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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 271-300 of 445 questions

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271 Medium
271 of 445
Application +4 | -1
A particle moves so that its acceleration a is given by a=−bxa = - bx where xx 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
2πb\frac{{2\pi }}{{\sqrt b }}
Option C (2πb\frac{{2\pi }}{{\sqrt b }}) 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 13\frac{1}{3} times of the initial value. The amplitude, when it completes 100 oscillations will be
View answer and explanation
Correct option: B
19\frac{1}{9} times of initial value
Option B (19\frac{1}{9} 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 RR and density ρ\rho is attached to one end of a massless spring of force constant kk. The other end of the spring is connected to another solid sphere of radius RR and density 3ρ3 \rho. The complete arrangement is placed in a liquid of density 2ρ2 \rho and is allowed to reach equilibrium. The correct statement is
View answer and explanation
Correct option: A
net elongation of spring is 4πR3ρg3k\frac{4 \pi R^3 \rho g}{3 k} and the light sphere is completely submerged
Option A (net elongation of spring is 4πR3ρg3k\frac{{4 \pi R^3 \rho g}}{{3 k}} 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 y1=10sin⁡(ωt+πT4),y2=25sin⁡(ωt+3πT4)y_1=10 \sin \left(\omega t+\frac{\pi T}{4}\right), y_2=25 \sin \left(\omega t+\frac{\sqrt{3} \pi T}{4}\right).
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
14\frac{1}{4}
Option B (14\frac{1}{4}) 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 y=4cos⁡2(t/2)sin⁡(1000t)y = 4{\cos ^2}\left( {t/2} \right)\sin \left( {1000t} \right) . 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
10m/s10m/s
Option C (10m/s10m/s) 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
T0{T_0} is the time period of a simple pendulum at a place. If the length of the pendulum is reduced to 116\frac{1}{{16}} times of its initial value, then the modified time period is
View answer and explanation
Correct option: D
14T0\frac{1}{4}{T_0}
Option D (14T0\frac{1}{4}{T_0}) 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 l.l. 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 mm
Option C is correct because the period of a simple pendulum is given by T=2πlgT = 2\pi\sqrt{\frac{l}{g}}, 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
ax=−2xa_x=-2 x
Option D (ax=−2xa_x=-2 x) 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 xx-axis is given as F=−kx3F=-k x^3, where xx 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.
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.
y=Asin⁡(πt+ϕ)y = A\sin \left( {\pi t + \phi } \right), 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, ℓ=gπ2\ell = \frac{g}{{{\pi ^2}}}.
286 Medium
286 of 445
Understanding +4 | -1
In the differential equation of linear simple harmonic motion, d2xdt2+ω2x=0\frac{d^2 x}{d t^2}+\omega^2 x=0, the term ω2\omega^2 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 ω2\omega^2 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 vv and displacement vector xx of a particle executing SHM are related as vdvdx=−ω2x\frac{v d v}{d x}=-\omega^2 x with the initial condition v=v0v=v_0 at x=0x=0. The velocity vv when displacement is xx, is
View answer and explanation
Correct option: B
v=v02−ω2x2v=\sqrt{v_0^2-\omega^2 x^2}
Option B (v=v02−ω2x2v=\sqrt{v_0^2-\omega^2 x^2}) 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 y=5(sin⁡ 4 π t+3cos⁡ 4 π t)y=5(\sin \,4 \,\pi\, t+\sqrt{3} \cos \,4 \,\pi \,t). 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 Asin⁡(ωt+ϕ)A \sin(\omega t + \phi) 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 y=Ay = A sin⁡  ωtory=A  cos⁡  ωt\sin \,\,\omega t\mathrm{ }or\mathrm{ }y = A\,\,\cos \,\,\omega t
View answer and explanation
Correct option: D
acceleration leads the displacement by a phase π\pi
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
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 mm attached to a spring of spring constant kk is stretched a distance x0x_0 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
kmx0⋅π2mk\sqrt{\frac{k}{m}} x_0 \cdot \frac{\pi}{2} \sqrt{\frac{m}{k}}
Option C (kmx0⋅π2mk\sqrt{\frac{k}{m}} x_0 \cdot \frac{\pi}{2} \sqrt{\frac{m}{k}}) 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.
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 x1,x2x_1, x_2 and x3x_3 from the centre at the end of three successive seconds. The time period of oscillation is
View answer and explanation
Correct option: A
2π/θ2 \pi / \theta
Option A (2π/θ2 \pi / \theta) 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 sin⁡ωt and y=a cos⁡ωt. 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 SHMSHM with an amplitude of 5 cm5\,cm. The percentage of its energy which is potential energy at a displacement of 1 cm1\,cm 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 10  rad s−110\,\,rad\,{s^{ - 1}} with a total mechanical energy of is
View answer and explanation
Correct option: C
10 cm10\,cm
Option C (10 cm10\,cm) 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
4T/34T/3
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.
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