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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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271
Medium
271 of 445
Application
+4 | -1
A particle moves so that its acceleration a is given by where is displacement from equilibrium position and b is a non-negative real constant. The time period of oscillation of the particle is
View answer and explanation
Correct option: C
Option C () is correct because it is the correct time period derived from the given acceleration equation and the SHM equation.
272
Medium
272 of 445
Application
+4 | -1
When an oscillator completes 50 oscillations, its amplitude reduces to times of the initial value. The amplitude, when it completes 100 oscillations will be
View answer and explanation
Correct option: B
times of initial value
Option B ( times of initial value) is correct because the amplitude decreases exponentially with time, and after 100 oscillations, it will be (1/3)*(1/3) of the initial amplitude.
273
Medium
273 of 445
Analyzing
+4 | -1
A solid sphere of radius and density is attached to one end of a massless spring of force constant . The other end of the spring is connected to another solid sphere of radius and density . The complete arrangement is placed in a liquid of density and is allowed to reach equilibrium. The correct statement is
View answer and explanation
Correct option: A
net elongation of spring is and the light sphere is completely submerged
Option A (net elongation of spring is and the light sphere is completely submerged) is correct because the buoyant force and weight balance out, leading to a specific spring elongation and complete submersion of the lighter sphere.
274
Medium
274 of 445
Understanding
+4 | -1
Two particles are executing S.H.M. The equations of their motion are .
What is the ratio of their amplitude?
What is the ratio of their amplitude?
View answer and explanation
Correct option: B
0.086805556
The correct answer is B (02:05) because the ratio of amplitudes is simply the ratio of the coefficients of the sine functions.
275
Medium
275 of 445
Application
+4 | -1
When the displacement is half the amplitude, the ratio of potential energy to the total energy is
View answer and explanation
Correct option: B
Option B () is correct because the potential energy is proportional to the square of the displacement.
276
Medium
276 of 445
Analyzing
+4 | -1
The displacement y of a particle executing periodic motion is given by . This expression may be considered to be a result of the superposition of…….. independent harmonic motions
View answer and explanation
Correct option: B
Three
The correct answer is Three because the given expression can be simplified into a sum of three sinusoidal functions with different frequencies.
277
Medium
277 of 445
Understanding
+4 | -1
A particle is executing simple harmonic motion with an amplitude A and time period T. The displacement of the particle after 2T period from its initial position is
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because after an integer number of periods, the particle returns to its initial position, resulting in zero displacement.
278
Medium
278 of 445
Application
+4 | -1
The kinetic energy of a particle executing S.H.M. is 16 J when it is at its mean position. If the mass of the particle is 0.32 kg, then what is the maximum velocity of the particle
View answer and explanation
Correct option: C
Option C () is correct because it is the maximum velocity calculated using the kinetic energy at the mean position.
279
Medium
279 of 445
Understanding
+4 | -1
The potential energy of a particle in SHM is:
View answer and explanation
Correct option: D
Proportional to the square of displacement from the centre
Option D is correct. The potential energy in SHM is proportional to the square of the displacement from the equilibrium position.
280
Medium
280 of 445
Application
+4 | -1
is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to times of its initial value, then the modified time period is
View answer and explanation
Correct option: D
Option D () is correct because the time period of a simple pendulum is proportional to the square root of its length.
281
Medium
281 of 445
Understanding
+4 | -1
A simple pendulum consists of a mass attached to a light string if the system is oscillating through small angles which of the following is true?
View answer and explanation
Correct option: C
The period is independent of mass
Option C is correct because the period of a simple pendulum is given by , which is independent of the mass of the pendulum.
282
Medium
282 of 445
Understanding
+4 | -1
The relation between acceleration and displacement of four particles are given below. Which one of the particles is executing simple harmonic motion?
View answer and explanation
Correct option: D
Option D () is correct because simple harmonic motion requires acceleration to be proportional to displacement and in the opposite direction.
283
Medium
283 of 445
Analyzing
+4 | -1
Force acting on a body moving along -axis is given as , where is the displacement of the body from its mean position. The motion of body is:
View answer and explanation
Correct option: B
Oscillatory but not SHM
Option B is correct. The force is proportional to -x3, which means it's oscillatory but not SHM because SHM requires the force to be proportional to -x.
284
Medium
284 of 445
Analyzing
+4 | -1
Statement-1: Time period of spring pendulum is the same whether in an accelerated or in an inertial frame of reference.
Statement-2: Mass of the bob of spring pendulum and spring constant of the spring are independent of the acceleration of the frame of reference.
Statement-2: Mass of the bob of spring pendulum and spring constant of the spring are independent of the acceleration of the frame of reference.
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1
Option A (Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1) is correct because the time period of a spring pendulum depends on mass and spring constant, which are independent of the frame's acceleration.
285
Medium
285 of 445
Application
+4 | -1
The motion of a simple pendulum executing S.H.M. is represented by the following equation.
, where time is measured in second. The length of pendulum is
, where time is measured in second. The length of pendulum is
View answer and explanation
Correct option: C
99.4 cm
Option C (99.4 cm) is correct because the length of the pendulum is calculated using the relation between angular frequency and length, .
