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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 331-360 of 445 questions

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331 Medium
331 of 445
Application +4 | -1
Time period of a simple pendulum of length ℓ\ell is T1T_1 and time period of uniform rod of same length ℓ\ell pivoted about one end and oscillating in a vertical plane is T2T_2. Amplitude of oscillations in both the cases is small, then T1T2\frac{T_1}{T_2} is
View answer and explanation
Correct option: C
32\sqrt{\frac{3}{2}}
Option C (32\sqrt{\frac{3}{2}}) is correct because it represents the ratio of the time periods of a simple pendulum and a uniform rod oscillating about one end, calculated using their respective formulas.
332 Medium
332 of 445
Application +4 | -1
A particle is executing SHM of amplitude ' AA ' and time period ' TT '. Find the time (T) taken by the particle to go from x=0x=0 to x=A/2x=A / 2.
View answer and explanation
Correct option: A
T/12\mathrm{T} / 12
Option A (T/12) is correct because the time taken to go from x=0 to x=A/2 in SHM is T/12, derived from the equation x = A sin(ωt).
333 Medium
333 of 445
Analyzing +4 | -1
The bob of a pendulum is in the lowest position and has a horizontal velocity of gL\sqrt {gL} where LL is length of the pendulum and gg is acceleration due to gravity is given to it. Then the bob
View answer and explanation
Correct option: C
Oscillates about the lowest position as mean position
Option C (Oscillates about the lowest position as mean position) is correct because the initial velocity is insufficient to complete a full circle, resulting in oscillation.
334 Medium
334 of 445
Application +4 | -1
A bob of mass m suspended by a thread of length ll undergoes simple harmonic oscillations with time period TT. If the bob is immersed in a liquid that has density 1/41/4 times that of the bob and the length of the thread is increased by 1/3  rd1\mathrm{/}3\,\,\mathrm{rd} of the original length, then the time period of the simple harmonic oscillations will be
View answer and explanation
Correct option: D
43T\frac{4}{3}T
Option D (43T\frac{4}{3}T) is correct because the time period of a simple pendulum changes due to the effective gravity in the liquid and the change in length.
335 Medium
335 of 445
Application +4 | -1
A particle is executing simple harmonic motion with frequency ff . The frequency at which its kinetic energy change into potential energy is
View answer and explanation
Correct option: C
2  f2\,\,f
Option C (2  f2\,\,f) is correct because the frequency of kinetic and potential energy interchange in SHM is twice the frequency of the oscillating particle.
336 Medium
336 of 445
Understanding +4 | -1
If a simple harmonic is represented by d2xdt2\frac{{{d^2}x}}{{d{t^2}}}+αx=0+ \alpha x = 0,its time period is
View answer and explanation
Correct option: B
2πα\frac{{2\pi }}{{\sqrt \alpha }}
Option B is correct because the time period of a simple harmonic motion is given by T=2πωT = \frac{{2\pi}}{{\omega}}, where ω=α\omega = \sqrt{{\alpha}} in this case.
337 Medium
337 of 445
Application +4 | -1
Two pendulums of length 121 cm and 100 cm start vibrating. At same instant the two are in the mean position in the same phase. After how many vibrations of the shorter pendulum, the two will be in phase in the mean position?
View answer and explanation
Correct option: B
11
Option B (11) is correct because it satisfies the condition where the shorter pendulum completes an integer number of vibrations while the longer pendulum completes one less to be in phase again.
338 Medium
338 of 445
Remembering +4 | -1
If a body oscillates at the angular frequency ωd{\omega _d} of the driving force, then the oscillations are called
View answer and explanation
Correct option: C
Forced oscillations
Option C (Forced oscillations) is correct because forced oscillations occur when a body oscillates at the driving force's frequency.
339 Medium
339 of 445
Understanding +4 | -1
Which one of the following is simple harmonic?
View answer and explanation
Correct option: D
Motion of oscillating liquid column in U-tube
Option D (Motion of oscillating liquid column in U-tube) is correct because the restoring force is proportional to the displacement from the equilibrium position.
340 Medium
340 of 445
Understanding +4 | -1
Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion
View answer and explanation
Correct option: D
When v is maximum, a is zero
Option D (When v is maximum, a is zero) is correct because in SHM, velocity is maximum at the equilibrium position where acceleration is zero.
