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 · Medium

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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 445 questions

1 Medium

1 of 445

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 445

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 445

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 Medium

4 of 445

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.

5 Medium

5 of 445

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.

6 Medium

6 of 445

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.

7 Medium

7 of 445

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.

8 Medium

8 of 445

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.

9 Medium

9 of 445

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.

10 Medium

10 of 445

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.

11 Medium

11 of 445

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.

12 Medium

12 of 445

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.

13 Medium

13 of 445

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.

14 Medium

14 of 445

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.

15 Medium

15 of 445

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.

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16 Medium

16 of 445

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.

17 Medium

17 of 445

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.

18 Medium

18 of 445

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.

19 Medium

19 of 445

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.

20 Medium

20 of 445

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}

21 Medium

21 of 445

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.

22 Medium

22 of 445

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

23 Medium

23 of 445

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.

24 Medium

24 of 445

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.

25 Medium

25 of 445

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.

26 Medium

26 of 445

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.

27 Medium

27 of 445

Understanding +4 | -1

The potential energy of a particle (

{U_x}

) executing SHM is given by

View answer and explanation

Correct option: A

{U_x} = \frac{k}{2}{\left( {x - a} \right)^2}

Option A (

{U_x} = \frac{k}{2}{\left( {x - a} \right)^2}

) is correct because it represents the potential energy of a particle in SHM, which is proportional to the square of the displacement from the equilibrium position.

28 Medium

28 of 445

Application +4 | -1

The length of a pendulum whose time period is 2 seconds at a place is increased by

3\,\%

and it is taken to a place where acceleration due to gravity is

1\,\%

less than the first place. Its new time period will be[in seconds]

View answer and explanation

Correct option: D

2.04

Option D (2.04) is correct because the new time period is calculated by considering the percentage change in length and acceleration due to gravity.

29 Medium

29 of 445

Application +4 | -1

A spring produces extension '

x

' by applying a force '

F

' N. A body of mass '

m

' suspended from spring oscillates vertically with a period '

T

', the mass of the suspended body is (neglect mass of spring)

View answer and explanation

Correct option: A

\frac{T^2 F}{4 \pi^2 x}

Option A (

\frac{T^2 F}{4 \pi^2 x}

) is correct because it correctly relates the mass of the suspended body to the spring constant, time period, and extension.

30 Medium

30 of 445

Application +4 | -1

A ring of radius

20 \mathrm{~cm}

is hinged from a point on its periphery. The time period of its oscillations will be

\left(g=9.8 \mathrm{~m} / \mathrm{s}^2\right)

View answer and explanation

Correct option: C

\frac{2 \sqrt{2} \pi}{7} \mathrm{~s}

Option C (

\frac{2 \sqrt{2} \pi}{7} \mathrm{~s}

) is correct because it is the calculated time period of oscillation for a ring hinged at a point on its periphery using the formula

T=2 \pi \sqrt{\frac{2R}{g}}

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 445 single-correct-answer MCQs on OSCILLATIONS.
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