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JEE • PHYSICS • OSCILLATIONS

Oscillations practice · Medium

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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 151-180 of 445 questions

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151 Medium
151 of 445
Application +4 | -1
A heavy sphere of mass mm is suspended by string of length ll. The sphere is made to revolve about a vertical line passing through the point of suspension in a horizontal circle such that the string always remains inclined to the vertical at an angle θ\theta. What is its period of revolution?
View answer and explanation
Correct option: B
T=2πlcos⁡θgT=2 \pi \sqrt{\frac{l \cos \theta}{g}}
Option B (T=2πlcos⁡θgT=2 \pi \sqrt{\frac{l \cos \theta}{g}}) is correct because it's derived from resolving tension and applying Newton's second law to circular motion.
152 Medium
152 of 445
Understanding +4 | -1
The acceleration of a particle in S.H.M. is
View answer and explanation
Correct option: C
Maximum at the extreme position
Option C (Maximum at the extreme position) is correct because the acceleration in SHM is directly proportional to the displacement from the equilibrium position, which is maximum at the extreme position.
153 Medium
153 of 445
Application +4 | -1
A block of mass one kg\mathrm{kg} is fastened to a spring with a spring constant 50 Nm−150 \mathrm{~Nm}^{-1}. The block is pulled to a distance x=10 cmx=10 \mathrm{~cm} from its equilibrium position at x=0x=0 on a frictionless surface from rest at t=0t=0. Write the expression for its x(t)x(t) and v(t)v(t).
View answer and explanation
Correct option: A
0.707cos⁡7.07t ms−10.707 \cos 7.07 t \mathrm{~ms}^{-1}
Option A (0.707cos⁡7.07t ms−10.707 \cos 7.07 t \text{{ ms}}^{-1}) is correct because it represents the velocity of the block as a function of time, derived from simple harmonic motion equations.
154 Medium
154 of 445
Understanding +4 | -1
Which one of the following is not a periodic motion?
View answer and explanation
Correct option: D
An arrow released from a bow
The correct answer is An arrow released from a bow because it only moves in one direction and does not repeat its motion.
155 Medium
155 of 445
Application +4 | -1
A particle starts S.H.M. from the mean position. Its amplitude is a and total energy E{E}. At one instant 5 its kinetic energy is 3 E/43 {~E} / 4, its displacement at this instant is
View answer and explanation
Correct option: B
y=a2y=\frac{a}{2}
Option B (y=a2y=\frac{a}{2}) is correct because it represents the displacement at the given kinetic energy, calculated using the energy conservation principle in SHM.
156 Medium
156 of 445
Understanding +4 | -1
Assertion: Every periodic motion is not always a simple harmonic motion.
Reason: The motion governed by the force law F=−kxF = −kx is simple harmonic.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of assertion
Option A (If both assertion and reason are true and reason is the correct explanation of assertion) is correct because simple harmonic motion is a specific type of periodic motion defined by a restoring force proportional to displacement.
157 Medium
157 of 445
Understanding +4 | -1
The acceleration of particle executing SHM
when it is at mean position is
View answer and explanation
Correct option: D
zero
Option D (zero) is correct because the acceleration in SHM is proportional to the displacement from the mean position, which is zero at the mean position.
158 Medium
158 of 445
Application +4 | -1
Two particles execute simple harmonic motion of the same amplitude and frequency along the same straight line. They pass one another when going in opposite directions and each time their displacement is half the amplitude. The phase difference between them is
View answer and explanation
Correct option: D
120∘120^{\circ}
Option D (120∘120^{\circ}) is correct because it represents the phase difference between two SHM particles when their displacement is half the amplitude and they are moving in opposite directions.
159 Medium
159 of 445
Application +4 | -1
A particle oscillates simple harmonically with a period of 16 s16 \mathrm{~s}. Two second after crossing the equilibrium position its velocity becomes 1 m/s1 \mathrm{~m} / \mathrm{s}. The amplitude is
View answer and explanation
Correct option: B
82πm\frac{8 \sqrt{2}}{\pi} \mathrm{m}
Option B (82πm\frac{8 \sqrt{2}}{\pi} \mathrm{m}) is correct because the amplitude is 82πm\frac{8 \sqrt{2}}{\pi} \mathrm{m}, calculated using the given period and velocity at a specific time.
160 Medium
160 of 445
Application +4 | -1
The equation of a simple harmonic motion is x=0.34cos⁡(3000t+0.74)x=0.34 \cos (3000 t+0.74), where xx and tt are in mm\mathrm{mm} and second, respectively. The frequency of the motion (in Hz\mathrm{Hz}) is
View answer and explanation
Correct option: B
3000/2π3000 / 2 \pi
Option B (3000/2π3000 / 2 \pi) is correct because frequency is angular frequency divided by 2π2\pi.