341 Medium
341 of 445
Application +4 | -1
The time period of a simple pendulum measured inside a stationary elevator is 4 s4 {~s}. If the lift start accelerating downwards with an acceleration 3g/43 g / 4, the time period will be
View answer and explanation
Correct option: B
8 s8 {~s}
Option B (8 s) is correct because the effective acceleration changes when the elevator accelerates downwards, affecting the time period of the pendulum.
342 Medium
342 of 445
Analyzing +4 | -1
Assertion: A heavy metal sphere is suspended by a spring. When pulled down a little and released, it oscillates up and down with a frequency f.f. If it is taken to the moon where the acceleration due to gravity is g/6,g/6, the frequency of oscillation is 6f\sqrt{6}f.
Reason: The frequency of oscillation of the spring depends only on the mass of the sphere and spring constant.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D (If both Assertion & Reason are false) is correct because the frequency of oscillation does not depend on gravity, and the frequency would remain the same on the moon.
343 Medium
343 of 445
Analyzing +4 | -1
The time period of a simple pendulum is TT. If its point of suspension is moved upward according to relation y=λt2y=\lambda t^2 where λ\lambda is a constant then its new time period T1T_1.
View answer and explanation
Correct option: C
is less than T{T}
Option C is correct because the effective acceleration due to gravity increases when the suspension point moves upwards, leading to a decrease in the time period of the pendulum.
344 Medium
344 of 445
Analyzing +4 | -1
A simple pendulum is made of bob which is a hollow sphere full of sand suspended by means of a wire. If all the sand is drained out, the period of the pendulum will
View answer and explanation
Correct option: C
Remain same
Option C (Remain same) is correct because the length of the pendulum, measured from the pivot to the center of mass, does not change when the sand is drained.
345 Medium
345 of 445
Application +4 | -1
A mass of 2 kg attached to a spring of force constant 260∘   Nm−1260^\circ \,\,\,N{m^{ - 1}} makes 200 oscillations. What is the time taken [in seconds] nearly?
View answer and explanation
Correct option: D
110
Option D (110) is correct because the time taken for 200 oscillations is calculated using the formula for the time period of a spring-mass system.
346 Medium
346 of 445
Understanding +4 | -1
A plate oscillates with time period ′T′'T'. Suddenly, another plate put on the first time, then time period
View answer and explanation
Correct option: C
Will be same
Option B (Will increase) is incorrect and option B (Will increase) is correct because adding another plate increases the mass, which affects the time period of oscillation.
347 Medium
347 of 445
Application +4 | -1
Two particles PP and QQ describe SHM of same amplitude aa and frequency vv along the same straight line. The maximum distance between two particle is 2a\sqrt{2} a. The initial phase difference between the particles is
View answer and explanation
Correct option: B
π/2\pi / 2
Option B (π/2\pi / 2) is correct because the maximum distance between the particles is related to the phase difference by 2a=a2(1−cos⁡ϕ)\sqrt{2}a = a\sqrt{2(1-\cos{\phi})}.
348 Medium
348 of 445
Application +4 | -1
For a SHO, if the magnitudes of maximum velocity and acceleration are equal, its frequency of oscillations is
View answer and explanation
Correct option: B
12π  Hz\frac{1}{{2\pi }}\,\,{\text{Hz}}
Option B (12π Hz\frac{{1}}{{2\pi }}\,{\text{{Hz}}}) is correct because equating the magnitudes of maximum velocity and acceleration in SHM leads to ω=1\omega = 1, and then using ω=2πf\omega = 2\pi f gives the frequency.
349 Medium
349 of 445
Understanding +4 | -1
The elongation of spring of length LL and of negligible mass due to a force is xx. the spring is cut into two pieces of length in ratio 1:n1: n. The ratio of the respective spring constants is:
View answer and explanation
Correct option: A
n:1{n}: 1
Option A is correct because the spring constant is inversely proportional to the length of the spring, so when the spring is cut into a ratio of 1:n, the spring constants are in the ratio of n:1.
350 Medium
350 of 445
Application +4 | -1
A particle executes SHM with a time period of 2 s2 \mathrm{~s} and amplitude 5 cm.5 \mathrm{~cm}. Find displacement after 1/31 / 3 second; starting from the mean position.
View answer and explanation
Correct option: A
4.33 cm4.33 \mathrm{~cm}
Option A (4.33 cm4.33 \mathrm{~cm}) is correct because it is the displacement calculated using the SHM equation x=Asin⁡ωtx=A \sin \omega t with the given time period, amplitude, and time.
