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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 61-90 of 445 questions
61 Medium
61 of 445
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.
62 Medium
62 of 445
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.
63 Medium
63 of 445
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.
64 Medium
64 of 445
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.
65 Medium
65 of 445
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.
66 Medium
66 of 445
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.
67 Medium
67 of 445
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.
68 Medium
68 of 445
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.
69 Medium
69 of 445
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.
70 Medium
70 of 445
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.
71 Medium
71 of 445
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.
72 Medium
72 of 445
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.
73 Medium
73 of 445
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}}.
74 Medium
74 of 445
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.
75 Medium
75 of 445
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.
76 Medium
76 of 445
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.
77 Medium
77 of 445
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.
78 Medium
78 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.
79 Medium
79 of 445
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.
80 Medium
80 of 445
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.
81 Medium
81 of 445
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.
82 Medium
82 of 445
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.
83 Medium
83 of 445
Understanding +4 | -1
A body is executing simple harmonic motion with
frequency 50 Hz, the frequency of its potential
energy is:
View answer and explanation
Correct option: A
100 Hz
Option A (100 Hz) is correct because the frequency of potential energy in SHM is twice the frequency of the SHM itself.
84 Medium
84 of 445
Application +4 | -1
The total energy of a particle executing S.H.M is 80 J. what is the potential energy when the particle is at a distance of 3/4 of amplitude from the mean position
View answer and explanation
Correct option: D
45 J
Option D (45 J) is correct because the potential energy is calculated as a fraction of the total energy based on the displacement from the mean position.
85 Medium
85 of 445
Understanding +4 | -1
A simple pendulum is taken from the equator to the pole. Its period
View answer and explanation
Correct option: A
Decreases
Option A (Decreases) is correct because the time period of a simple pendulum is inversely proportional to the square root of the acceleration due to gravity, which increases from the equator to the pole.
86 Medium
86 of 445
Application +4 | -1
The time period of a body suspended by a spring be TT. What will be the new period, if the spring is cut into two equal parts and when (i) the body is suspended from one part (ii) the body is suspended from both the parts connected in parallel.
View answer and explanation
Correct option: A
T/2;T/2T / \sqrt{2}; T / 2
Option A (T/2;T/2T / \sqrt{2}; T / 2) is correct because cutting the spring changes the spring constant, affecting the time period of the oscillation.
87 Medium
87 of 445
Application +4 | -1
If a spring has time period TT, and is cut into nn equal parts, then the time period of each part will be-
View answer and explanation
Correct option: B
T/nT / \sqrt{n}
Option B (T/nT / \sqrt{n}) is correct because the time period is inversely proportional to the square root of the spring constant, and cutting the spring into n parts increases the spring constant of each part by a factor of n.
88 Medium
88 of 445
Application +4 | -1
A simple pendulum has time period T1{T_1}.The point of suspension is now moved upward according to the relation y=kt2,(k=1ms−2)y = k{t^2},\left( {k = 1\mathrm{m}{\mathrm{s}^{ - 2}}} \right) where y is the vertical displacement. The time period now becomes T2{T_2} The ratio of T12T22\frac{{T_1^2}}{{T_2^2}} is (g=10ms−2)\left( {\mathrm{g} = 10\mathrm{m}{\mathrm{s}^{ - 2}}} \right)
View answer and explanation
Correct option: A
6⁄5
Option A (6⁄5) is correct because the effective acceleration due to gravity changes when the suspension point moves upwards, affecting the time period.
89 Medium
89 of 445
Application +4 | -1
A person of mass 80 Kg80 \mathrm{~Kg} sits on a spring and the spring compresses by 2 m2 \mathrm{~m}. If the person himself is projected at 20 ms−120 \mathrm{~ms}^{-1} velocity by the spring then the spring constant of the spring in Nm−1\mathrm{Nm}^{-1} is
View answer and explanation
Correct option: D
8000
Option D (8000) is correct because the spring constant is calculated using the conservation of energy principle, equating the spring's potential energy to the kinetic energy of the person.
90 Medium
90 of 445
Application +4 | -1
A particle is oscillating according to the equation V(t)=4sin⁡0.4πtV(t)=4 \sin 0.4 \pi t metres. In how much time after the motion begins will that particle reach the extreme position form the mean position?
View answer and explanation
Correct option: A
1.25 s1.25 \mathrm{~s}
Option A (1.25 s1.25 \mathrm{~s}) is correct because it is the time taken to reach the extreme position, calculated as T/4.
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