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JEE • PHYSICS • OSCILLATIONS

Oscillations practice · Mixed difficulty

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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 61-90 of 483 questions
61 Medium
61 of 483
Understanding +4 | -1
Two equations of two S.H.M. are y=asin⁡(ωt−α)y = a\sin \left( {\omega t - \alpha } \right) and y=bcos⁡(ωt−α)y = b\cos \left( {\omega t - \alpha } \right) The phase difference between the two is
View answer and explanation
Correct option: C
90∘90^\circ
Option C (90∘90^\circ) is correct because converting sine to cosine reveals a phase difference of π2\frac{\pi}{2} between the two SHM equations.
62 Medium
62 of 483
Understanding +4 | -1
The equation of linear simple harmonic motion is x=8cos⁡(12πt)x=8 \cos (12 \pi t) where xx is in cm\mathrm{cm} and tt is in second. The initial phase angle is:
View answer and explanation
Correct option: C
π/2 rad\pi / 2 \mathrm{~rad}
Option C (π/2 rad\pi / 2 \mathrm{~rad}) is correct because the given equation can be rewritten in terms of sine with an initial phase of π/2\pi/2.
63 Medium
63 of 483
Application +4 | -1
Two pendulums of time periods 3 s3 \mathrm{~s} and 7 s7 \mathrm{~s} respectively, start oscillating simultaneously from two opposite extreme positions. After how much time they will be in same phase?
View answer and explanation
Correct option: A
218 s\frac{21}{8} \mathrm{~s}
Option A (218 s\frac{21}{8} \mathrm{~s}) is correct because it is the time calculated for the two pendulums to be in the same phase.
64 Easy
64 of 483
Analyzing +4 | -1
Statement I When a wave goes from one medium to other, average power transmitted by the wave may change.
Statement II Due to a change in the medium, amplitude, speed, wavelength and frequency of the wave may change.
View answer and explanation
Correct option: A
Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I
Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I because the average power transmitted by a wave depends on the medium's properties, which can change the wave's amplitude, speed, wavelength, and frequency.
65 Medium
65 of 483
Understanding +4 | -1
It is impossible for two oscillators, each
executing simple harmonic motion, to remain
in phase with each other if they have different
View answer and explanation
Correct option: A
time periods
Option A is correct because if two oscillators have different time periods, they cannot maintain a constant phase relationship.
66 Medium
66 of 483
Understanding +4 | -1
During the swinging of simple pendulum,
View answer and explanation
Correct option: A
the work done by tension force is always zero
Option A (the work done by tension force is always zero) is correct because the tension force is always perpendicular to the displacement of the pendulum bob.
67 Medium
67 of 483
Application +4 | -1
The motion of a particle is expressed by the equation acc. a =−=- bx where x\mathrm{x} is displacement from equilibrium position and bb is constant. What is the periodic time?
View answer and explanation
Correct option: B
2πb\frac{2 \pi}{\sqrt{b}}
Option B (2πb\frac{2 \pi}{\sqrt{b}}) is correct because the periodic time is calculated using the formula T=2πωT = \frac{2\pi}{\omega}, where ω=b\omega = \sqrt{b} from the given acceleration equation.
68 Medium
68 of 483
Application +4 | -1
Displacement-time equation of a particle executing SHM is, x=4sin⁡ωt+3sin⁡  (ωt+π/3)x = 4\sin \omega t + 3\sin \,\,(\omega t + \pi /3) Here xx is in centimeter and t in second. The amplitude of oscillation of the particle is approximately
View answer and explanation
Correct option: B
6cm6\mathrm{ }cm
Option B (6cm6\mathrm{ }cm) is correct because the amplitude of the resulting SHM is found by vector addition of the individual SHMs.
69 Medium
69 of 483
Application +4 | -1
A particle of mass 1 kg is moving in SHM with an amplitude 0.02 m and a frequency of 60 Hz. The maximum force in newton acting on the particle is
View answer and explanation
Correct option: C
288  π2288\,\,{\pi ^2}
Option C (288π2288\pi ^2) is correct because the maximum force is calculated using the formula Fmax=mω2a=m4π2f2aF_{max} = m\omega^2a = m4\pi^2f^2a.
