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JEE • PHYSICS • OSCILLATIONS

Oscillations practice · Easy

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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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1 Easy

1 of 30

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.

2 Easy

2 of 30

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.

3 Easy

3 of 30

Remembering +4 | -1

The value of phase at maximum displacement from the mean position of a particle in S.H.M. is:

View answer and explanation

Correct option: A

\pi / 2

Option A (

\pi / 2

) is correct because at maximum displacement in SHM, the phase is

\pi/2

3\pi/2

4 Easy

4 of 30

Analyzing +4 | -1

Statement I When a wave goes from one medium to other, average power transmitted by the wave may change.
Statement II Due to a change in the medium, amplitude, speed, wavelength and frequency of the wave may change.

View answer and explanation

Correct option: A

Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I

Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I because the average power transmitted by a wave depends on the medium's properties, which can change the wave's amplitude, speed, wavelength, and frequency.

5 Easy

5 of 30

Remembering +4 | -1

The mass of a bob, suspended in a simple pendulum, is halved from the initial mass, its time period will:

View answer and explanation

Correct option: C

Remain unchanged

Option C (Remain unchanged) is correct because the time period of a simple pendulum is independent of the mass of the bob.

6 Easy

6 of 30

Understanding +4 | -1

When a particle executing SHM oscillates with a frequency

v

, then the kinetic energy of the particle

View answer and explanation

Correct option: B

changes periodically with a frequency of

2 \mathrm{v}

Option B is correct. The kinetic energy of a particle in SHM oscillates with twice the frequency of the SHM itself.

7 Easy

7 of 30

Understanding +4 | -1

In a seconds pendulum, the mass of the bob is

30 \mathrm{~g}

. If it is replaced by

90 \mathrm{~g}

mass, then its time period will be

View answer and explanation

Correct option: B

2 \mathrm{~s}

Option B (2 s) is correct because the time period of a simple pendulum depends only on its length and the acceleration due to gravity, not on the mass of the bob.

8 Easy

8 of 30

Understanding +4 | -1

Assertion(A): In SHM,

v-x

graph is an ellipse, where

v

is velocity and

x

is displacement from mean position.

Reason(R): Equation between

v

and

x

\frac{v^2}{\omega^2}+\frac{x^2}{1}=A^2

View answer and explanation

Correct option: A

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

Option A (If both Assertion and Reason are correct and Reason is the correct explanation of Assertion) is correct because the equation relating velocity and displacement in SHM represents an ellipse, and the given equation is the correct form.

9 Easy

9 of 30

Application +4 | -1

The time period of oscillation of a simple pendulum of length

L

suspended from the roof a vehicle, which moves without friction down an inclined plane of inclination

\alpha

, is given by:

View answer and explanation

Correct option: A

2 \pi \sqrt{L /(g \cos \alpha)}

Option A (

2 \pi \sqrt{L /(g \cos \alpha)}

) is correct because the effective acceleration due to gravity is reduced to

g \cos \alpha

on the inclined plane.

10 Easy

10 of 30

Understanding +4 | -1

Two particles are executing SHMs. The equations of their motions are

y_1=10 \sin \left(\omega t+\frac{\pi}{4}\right) \text { and } y_2=5 \sin \left(\omega t+\frac{\sqrt{3} \pi}{4}\right)

What is the ratio of their amplitudes?

View answer and explanation

Correct option: B

2: 1

Option B is correct because the ratio of the amplitudes is simply the ratio of the coefficients of the sine functions, which is 10/5 = 2/1.

11 Easy

11 of 30

Understanding +4 | -1

If a simple pendulum executing SHM falls freely along with the support, then

View answer and explanation

Correct option: A

it does not oscillate at all

Option A (it does not oscillate at all) is correct because in free fall, the effective gravitational acceleration becomes zero, thus preventing oscillation.

12 Easy

12 of 30

Analyzing +4 | -1

A vertical spring block system is made to oscillate.
Statement-1: Its time period on earth is more than that on the moon.
Statement-2: Its extension on earth (in equilibrium) is more than that on the moon.

