Medium

8 of 483

Analyzing +4 | -1

A clock 'S' is working on oscillations of a spring and a clock 'P' is working on pendulum motion. Both clocks are running at the same rate on earth. What will happen to their functioning on a planet which has the same density as that of earth, but the radius is twice that of earth?

View answer and explanation

Correct option: D

S' will run faster than 'P'

Option D is correct because the time period of a spring clock is independent of gravity, while the time period of a pendulum clock depends on gravity. Since g decreases, the pendulum clock will run slower, making the spring clock run faster relative to it.

9 Medium

9 of 483

Understanding +4 | -1

In SHM restoring force is

F = - k\mathrm{ }x

where k is force constant ,

x

is displacement and

A

is amplitude of motion, then total energy depends upon

View answer and explanation

Correct option: C

k,A

Option C (

k,A

) is correct because the total energy in SHM is proportional to the force constant and the square of the amplitude.

10 Medium

10 of 483

Application +4 | -1

A point particle of mass 0.1 kg is executing SHM of amplitude 0.1

m

. When the particle passes through the mean position, its KE is

8 \times {10^{ - 3}}\mathrm{J}

. The equation of motion of this particle, if its initial phase of oscillation is

{45^0}

View answer and explanation

Correct option: D

y = 0.1\sin \left( {4t + \frac{\pi }{4}} \right)

Option D (

y = 0.1\sin \left( {4t + \frac{\pi }{4}} \right)

) is correct because the equation represents SHM with the calculated angular frequency and given initial phase.

11 Easy

11 of 483

Analyzing +4 | -1

Assertion(A): If a pendulum is suspended in a lift and lift accelerates upwards, then its time period will decrease.

Reason(R): Effective value of

g

will be

g_{\text {eff }}=g+a

View answer and explanation

Correct option: A

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Option A is correct. When a lift accelerates upwards, the effective acceleration due to gravity increases, which decreases the time period of a pendulum. The reason correctly explains this.

12 Medium

12 of 483

Application +4 | -1

A particle under the action of a SHM has a period of

3 \mathrm{~s}

and under the effect of another it has a period

4 \mathrm{~s}

. What will be its period under the combined action of both the SHM's in the same direction?

View answer and explanation

Correct option: C

2.4 \mathrm{~s}

Option C (2.4 s) is correct because the period under the combined action of SHMs is calculated by considering the effective spring constant.

13 Medium

13 of 483

Understanding +4 | -1

A weakly damped harmonic oscillator of frequency

n_1

is driven by an external periodic force of frequency

n_2

. When the steady state is reached, the frequency of the oscillator will be

View answer and explanation

Correct option: B

n_2

Option B (

n_2

) is correct because when a weakly damped harmonic oscillator is driven by an external force, it will eventually oscillate at the frequency of the driving force (

n_2

) once it reaches steady state.

14 Medium

14 of 483

Understanding +4 | -1

A particle is performing SHM. When acceleration increases, its time period:

View answer and explanation

Correct option: B

Decreases

Option B (Decreases) is correct because the time period of a simple pendulum is inversely proportional to the square root of the effective acceleration due to gravity.

15 Medium

15 of 483

Understanding +4 | -1

Statement-1: In simple harmonic motion, the graph between velocity and the displacement is an ellipse.
Statement-2: In simple harmonic motion the phase difference between velocity and displacement is

\pi / 2.

View answer and explanation

Correct option: B

Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation for Statement-1

Option D (Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation for Statement-1) is correct because both statements are true, but the phase difference doesn't directly explain the elliptical graph.

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16 Easy

16 of 483

Understanding +4 | -1

x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \mathrm{m}

represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are

View answer and explanation

Correct option: A

5 \mathrm{~m}, 2 \mathrm{~s}

Option A (5 m, 2 s) is correct because the amplitude is directly read from the equation, and the time period is calculated from the angular frequency.

17 Medium

17 of 483

Understanding +4 | -1

Assertion: If the amplitude of a simple harmonic oscillator is doubled, its total energy becomes four times.
Reason: The total energy is directly proportional to the square of amplitude of vibration of the harmonic oscillator.

View answer and explanation

Correct option: A

If both assertion and reason are true and reason is the correct explanation of the assertion

Option A (If both assertion and reason are true and reason is the correct explanation of the assertion) is correct because the total energy of a simple harmonic oscillator is proportional to the square of its amplitude.

18 Medium

18 of 483

Understanding +4 | -1

In a simple oscillating pendulum, the work done by the pendulum in one oscillation will be

View answer and explanation

Correct option: D

Zero

Option D (Zero) is correct because in one complete oscillation, the pendulum returns to its starting point, resulting in zero net displacement and thus zero work done by the pendulum.

19 Medium

19 of 483

Application +4 | -1

The time period of a particle performing linear SHM is 18s. What is the time taken by it to make a displacement equal to half its amplitude from its equilibrium position ?

View answer and explanation

Correct option: C

\frac{3}{2}s

Option C (

\frac{3}{2}s

) is correct because it's calculated by finding the time taken to reach half the amplitude in SHM, which is T/12.