70 Medium
70 of 483
Understanding +4 | -1
Which of the following characteristics does not change� due to the damping of simple harmonic motion?
View answer and explanation
Correct option: C
Initial phase
Option C (Initial phase) is correct because damping primarily affects amplitude, frequency, and time period, not the initial phase.
71 Medium
71 of 483
Application +4 | -1
The maximum velocity of a simple harmonic motion represented by y=3sin⁡[100t+π6]y = 3\sin \left[ {100t + \frac{\pi }{6}} \right] is given by
View answer and explanation
Correct option: A
300
Option A (300) is correct because the maximum velocity in SHM is the product of amplitude and angular frequency.
72 Medium
72 of 483
Understanding +4 | -1
The quantity which does not vary periodically
for a particle performing S.H.M. is
View answer and explanation
Correct option: B
total energy
Option B (total energy) is correct because the total energy of a particle in SHM remains constant and does not vary periodically.
73 Medium
73 of 483
Application +4 | -1
A block whose mass is 650 g650 \mathrm{~g} is fastened to a spring whose spring constant is 65 N/m65 \mathrm{~N} / \mathrm{m}. The block is pulled a distance x=11 cmx=11 \mathrm{~cm} from its equilibrium position at x=0x=0 on a frictionless surface and released from rest at t=0t=0. The maximum velocity of the vibrating block is
View answer and explanation
Correct option: A
1.1 m/s1.1 \mathrm{~m} / \mathrm{s}
Option A (1.1 m/s1.1 \mathrm{~m} / \mathrm{s}) is correct because the maximum velocity is calculated by equating the potential energy stored in the spring to the kinetic energy of the block.
74 Medium
74 of 483
Understanding +4 | -1
The average acceleration of a particle performing SHM over one complete oscillation is
View answer and explanation
Correct option: C
Zero
Option C (Zero) is correct because in SHM, the average acceleration over one complete cycle is zero as the particle's velocity changes direction.
75 Easy
75 of 483
Remembering +4 | -1
The mass of a bob, suspended in a simple pendulum, is halved from the initial mass, its time period will:
View answer and explanation
Correct option: C
Remain unchanged
Option C (Remain unchanged) is correct because the time period of a simple pendulum is independent of the mass of the bob.
76 Medium
76 of 483
Application +4 | -1
Two springs have spring constants KA{K_A} and KB{K_B} and   KA>KB\;{K_A} > {K_B} The work required to stretch them by same extension will be
View answer and explanation
Correct option: A
More in spring A
Option A (More in spring A) is correct because the work done to stretch a spring is directly proportional to its spring constant.
77 Medium
77 of 483
Understanding +4 | -1
A simple harmonic wave having an amplitude a and time period T is represented by the equation y=5sin⁡π(t+4)my = 5\sin \pi \left( {t + 4} \right)m then the value of amplitude (a) in (m) and time period (T) in second are
View answer and explanation
Correct option: D
a = 5, T = 2
Option D is correct because by comparing the given equation to the standard form, we can identify the amplitude as 5 and the time period as 2.
78 Medium
78 of 483
Application +4 | -1
The height of a swing changes during its motion from 0.1 mm to 2.5 mm. The minimum velocity of a boy who swings in this swing is
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because the minimum velocity of the swing occurs at the highest point of its motion, which is momentarily zero.
79 Medium
79 of 483
Analyzing +4 | -1
Assertion: In x=Acos⁡ωt,xx=A \cos \omega t, x is the displacement measured from extreme position.

Reason: In the above equation x=Ax=A at time t=0t=0.
View answer and explanation
Correct option: D
If Assertion is false but the Reason is true
Option D is correct because the assertion is false, as x is measured from the mean position, while the reason is true, as x=A at t=0 in the given equation.
80 Medium
80 of 483
Remembering +4 | -1
How does time period of a pendulum very with length?