View answer and explanation

Correct option: D

Statement-1 is false, Statement-2 is true

Option D (Statement-1 is false, Statement-2 is true) is correct because the time period of a spring-block system is independent of gravity, while the extension depends on gravity.

13 Easy

13 of 30

Understanding +4 | -1

In a simple pendulum, the period of oscillation

T

is related to length of the pendulum

l

View answer and explanation

Correct option: C

\frac{l}{T^2}=

constant

The correct answer is

\frac{l}{T^2}=

constant because the period of a simple pendulum is proportional to the square root of its length.

14 Easy

14 of 30

Analyzing +4 | -1

Molten-wax of mass

m

drops on a block of mass

M

, which is oscillating on a frictionless table. Select the correct option.

View answer and explanation

Correct option: D

time period increases

Option D (time period increases) is correct because the time period of a simple harmonic oscillator is proportional to the square root of the mass, and the mass increases when wax drops on the block.

15 Easy

15 of 30

Understanding +4 | -1

Two particles are executing simple harmonic motion. At an instant of time

t

, their displacements are

y_1=A \cos (\omega t)

and

y_2=A \sin (\omega t)

Then, the phase difference between

y_1

and

y_2

View answer and explanation

Correct option: B

90^{\circ}

Option B (

90^{\circ}

) is correct because the phase difference between cosine and sine functions with the same argument is 90 degrees.

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16 Easy

16 of 30

Application +4 | -1

To make the frequency of an oscillator double, one has to

View answer and explanation

Correct option: D

reduce the mass to one-fourth

Option D (reduce the mass to one-fourth) is correct because the frequency of an oscillator is inversely proportional to the square root of the mass.

17 Easy

17 of 30

Understanding +4 | -1

In SHM, phase difference between velocity and acceleration is

View answer and explanation

Correct option: A

\frac{\pi }{2}

Option A is correct because in SHM, velocity leads displacement by π/2, and acceleration leads velocity by π/2, making the phase difference between velocity and acceleration π/2.

18 Easy

18 of 30

Understanding +4 | -1

The time period of a simple pendulum on earth is

T.

If it is taken on the moon, and made to oscillate, the period of vibration will be:

View answer and explanation

Correct option: C

More than

T

Option C (More than

T

) is correct because the acceleration due to gravity on the moon is less than that on Earth, leading to a longer period.

19 Easy

19 of 30

Application +4 | -1

Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is :

View answer and explanation

Correct option: B

Option B (11) is correct because the two pendulums are in phase at the mean position after the shorter pendulum completes 11 vibrations.

20 Easy

20 of 30

Analyzing +4 | -1

Under the action of a force

F=-k x^3

, the motion of a particle is

(k=

a positive constant

)

View answer and explanation

Correct option: D

periodic but not simple harmonic

Option D is correct because the force is proportional to

x^3

, which means the motion is periodic but not simple harmonic.

21 Easy

21 of 30

Understanding +4 | -1

Assertion: In simple harmonic motion, the motion is to and fro and periodic
Reason: Velocity of the particle in simple harmonic motion

(v)=\omega \sqrt{{k}^2-{x}^2}

(where

{x}

is the displacement and

{k}

is spring factor).

View answer and explanation

Correct option: C

If Assertion is true but Reason is false

Option C is correct because the assertion is true (SHM is to and fro and periodic), but the reason provides an incorrect formula for the velocity in SHM.

22 Easy

22 of 30

Analyzing +4 | -1

Statement-1: A particle is moving along

x

-axis. The resultant force

F

acting on it is given by

F=-a x-b

where

a

and

b

are both positive constants. The motion of this particle is not S.H.M.
Statement-2: In S.H.M. resultant force must be proportional to the displacement from mean position.