20 Medium

20 of 483

Application +4 | -1

Two simple pendulums of length

0.5 \mathrm{~m}

and

20 \mathrm{~m}

respectively are given small linear displacement in one direction at the same time. They will again be in the phase when the pendulum of shorter length has completed oscillations

\left[n T_1=(n-1) T_2\right.

where

T_1

is time period of shorter length &

T_2

be time period of longer length and

n

are no. of oscillations completed]

View answer and explanation

Correct option: B

Option A (5) is correct because the shorter pendulum completes approximately 5 oscillations when they are again in phase.

21 Medium

21 of 483

Application +4 | -1

A uniform spring of normal length

\lambda

has a force constant

\mathrm{k}

. It is cut into two pieces of lengths

\lambda_1

and

\lambda_2

such that

\lambda_1=

\mathrm{n} \lambda_2

, where

\mathrm{n}

is an integer. Then the value of

\mathrm{k}_1

(force constant of spring of length

\lambda_1

) is -

View answer and explanation

Correct option: B

\frac{\mathrm{k}(\mathrm{n}+1)}{\mathrm{n}}

Option B (

\frac{\mathrm{k}(\mathrm{n}+1)}{\mathrm{n}}

) is correct because the force constant of a spring is inversely proportional to its length, and the total length is divided proportionally.

22 Medium

22 of 483

Understanding +4 | -1

Y=A \sin (2 \pi m t+\phi)

represents SHM of a particle then its phase at time

t

View answer and explanation

Correct option: C

(2 \pi m t+\phi)

Option C (

(2 \pi m t+\phi)

) is correct because the phase of a SHM is the argument of the sinusoidal function.

23 Medium

23 of 483

Application +4 | -1

A particle executes SHM of amplitude

5 \mathrm{~cm}

and period

3 \mathrm{~s}

. The velocity of the particle at a distance

4 \mathrm{~cm}

from the mean position -

View answer and explanation

Correct option: D

6 \mathrm{~cm} / \mathrm{s}

The correct answer is D (

6 \mathrm{~cm} / \mathrm{s}

) because the velocity of a particle in SHM is calculated using the formula

v = \omega \sqrt{a^2 - x^2}

24 Medium

24 of 483

Analyzing +4 | -1

To study the dissipations of the energy of a simple pendulum, student plots a graph between square root of time and amplitude. The graph would be a

View answer and explanation

Correct option: B

Hyperbola

Option B (Hyperbola) is correct because the amplitude decreases over time due to energy dissipation, leading to a hyperbolic relationship between amplitude and the square root of time.

25 Medium

25 of 483

Application +4 | -1

A horizontal platform with an object placed on it is executing SHM in the vertical direction. The amplitude of oscillation is

30.92\mathrm{ \times }{10^{ - 3}}

m. What must be the least period of these oscillations, so that the object is not detached from the platform?

View answer and explanation

Correct option: A

0.1256 s

Option A (0.1256 s) is correct because it represents the minimum period required to prevent detachment, calculated using the condition

g = \omega^2 r

26 Medium

26 of 483

Application +4 | -1

The length of a simple pendulum executing simple harmonic motion is increased by

21 \%

. The percentage increase in the time period of the pendulum of increased length is

View answer and explanation

Correct option: D

10 \%

Option D (

10 \%

) is correct because it is the percentage increase in the time period calculated using the formula for the time period of a simple pendulum.

27 Medium

27 of 483

Application +4 | -1

A body of weight

30 \mathrm{~N}

hangs from one end of a long, vertical spring whose other end is attached to a rigid support. The spring extends by

5 \mathrm{~cm}

when a force of

10 \mathrm{~N}

is applied to it. The time period of the vertical oscillations of the body is

View answer and explanation

Correct option: B

(\pi / 5) \sqrt{(3 / 2)} \,\mathrm{sec}

Option B (

\frac{\pi}{5} \sqrt{\frac{3}{2}} \,\mathrm{sec}

) is correct because the time period of vertical oscillations can be calculated using the spring constant and the mass of the body.

28 Hard

28 of 483

Understanding +4 | -1

Given below are two statements:
Statement I A second’s pendulum has a time period of

1\,s

.
Statement II It takes precisely one second to move between the two extreme positions.
In the light of the above statements, choose the correct answer from the option given below.

View answer and explanation

Correct option: B

Statement I is false but statement II is true.

Statement B is correct because Statement I is false (second's pendulum has a period of 2s) and Statement II is true (time between extremes is half the period, or 1s).

29 Medium

29 of 483

Analyzing +4 | -1

A student says that he had applied a force

F=-k \sqrt{x}

on a particle and the particle moved in simple harmonic motion. He refuses to tell whether

{k}

is constant or not. Assume that he has worked only with positive

{x}

and no other force acted on the particle

View answer and explanation

Correct option: A

{x}

increases

{k}

increases

Option A (as

{x}

increases

{k}

increases) is correct because for SHM, the force is proportional to x, and comparing with the given force, k must increase with x.

30 Medium

30 of 483

Application +4 | -1

The co-ordinates of a moving particle at a time

t,

are given by,

x=5 \sin 10 t,

y=5 \cos 10 t.

Considering all quantities in SI units, the speed of the particle is

View answer and explanation

Correct option: B

Option B (50) is correct because the speed is calculated by finding the derivatives of x and y with respect to time, then using the Pythagorean theorem.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 483 single-correct-answer MCQs on OSCILLATIONS.
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