351 Medium
351 of 445
Application +4 | -1
F−xF-x equation of a body in SHM along x{x}-axis is F+12x=0F+12 x=0. Here, FF is in newton and xx in metre. If the mass of the body is 3 kg3 {~kg}, then the time period of oscillations is;
View answer and explanation
Correct option: C
πs\pi {s}
Option C (π\pi s) is correct because it is calculated using the formula for the time period of SHM, T=2πmKT = 2\pi \sqrt{{\frac{{m}}{{K}}}} after finding K from the given equation.
352 Medium
352 of 445
Application +4 | -1
The force constant of two springs are  K1\;{K_1} and K2{K_2} Both are stretched till their elastic energies are equal. If the stretching forces are F1{F_1} and F2,{F_2}, then F1:F2{F_1}:{F_2} is
View answer and explanation
Correct option: C
K1:K2\sqrt {{K_1}} :\sqrt {{K_2}}
Option C (K1:K2\sqrt {{K_1}} :\sqrt {{K_2}}) is correct because it is derived from equating the elastic potential energies and relating them to the stretching forces.
353 Medium
353 of 445
Application +4 | -1
The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement
View answer and explanation
Correct option: C
a2\frac{a}{{\sqrt 2 }}
Option C (a2\frac{a}{{\sqrt 2 }}) is correct because at this displacement, the potential energy and kinetic energy of the particle in SHM are equal.
354 Medium
354 of 445
Understanding +4 | -1
What is the phase difference between velocity and acceleration of a particle executing SHM?
View answer and explanation
Correct option: C
π/2\pi / 2
Option C is correct because in SHM, velocity leads displacement by π/2\pi/2 and acceleration leads velocity by π/2\pi/2, hence the phase difference between velocity and acceleration is π/2\pi / 2.
355 Medium
355 of 445
Understanding +4 | -1
A body is executing simple harmonic motion with frequency 'nn' the frequency of its potential energy is
View answer and explanation
Correct option: D
2n2n
Option D (2n2n) is correct because the frequency of potential energy in SHM is twice the frequency of the displacement.
356 Medium
356 of 445
Application +4 | -1
A particle performs Simple Harmonic Motion of amplitude A{A} and angular frequency ω\omega along a straight line. When it is at a distance 32 A\frac{\sqrt{3}}{2} {~A} from mean position, its kinetic energy gets increased suddenly by an amount 12mω2A2\frac{1}{2} m \omega^2 A^2, due to an impulsive force. Then new amplitude of oscillations of the particle will be :
View answer and explanation
Correct option: C
2 A\sqrt{2} {~A}
Option C (2 A\sqrt{2} {~A}) is correct because the new amplitude is calculated by equating the new total energy after the impulse to the potential energy at the new amplitude.
357 Medium
357 of 445
Application +4 | -1
A second's pendulum is taken in a carriage. Find the period of oscillation when the carriage moves with an acceleration of 4 ms−24 \mathrm{~ms}^{-2} (i) vertically upwards (ii) vertically downwards, and (iii) in a horizontal direction.
View answer and explanation
Correct option: A
1.69 s;2.59 s;1.92 s1.69 \mathrm{~s}; 2.59 \mathrm{~s}; 1.92 \mathrm{~s}
Option A (1.69 s;2.59 s;1.92 s1.69 \mathrm{~s}; 2.59 \mathrm{~s}; 1.92 \mathrm{~s}) is correct because it correctly calculates the period of oscillation of a second's pendulum in a carriage moving with different accelerations.
358 Medium
358 of 445
Remembering +4 | -1
The differential equation of a particle executing simple harmonic motion along yy-axis is
View answer and explanation
Correct option: A
d2ydt2+ω2y=0\frac{d^2 y}{d t^2}+\omega^2 y=0
Option A (d2ydt2+ω2y=0\frac{{d^2 y}}{{d t^2}}+\omega^2 y=0) is the correct differential equation for a particle executing simple harmonic motion along the y-axis because it represents the relationship between acceleration and displacement in SHM.
359 Medium
359 of 445
Remembering +4 | -1
Spring force is a
View answer and explanation
Correct option: B
Conservative force
Option B (Conservative force) is correct because spring force is a conservative force, meaning the work done by it is path-independent and can be expressed as a potential energy.
360 Medium
360 of 445
Understanding +4 | -1
A spring of spring constant k{k} is cut into 3 equal part find kk of each
View answer and explanation
Correct option: A
3k3 k
Option A (3k3 k) is correct because when a spring is cut into equal parts, the spring constant of each part increases proportionally.
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