161 Medium
161 of 445
Application +4 | -1
A particle is executing SHM of period 24 xx and of amplitude 41 cm with O as equilibrium position. The minimum time in seconds taken by the particle to go from P to Q. where OP = -9 cm and OQ =40 cm is
View answer and explanation
Correct option: B
6
Option B (6) is correct because the minimum time taken to go from P to Q is calculated using the SHM equation and the given displacements.
162 Medium
162 of 445
Understanding +4 | -1
S-I: In SHM, total mechanical energy can be negative also.
S-II: Potential energy is always negative.
View answer and explanation
Correct option: B
S-I &\& S-II are false
Option B (S-I &S−IIarefalse)iscorrectbecauseinSHM,totalmechanicalenergyisalwayspositive,andpotentialenergycanbezeroorpositive,butnotalwaysnegative.\& S-II are false) is correct because in SHM, total mechanical energy is always positive, and potential energy can be zero or positive, but not always negative.
163 Medium
163 of 445
Application +4 | -1
The equation of a simple harmonic motion is given by y=6sin⁡10πt+8cos⁡10πty=6 \sin 10 \pi t+8 \cos 10 \pi t, where yy is in cm and tt in sec. Determine the period and initial phase.
View answer and explanation
Correct option: A
0.2 s,53∘8′0.2 \mathrm{~s}, 53^{\circ} 8^{\prime}
Option A (0.2 s,53∘8′0.2 \mathrm{~s}, 53^{\circ} 8^{\prime}) is correct because it correctly calculates the period and initial phase by comparing the given equation with the standard SHM equation.
164 Medium
164 of 445
Application +4 | -1
A particle executing simple harmonic motion has an amplitude of 6 cm6 \mathrm{~cm}. Its acceleration at a distance of 2 cm2 \mathrm{~cm} from the mean position is 8cms−28 \mathrm{cms}^{-2}. The maximum speed of the particle is
View answer and explanation
Correct option: B
12 cms−112 \mathrm{~cms}^{-1}
Option B (12 cms−112 \mathrm{~cms}^{-1}) is correct because it represents the maximum speed of the particle, calculated using the given acceleration and amplitude.
165 Medium
165 of 445
Understanding +4 | -1
Which of the following expressions corresponds to simple harmonic motion along a straight line, where xx is the displacement and a,b,ca, b, c are positive constants?
View answer and explanation
Correct option: D
−bx-b x
Option D (−bx-b x) is correct because it represents a restoring force that is proportional to the displacement and directed towards the equilibrium position, which is the condition for simple harmonic motion.
166 Medium
166 of 445
Analyzing +4 | -1
Statement-I: In simple harmonic motion AA is the amplitude of oscillation. If t1t_1 be the time to reach the particle from mean position to A2\frac{A}{\sqrt{2}} and t2t_2 the time to reach from A2\frac{A}{\sqrt{2}} to AA. Then t1=t22t_1=\frac{t_2}{\sqrt{2}}

Statement-II: Equation of motion for the particle starting from mean position is given by x=±Asin⁡ωtx= \pm \mathrm{A} \sin \omega \mathrm{t} and of the particle starting from extreme position is given by x=±x= \pm Acos⁡ωtA\cos \omega t
View answer and explanation
Correct option: D
If Statement - I is false but Statement - II is true.
Option D (If Statement - I is false but Statement - II is true) is correct because the time to reach A/sqrt(2) is not t2/sqrt(2), and the equations of motion are correctly stated.
167 Medium
167 of 445
Analyzing +4 | -1
The equation y=acos⁡2ωt+asin⁡2ωty={a \cos} 2 \omega t+a \sin 2 \omega t is
View answer and explanation
Correct option: C
SHM
Option C (SHM) is correct because the given equation can be simplified to a sinusoidal function, which represents Simple Harmonic Motion.
168 Medium
168 of 445
Understanding +4 | -1
For a body executing SHM
A. potential energy is always equal to its kinetic energy.
B. average potential and kinetic energy over any given time interval are always equal.
C. sum of the kinetic and potential energy at any point of time is constant.
D. average kinetic energy is one time period is equal to average potential energy in one time period.
Choose the most appropriate option from the options given below.
View answer and explanation
Correct option: A
(C) and (D)
Option A ((C) and (D)) is correct because for SHM, the sum of kinetic and potential energy is constant, and the average kinetic and potential energies over one period are equal.
