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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 91-120 of 445 questions

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91 Medium
91 of 445
Analyzing +4 | -1
A sealed spherical ball half filled with a volatile liquid is used as the bob of a pendulum executing simple harmonic motion with period T1T_1. The bob is heated slightly so that the liquid evaporates completely and the period is now found to be T2T_2. Which one of the following is true?
View answer and explanation
Correct option: C
T2<T1T_2 < T_1
Option C (T2<T1T_2 < T_1) is correct because when the liquid evaporates, the center of mass shifts closer to the pivot point, effectively shortening the pendulum's length and thus decreasing the period.
92 Medium
92 of 445
Understanding +4 | -1
In simple harmonic motion, loss of kinetic energy is proportional to
View answer and explanation
Correct option: D
x2x^2
Option D (x2x^2) is correct because the potential energy, and thus the loss of kinetic energy, in SHM is proportional to the square of the displacement.
93 Medium
93 of 445
Application +4 | -1
A particle which is attached to a spring oscillates horizontally with simple harmonic motion with a frequency of 1/π Hz1 / \pi \mathrm{~Hz} and total energy of 10 J10 \mathrm{~J}. If the maximum speed of the particle is 0.4 ms−10.4 \mathrm{~ms}^{-1}, what is the force constant of the spring? What will be the maximum potential energy of the spring during the motion?
View answer and explanation
Correct option: A
500 Nm−1,10 J500 \mathrm{~Nm}^{-1}, 10 \mathrm{~J}
Option A (500 Nm−1,10 J500 \text{ Nm}^{-1}, 10 \text{ J}) is correct because the force constant and maximum potential energy are calculated correctly using the given frequency, total energy, and maximum speed.
94 Medium
94 of 445
Understanding +4 | -1
Which one of the following is an example of the
force of gravity of the earth acting on a
vibrating pendulum bob?
View answer and explanation
Correct option: C
Restoring force
Option C (Restoring force) is correct because the force of gravity acting on a vibrating pendulum bob pulls it towards the equilibrium position, thus acting as a restoring force.
95 Medium
95 of 445
Understanding +4 | -1
The motion which is not simple harmonic is
View answer and explanation
Correct option: C
Motion of a planet around the sun
Option C (Motion of a planet around the sun) is correct because planetary motion is elliptical and does not follow the restoring force principle of simple harmonic motion.
96 Medium
96 of 445
Understanding +4 | -1
S-I: In a simple harmonic oscillation the average potential energy in one time period is equal to the average kinetic energy in this period.
S-II: The average kinetic energy for one oscillation is equal to half of total mechanical energy.
View answer and explanation
Correct option: A
S-I &\& S-II are true
Option A (S-I & S-II are true) is correct because in SHM, the average potential and kinetic energies over one time period are equal, and each is half the total mechanical energy.
97 Medium
97 of 445
Application +4 | -1
A body oscillates with a simple harmonic motion having amplitude 0.05 m0.05 \mathrm{~m}. At a certain instant, its displacement is 0.01 m0.01 \mathrm{~m} and acceleration is 1.0 m/s21.0 \mathrm{~m} / \mathrm{s}^2. The period of oscillation is
View answer and explanation
Correct option: D
π5 s\frac{\pi}{5} \mathrm{~s}
Option D (π5 s\frac{\pi}{5} \mathrm{~s}) is correct because the period is calculated from the angular frequency, which is derived from the given displacement and acceleration.
98 Medium
98 of 445
Application +4 | -1
Two simple harmonic motions are represented by y1=4sin⁡(4πt+π2){y_1} = 4\sin \left( {4\pi t + \frac{\pi }{2}} \right) and   y2=3cos⁡(4πt)\;{y_2} = 3\cos \left( {4\pi t} \right) The resultant amplitude is
View answer and explanation
Correct option: A
7
Option A (7) is correct because the two SHM equations are in phase, so the amplitudes add directly.
99 Medium
99 of 445
Analyzing +4 | -1
The displacement of a particle along the xx axis is given by x=asin⁡2ωtx = a{\sin ^2}\omega t The motion of the particle corresponds to
View answer and explanation
Correct option: D
Non simple harmonic motion
Option D (Non simple harmonic motion) is correct because the given displacement equation does not represent simple harmonic motion.
100 Medium
100 of 445
Application +4 | -1
A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time (in seconds) is
View answer and explanation
Correct option: B
8π3\frac{{8\pi }}{3}
Option B (8π3\frac{{8\pi }}{{3}}) is correct because the periodic time is calculated using the relationship between velocity, acceleration, amplitude, and angular frequency in SHM.
101 Medium
101 of 445
Application +4 | -1
A particle moving on xx-axis has potential energy U=2−20x+5x2U=2-20 x+5 x^2 joule along xx-axis. The particle is released at x=−3x=-3. The maximum value of xx will be (xx is in metre)
View answer and explanation
Correct option: C
7 m7 \mathrm{~m}
Option C (7 m7 \mathrm{~m}) is correct because the maximum value of x is found by determining the equilibrium position and using the amplitude of the oscillation.
102 Medium
102 of 445
Application +4 | -1
If the maximum acceleration of a SHMSHM is α\alpha and the maximum velocity is β\beta, then the amplitude of vibration is
View answer and explanation
Correct option: C
β2α\frac{{{\beta ^2}}}{\alpha }
Option C (β2α\frac{{{\beta ^2}}}{\alpha }) is correct because it is derived from the relationship between maximum acceleration, maximum velocity, angular frequency, and amplitude in SHM.
103 Medium
103 of 445
Understanding +4 | -1
In a linear Simple Harmonic Motion (SHM)