View answer and explanation
Correct option: A
l\sqrt l
Option A (l\sqrt l) is correct because the time period of a pendulum is directly proportional to the square root of its length, as given by T=2πlgT = 2\pi\sqrt{\frac{l}{g}}.
81 Medium
81 of 483
Analyzing +4 | -1
Which of the following quantities are always positive in a simple harmonic motion?
View answer and explanation
Correct option: A
F⃗⋅a⃗\vec{F} \cdot \vec{a}
Option A (F⃗⋅a⃗\vec{F} \cdot \vec{a}) is correct because in SHM, force and acceleration are always in the same direction.
82 Medium
82 of 483
Application +4 | -1
The amplitude of a simple harmonic oscillator is 2 cm. At certain instant its PE and KE are equal. The displacement from mean position at that instant is
View answer and explanation
Correct option: A
2 cm\sqrt 2 \,\mathrm{cm}
Option A (2 cm\sqrt 2 \,\mathrm{cm}) is correct because at the point where potential and kinetic energies are equal, the displacement is A/2\sqrt{2}, where A is the amplitude.
83 Easy
83 of 483
Understanding +4 | -1
When a particle executing SHM oscillates with a frequency vv, then the kinetic energy of the particle
View answer and explanation
Correct option: B
changes periodically with a frequency of 2v2 \mathrm{v}
Option B is correct. The kinetic energy of a particle in SHM oscillates with twice the frequency of the SHM itself.
84 Medium
84 of 483
Application +4 | -1
Ratio of kinetic energy at mean position to potential energy at A⁄2 of a particle performing SHM
View answer and explanation
Correct option: B
  4:1\;4:1
Option B (  4:1\;4:1) is correct because it represents the ratio of kinetic energy at the mean position to potential energy at A/2 in SHM.
85 Medium
85 of 483
Understanding +4 | -1
A simple pendulum is vibrating in an
evacuated chamber. It will oscillate with ––––
View answer and explanation
Correct option: A
Constant amplitude
Option A (Constant amplitude) is correct because in an evacuated chamber, there is no air resistance to dampen the oscillations, so the pendulum will swing with a constant amplitude.
86 Medium
86 of 483
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.
87 Medium
87 of 483
Understanding +4 | -1
Assertion: When a particle is at extreme position performing SHM, its momentum is equal to zero.
Reason: At extreme position the velocity of particle performing SHM is equal to zero.
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 at the extreme position in SHM, the velocity is zero, hence the momentum is also zero.
88 Medium
88 of 483
Application +4 | -1
A simple harmonic oscillator has an amplitude of 0.5 m0.5 \mathrm{~m} and a time period of 2 seconds. What is the magnitude of acceleration when it is displaced from the mean position by 0.25 m0.25 \mathrm{~m} ?
View answer and explanation
Correct option: C
π24 m/s2\frac{\pi^2}{4} \mathrm{~m} / \mathrm{s}^2
Option C (π24 m/s2\frac{\pi^2}{4} \mathrm{~m} / \mathrm{s}^2) is correct because the acceleration in SHM is proportional to the displacement and can be calculated using the angular frequency and displacement.
89 Medium
89 of 483
Application +4 | -1
A simple pendulum has a length ll . The inertial and gravitational masses of the bob are mi{m_i} and mg{m_g} respectively. Then the time period TT is given by
View answer and explanation
Correct option: B
T=2πmilmggT = 2\pi \sqrt {\frac{{{m_i}l}}{{{m_g}g}}}
The correct answer is B (T=2πmilmggT = 2\pi \sqrt {\frac{{{{m_i}l}}}{{{{m_g}g}}}}) because the time period of a simple pendulum depends on the ratio of inertial mass to gravitational mass.
90 Medium
90 of 483
Application +4 | -1
A block of mass mm attached to a massless spring is performing oscillatory motion of amplitude AA on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system becomes fA.fA. The value of ff is
View answer and explanation
Correct option: C
12\frac{1}{{\sqrt 2 }}
Option C, 12\frac{1}{{\sqrt 2 }}, is correct because the energy remains the same, but the mass is halved, which affects the amplitude.
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