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286
Medium
286 of 445
Understanding
+4 | -1
In the differential equation of linear simple harmonic motion, , the term represents
View answer and explanation
Correct option: B
Restoring force per unit mass per unit displacement.
Option B (Restoring force per unit mass per unit displacement) is correct because represents the ratio of restoring force per unit displacement to the mass, which is restoring force per unit mass per unit displacement.
287
Medium
287 of 445
Application
+4 | -1
The velocity vector and displacement vector of a particle executing SHM are related as with the initial condition at . The velocity when displacement is , is
View answer and explanation
Correct option: B
Option B () is correct because it is the result of integrating the given differential equation with the specified initial condition.
288
Medium
288 of 445
Application
+4 | -1
A simple wave motion is represented by . Its amplitude is
View answer and explanation
Correct option: D
10 units
Option D (10 units) is correct because the amplitude is found by converting the equation into the form and calculating A.
289
Medium
289 of 445
Understanding
+4 | -1
What is the effect on the time period of a simple pendulum if the mass of the bob is doubled
View answer and explanation
Correct option: D
No effect
The correct answer is D because the time period of a simple pendulum is independent of the mass of the bob, depending only on the length and gravitational acceleration.
290
Medium
290 of 445
Analyzing
+4 | -1
Starting from
View answer and explanation
Correct option: D
acceleration leads the displacement by a phase
Option D is correct. In simple harmonic motion, acceleration leads the displacement by a phase of π (180 degrees).
291
Medium
291 of 445
Understanding
+4 | -1
The physical quantity conserved in simple
harmonic motion is
harmonic motion is
View answer and explanation
Correct option: B
total energy
Option B (total energy) is correct because in simple harmonic motion, the total energy remains constant as potential and kinetic energy are interconverted.
292
Medium
292 of 445
Application
+4 | -1
A mass attached to a spring of spring constant is stretched a distance from its equilibrium position and released with no initial velocity. The maximum speed attained by the mass in its subsequent motion and the time at which this speed would be attained are respectively,
View answer and explanation
Correct option: C
Option C () is correct because the maximum speed is achieved at the mean position and the time to reach the mean position from the extreme is T/4.
293
Medium
293 of 445
Analyzing
+4 | -1
Assertion: Amplitude of a forced vibration can never reach infinity.
Reason: The driving frequency cannot be increased beyond a certain limit.
Reason: The driving frequency cannot be increased beyond a certain limit.
View answer and explanation
Correct option: C
If assertion is true but reason is false
Option C (If assertion is true but reason is false) is correct because while the amplitude of forced vibration cannot reach infinity due to damping, the driving frequency can be increased beyond a certain limit.
294
Medium
294 of 445
Understanding
+4 | -1
Consider the following statements : The total energy of a particle executing simple harmonic motion depends on its I. Amplitude II. Period III. Displacement Of these statements
View answer and explanation
Correct option: A
I and II are correct
Option A (I and II are correct) is correct because the total energy in SHM depends on amplitude and period.
295
Medium
295 of 445
Understanding
+4 | -1
For Simple Harmonic Oscillator, the potential energy is equal to kinetic energy
View answer and explanation
Correct option: B
twice during each cycle
Option B (twice during each cycle) is correct because the potential and kinetic energies are equal at two points in each oscillation cycle.
296
Medium
296 of 445
Analyzing
+4 | -1
A particle executing SHM while moving from one extremity is found at distances and from the centre at the end of three successive seconds. The time period of oscillation is
View answer and explanation
Correct option: A
Option A () is correct, but the provided solution is incomplete and doesn't define theta. The correct approach involves using trigonometric identities and solving for omega.
297
Medium
297 of 445
Understanding
+4 | -1
A particle moves in x-y plane according to rule x=a sinωt and y=a cosωt. The particle follows
View answer and explanation
Correct option: B
A circular path
Option B (A circular path) is correct because the equations x = a sin(ωt) and y = a cos(ωt) describe parametric equations of a circle.
298
Medium
298 of 445
Application
+4 | -1
A particle executes with an amplitude of . The percentage of its energy which is potential energy at a displacement of from the mean position is
View answer and explanation
Correct option: D
4
Option D (4) is correct because the percentage of potential energy is calculated by dividing the square of the displacement by the square of the amplitude and multiplying by 100.
299
Medium
299 of 445
Application
+4 | -1
The amplitude of oscillations of a simple harmonic oscillator of mass 100 gram having an angular frequency of with a total mechanical energy of is
View answer and explanation
Correct option: C
Option C () is correct because it calculates the amplitude using the formula for total mechanical energy of a simple harmonic oscillator.
300
Medium
300 of 445
Analyzing
+4 | -1
Two pendulums of length 1 m and 16 m start vibrating one behind the other from the same stand. At some instant, the two are in the mean position in the same phase. The time period of shorter pendulum is T. the minimum time after which the two threads of the pendulum will be one behind the other is?
View answer and explanation
Correct option: C
Option C (4T/3) is correct. It is derived by finding the time when the two pendulums are in phase again, considering their different time periods.
Revision notes and key takeaways
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Showing 271-300 of 445 questions