(A) Restoring force is directly proportional to the
displacement.

(B) The acceleration and displacement are
opposite in direction.

(C) The velocity is maximum at mean position.

(D) The acceleration is minimum at extreme points.
Choose the correct answer from the options given
below:
View answer and explanation
Correct option: C
(A), (B) and (C) only
Option C ((A), (B) and (C) only) is correct because it accurately describes the properties of Simple Harmonic Motion (SHM).
104 Medium
104 of 445
Understanding +4 | -1
Assertion: Pendulum clocks go slow in summer and fast in winter.
Reason: The length of the pendulum used in clock increases in summer.
View answer and explanation
Correct option: A
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion
Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because the length of a pendulum increases in summer due to thermal expansion, which increases the time period and makes the clock go slow.
105 Medium
105 of 445
Analyzing +4 | -1
The displacement of a particle along the x{x}-axis is given by The motion of the particle corresponds to
View answer and explanation
Correct option: C
Non simple harmonic motion
Option C (Non simple harmonic motion) is correct because the given displacement equation, x=asin⁡2ωtx = a \sin^2 \omega t, does not satisfy the condition for simple harmonic motion, which requires acceleration to be proportional to the negative of displacement.
106 Medium
106 of 445
Understanding +4 | -1
There is a large difference between the frequencies of external periodic force and natural frequency of a body. The body will:
View answer and explanation
Correct option: D
Vibrate with small amplitude and with frequency of force
Option D (Vibrate with small amplitude and with frequency of force) is correct because when the external force frequency is very different from the natural frequency, the amplitude is small, and the body vibrates at the driving frequency.
107 Medium
107 of 445
Understanding +4 | -1
The principle involved in the performance of a
spinning-chair circus acrobat is
View answer and explanation
Correct option: A
conservation of angular momentum
Option A (conservation of angular momentum) is correct because the acrobat changes their moment of inertia, leading to a change in angular velocity to conserve angular momentum.
108 Medium
108 of 445
Understanding +4 | -1
S-I: In a simple harmonic oscillation the average potential energy in one time period is equal to the average kinetic energy in this period.
S-II: The average kinetic energy for one oscillation is equal to half of total mechanical energy.
View answer and explanation
Correct option: A
S-I &\& S-II are true
Option A is correct because in SHM, the average potential energy and average kinetic energy over one time period are equal, and each is half the total mechanical energy.
109 Medium
109 of 445
Analyzing +4 | -1
A particle, with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force. F=F0sin⁡ωtF = {F_0}\sin \omega t If the amplitude of the particle is maximum for &omega;=ω1{\omega _1} and the energy of the particle is maximum for &omega;=ω2{\omega _2} then
View answer and explanation
Correct option: C
ω1≠ω0  and  ω2=ω0{\omega _1} \ne {\omega _0}\,\,\mathrm{and}\,\,{\omega _2} = {\omega _0}
Option C (ω1≠ω0  and  ω2=ω0{\omega _1} \ne {\omega _0}\,\,\mathrm{and}\,\,{\omega _2} = {\omega _0}) is correct because amplitude resonance occurs at a frequency slightly less than the natural frequency, while energy (or velocity) resonance occurs at the natural frequency.
110 Medium
110 of 445
Application +4 | -1
A particle is attached to a vertical spring and is pulled down a distance 4 cm4 \mathrm{~cm} below its equilibrium and is released from rest. The initial upward acceleration is 0.5 ms−20.5 \mathrm{~ms}^{-2}. The angular frequency of oscillation is
View answer and explanation
Correct option: A
3.53 rad/s3.53 \mathrm{~rad} / \mathrm{s}
Option A (3.53 rad/s) is correct because the angular frequency is calculated from the initial upward acceleration and displacement.
111 Medium
111 of 445
Analyzing +4 | -1
Statement-I: The graph of potential energy and kinetic energy of a particle in SHM with respect to position is a parabola.