View answer and explanation

Correct option: D

Statement-1 is false, Statement-2 is true

Option D (Statement-1 is false, Statement-2 is true) is correct because the force equation indicates SHM about a shifted equilibrium, and the definition of SHM requires proportionality to displacement from the *mean* position.

23 Easy

23 of 30

Application +4 | -1

The acceleration of a particle in SHM at 5 cm from its mean position is

80 \mathrm{~cm} / \mathrm{sec}^2.

The value of angular frequency in radian/sec will be:

View answer and explanation

Correct option: B

Option B (4) is correct because the angular frequency can be found using the formula a = ω²y.

24 Easy

24 of 30

Analyzing +4 | -1

Which of the following quantities is always negative in SHM?
Here,

\mathbf{s}

is displacement from mean position.

View answer and explanation

Correct option: C

\mathbf{a} \cdot \mathbf{s}

Option C (

\mathbf{{a}} \cdot \mathbf{{s}}

) is correct because in SHM, acceleration and displacement are always in opposite directions, resulting in a negative dot product.

25 Easy

25 of 30

Analyzing +4 | -1

A small ball is dropped from a certain height on the surface of a non-viscous liquid of density less than the density of ball. The motion of the ball is

View answer and explanation

Correct option: B

periodic but not SHM

Option B (periodic but not SHM) is correct because after attaining terminal velocity, the ball moves with constant velocity, resulting in periodic but non-SHM motion.

26 Easy

26 of 30

Understanding +4 | -1

A horizontally placed spring mass system has time period

T

. The same system is now placed on a car moving with acceleration

a

in horizontal direction. Then,

View answer and explanation

Correct option: C

time period will remain constant

Option C (time period will remain constant) is correct because the time period of a spring-mass system depends only on the mass and spring constant, not on external acceleration.

27 Easy

27 of 30

Understanding +4 | -1

A particle executing simple harmonic motion along

Y

-axis has its motion described by the equation

y=A \sin (\omega t)+B

. The amplitude of the simple harmonic motion is

View answer and explanation

Correct option: A

A

Option A (

A

) is correct because in the given equation

y=A \sin (\omega t)+B

A

represents the amplitude of the simple harmonic motion, while

B

is just a vertical shift.

28 Easy

28 of 30

Understanding +4 | -1

I. In SHM to find time taken in moving from one point to another point, we cannot apply the relation, Time

=\frac{\text { Distance }}{\text { Speed }}

II. In SHM, speed is constant.
Which of the following statement(s) is/are correct?

View answer and explanation

Correct option: B

Only I

Option B (Only I) is correct because speed is not constant in SHM, so Time = Distance/Speed cannot be directly applied.

29 Easy

29 of 30

Analyzing +4 | -1

Statement-1: For a particle of mass

1 \mathrm{~kg}

, executing S.H.M., if slope of restoring force vs displacement graph is

=-1

, then the time period of oscillation will be

6.28 \mathrm{~s}

.
Statement-2: If

1 \mathrm{~kg}

mass is replaced by

2 \mathrm{~kg}

mass and rest of the information remains same as in statement-1, then the time period of oscillation will remain

6.28 \mathrm{~s}

View answer and explanation

Correct option: C

Statement-1 is true, Statement-2 is false

Option C (Statement-1 is true, Statement-2 is false) is correct because the time period depends on the mass and the slope of the restoring force vs displacement graph, and changing the mass while keeping the slope the same will change the time period.

30 Easy

30 of 30

Application +4 | -1

The maximum acceleration of a particle in SHM is made two times keeping the maximum speed to be constant. It is possible when

View answer and explanation

Correct option: C

frequency is doubled while amplitude is halved

Option C (frequency is doubled while amplitude is halved) is correct because maximum acceleration is

\omega^2 A

and maximum speed is

\omega A

. To double acceleration while keeping speed constant,

\omega

must double and

A

must halve.

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Key Takeaways

✓ This set contains 30 single-correct-answer MCQs on OSCILLATIONS.
✓ Try each question before opening its answer. Read the explanation where one is available.
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