169 Medium
169 of 445
Understanding +4 | -1
Periodic motion is one which:
View answer and explanation
Correct option: C
Repeats in both position and phase after regular intervals of time
Option C is correct because periodic motion is defined by the repetition of both position and phase at regular time intervals.
170 Medium
170 of 445
Understanding +4 | -1
When the bob of a simple pendulum swings, the work done by tension in the string is
View answer and explanation
Correct option: C
zero
Option C (zero) is correct because the tension in the string is always perpendicular to the displacement of the bob, resulting in zero work done.
171 Medium
171 of 445
Application +4 | -1
A pendulum clock gains 5 seconds per day at a temperature of 15∘C15^{\circ} \mathrm{C} and loses 10 seconds per day at a temperature 30∘C30^{\circ} \mathrm{C}. At what temperature, the pendulum clock will neither gain nor lose time?
View answer and explanation
Correct option: B
20∘C20^{\circ} \mathrm{C}
Option A (200  J200\,\,J) is correct because the total mechanical energy is calculated using the formula 12KA2\frac{1}{2}KA^2.
172 Medium
172 of 445
Understanding +4 | -1
The necessary condition for the bob of a
pendulum to execute SHM is
View answer and explanation
Correct option: C
amplitude of oscillations should be small
Option C (amplitude of oscillations should be small) is correct because SHM requires small angular displacement and thus small amplitude.
173 Medium
173 of 445
Application +4 | -1
Two pendulums of lengths 1m and 1.21m respectively start swinging together with same amplitude. The number of vibrations that will be executed by the longer pendulum before the two will swing together again are
View answer and explanation
Correct option: B
10
Option B (10) is correct because the number of vibrations executed by the longer pendulum before they swing together again is calculated based on the ratio of their time periods.
174 Medium
174 of 445
Application +4 | -1
A simple pendulum is hanging from a peg inserted in a vertical wall. Its bob is stretched in horizontal position from the wall and is left free to move. The bob hits on the wall the coefficient of restitution is 25\frac{2}{{\sqrt 5 }} After how many collisions the amplitude of vibration will become less than 60°
View answer and explanation
Correct option: B
3
Option B (3) is correct because after 3 collisions, the amplitude of vibration will become less than 60 degrees.
175 Medium
175 of 445
Understanding +4 | -1
Which type of oscillations are damped oscillations?
View answer and explanation
Correct option: D
Oscillations whose amplitude goes on decreasing and finally becomes zero
Option D is correct because damped oscillations are characterized by a decreasing amplitude over time due to energy loss.
176 Medium
176 of 445
Understanding +4 | -1
In the case of conical pendulum, if T\mathrm{T} is the tension in the string and θ\theta is the semi vertical angle of cone, then the component of tension which balances the centrifugal force in equilibrium position is
View answer and explanation
Correct option: C
Tsin⁡θ\mathrm{T} \sin \theta
Option C is correct because the horizontal component of the tension, T sin θ, balances the centrifugal force in a conical pendulum.
177 Medium
177 of 445
Understanding +4 | -1
The time period of a simple pendulum, when it is made to oscillate on the surface of moon
View answer and explanation
Correct option: A
Increases
Option A (Increases) is correct because the time period of a simple pendulum is inversely proportional to the square root of the acceleration due to gravity, which is less on the moon.
178 Medium
178 of 445
Understanding +4 | -1
A particle is executing S.H.M. Then the graph of acceleration as a function of displacement is
View answer and explanation
Correct option: A
A straight line
Option A (A straight line) is correct because in SHM, acceleration is directly proportional to displacement with a negative sign (a=−ω2xa = -\omega^2 x).
179 Medium
179 of 445
Application +4 | -1
A simple pendulum performs simple harmonic motion about x=0x=0 with an amplitude AA and time period TT. The speed of the pendulum at x=A3x=\frac{A}{\sqrt{3}} will be
View answer and explanation
Correct option: D
πAT83\frac{\pi A}{T} \sqrt{\frac{8}{3}}
Option D (πAT83\frac{\pi A}{T} \sqrt{\frac{8}{3}}) is correct because it is the correct application of the velocity formula in SHM at the given displacement.
180 Medium
180 of 445
Analyzing +4 | -1
Which of the following equation does not represent a SHM?
View answer and explanation
Correct option: C
1−sin⁡2ωt1-\sin 2 \omega t
Option D (1−sin⁡2ωt1-\sin 2 \omega t) is correct because it is a periodic function but does not satisfy the condition for SHM, which requires the form Asin⁡(ωt+ϕ)A\sin(\omega t + \phi).
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