Statement-II: Potential energy and kinetic energy do not vary linearly with position.
View answer and explanation
Correct option: B
If both Statement - I and Statement - II are true but Statement - II is not the correct explanation of Statement - I.
The correct answer is C because S-I is correct (a straight conductor has practically zero inductance), but S-II is incorrect (an emf is induced only with changing magnetic flux).
112 Medium
112 of 445
Understanding +4 | -1
Identify the function which represents a periodic motion.
View answer and explanation
Correct option: C
sin⁡ωt+cos⁡ωt\sin \omega t+\cos \omega t
Option C (sin⁡ωt+cos⁡ωt\sin \omega t+\cos \omega t) is correct because it represents a periodic function, as its value repeats after a fixed interval.
113 Medium
113 of 445
Analyzing +4 | -1
The displacement of a particle is represented by the equation y=sin⁡3ωty=\sin^3 \omega t. The motion is
View answer and explanation
Correct option: B
Periodic but not simple harmonic
Option B (Periodic but not simple harmonic) is correct because the function repeats itself but cannot be represented by a single sine or cosine function.
114 Medium
114 of 445
Analyzing +4 | -1
Assertion: If a pendulum falls freely, then its time period becomes infinite.
Reason: Free falling body has acceleration, equal to gg.
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 in free fall, the effective acceleration due to gravity becomes zero, leading to an infinite time period for the pendulum.
115 Medium
115 of 445
Understanding +4 | -1
The velocity of a particle performing simple
harmonic motion, when it passes through its
mean position is
View answer and explanation
Correct option: D
maximum
Option D (maximum) is correct because the velocity of a particle in SHM is maximum at the mean position, where the displacement is zero.
116 Medium
116 of 445
Application +4 | -1
A light spiral spring supports a 200 g weight at its lower end. It oscillates up and down with a period of 1 s. How much weight (in gram) must be removed from the lower end to reduce the period to 0.5 s?
View answer and explanation
Correct option: C
150
Option C (150) is correct because the period of oscillation of a spring-mass system is proportional to the square root of the mass, and the calculation shows that 150g must be removed to halve the period.
117 Medium
117 of 445
Analyzing +4 | -1
Assertion (A): The bob of a simple pendulum is a ball full of water. If a fine hole is made at the bottom of the ball, then the time period will no more remain constant.
Reason (R): The time period of simple pendulum does not depend upon mass.
View answer and explanation
Correct option: B
If Both A &\& R are true but R is not correct explanation of A
Option B (If Both A & R are true but R is not correct explanation of A) is correct because while the time period does change, it is due to the changing effective length of the pendulum, not directly because the mass is changing.
118 Medium
118 of 445
Understanding +4 | -1
In resonance:
View answer and explanation
Correct option: B
Energy absorbed by the vibrating body is maximum
Option B is correct because at resonance, the energy absorbed by the vibrating body is maximum.
119 Medium
119 of 445
Analyzing +4 | -1
If a body is performing SHM, then its

A) average kinetic energy per cycle is equal to half of its maximum kinetic energy

B) mean speed over a complete cycle is equal to 2π\frac{2}{\pi} times of its maximum speed
View answer and explanation
Correct option: C
Both A and B are correct
Option C is correct because both statements A and B are true. The average kinetic energy in SHM is half the maximum kinetic energy, and the mean speed over a complete cycle is 2π\frac{{2}}{{\pi}} times the maximum speed.
120 Medium
120 of 445
Analyzing +4 | -1
Assertion: All small oscillations are simple harmonic in nature.

Reason: Oscillations of spring block system are always simple harmonic whether amplitude is small or large.
View answer and explanation
Correct option: D
If Assertion is false but the Reason is true
Option D (If Assertion is false but the Reason is true) is correct because not all small oscillations are simple harmonic, but oscillations of a spring-block system are always simple harmonic.
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