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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 421-445 of 445 questions

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421 Medium
421 of 445
Understanding +4 | -1
(A): Simple harmonic motion is not a uniform motion.
(R): It is the projection of uniform circular motion
View answer and explanation
Correct option: B
If Both A &\& R are true but R is not correct explanation of A
Option B (If Both A &RaretruebutRisnotcorrectexplanationofA)iscorrectbecausewhilebothstatementsaretrue,thereasondoesn′texplainwhySHMisn′tuniformmotion.\& R are true but R is not correct explanation of A) is correct because while both statements are true, the reason doesn't explain why SHM isn't uniform motion.
422 Medium
422 of 445
Application +4 | -1
The length of the seconds pendulum is 1 m on
the earth. If the mass and diameter of the
planet is half that of the earth then the length
of the seconds pendulum on the planet will be
View answer and explanation
Correct option: B
2 m
Option B (2 m) is correct because the length of the seconds pendulum is proportional to the acceleration due to gravity.
423 Medium
423 of 445
Analyzing +4 | -1
A hollow sphere is filled with water through the small hole in it. It is then hung by a long thread and made to oscillate. As the water slowly flow out of the hole at the bottom, the period of oscillation will
View answer and explanation
Correct option: C
First decrease then increase
Option C (First decrease then increase) is correct because as water flows out, the center of mass initially lowers, increasing the effective length and period, then the center of mass rises, decreasing the effective length and period.
424 Medium
424 of 445
Application +4 | -1
A particle executes SHMSHM. with an amplitude AA. The ratio of its potential energy to kinetic energy at displacement y=A2y = \frac{A}{2} from mean position is
View answer and explanation
Correct option: C
13\frac{1}{3}
Option C (13\frac{1}{3}) is correct because the ratio of potential energy to kinetic energy in SHM is calculated using the displacement from the mean position.
425 Medium
425 of 445
Application +4 | -1
Two masses m1{m_1} and m2{{m_2}} are suspended together by a massless spring of constant k. When the masses are in equilibrium, m1{{m_1}} is removed without disturbing the system. Then the angular frequency of oscillation of m2{{m_2}} is
View answer and explanation
Correct option: B
k/m2\sqrt {k/{m_2}}
Option B (k/m2\sqrt {k/{m_2}}) is correct because the angular frequency of oscillation depends only on the remaining mass and the spring constant after removing m1.
426 Medium
426 of 445
Understanding +4 | -1
Assertion: The amplitude of oscillation can never be infinite.
Reason: The energy of oscillator is continuously dissipated.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of the assertion
Option A (If both assertion and reason are true and reason is the correct explanation of the assertion) is correct because energy dissipation limits the amplitude of oscillation.
427 Medium
427 of 445
Application +4 | -1
A body starting at t=0t=0 from origin and oscillates simple harmonically with a period of 4s. After what time will its kinetic energy by 75%75 \% of its total energy.
View answer and explanation
Correct option: B
1/3 s1 / 3 \mathrm{~s}
Option D (1/3 s) is correct because it is calculated using the formula for kinetic energy in SHM and the given condition.
428 Medium
428 of 445
Understanding +4 | -1
The acceleration d2x/dt2d^2 x / d t^2 of a particle varies with displacement xx as d2xdt2=−kx\frac{d^2 x}{d t^2}=-k x where kk is a constant of the motion. The time period TT of the motion is equal to
View answer and explanation
Correct option: C
2π/k2 \pi / \sqrt{k}
Option C is correct because the time period T is calculated using T=2π/kT = 2 \pi / \sqrt{k} based on the given acceleration equation.
429 Medium
429 of 445
Understanding +4 | -1
A simple pendulum oscillates in air with time period TT and amplitude AA. As the time passes:
View answer and explanation
Correct option: C
TT remains same and AA decreases
Option C (TT remains same and AA decreases) is correct because air resistance causes the amplitude to decrease while the period is largely unaffected for small damping.
430 Medium
430 of 445
Understanding +4 | -1
A pendulum clock has time period TT at normal room temperature. If TsT_s and TwT_w are the time periods of the same pendulum in summers and winters respectively, then;
View answer and explanation
Correct option: D
Ts>TT_s>T and Tw<TT_w < T
Option D (Ts>TT_s>T and Tw<TT_w < T) is correct because the time period increases with temperature due to thermal expansion and decreases with lower temperature due to thermal contraction.
431 Medium
431 of 445
Analyzing +4 | -1
Assertion: The graph between velocity and displacement for a harmonic oscillation is a straight line.
Reason: Velocity changes uniformly with displacement in simple harmonic motion.
View answer and explanation
Correct option: D
If both Assertion & Reason are false
Option D is correct because both the assertion and reason are false. The graph between velocity and displacement for a harmonic oscillation is an ellipse, not a straight line, and velocity does not change uniformly with displacement in simple harmonic motion.
432 Medium
432 of 445
Understanding +4 | -1
What is the number of degrees of freedom for an oscillating simple pendulum?
View answer and explanation
Correct option: A
One
Option A (One) is correct because the position of a simple pendulum can be described by a single variable, the angle.
433 Medium
433 of 445
Application +4 | -1
A man measures the period of simple pendulum inside a stationary lift and finds it to be T\mathrm{T} second. If the lift accelerates downwards with acceleration of g/4\mathrm{g} / 4, the period of oscillation will be-
View answer and explanation
Correct option: B
T×23 s\mathrm{T} \times \frac{2}{\sqrt{3}} \mathrm{~s}
Option B (T ×23\times \frac{2}{\sqrt{3}} s) is correct because the time period increases when the lift accelerates downwards, reducing the effective gravity.
434 Medium
434 of 445
Application +4 | -1
A pendulum clock is placed on the moon, where object weighs only one-sixth as much as on earth, how many seconds the clock tick out in an actual time of 1 minute the clock keeps good time on earth?
View answer and explanation
Correct option: B
24.5
Option B (24.5) is correct because the time the clock ticks out is inversely proportional to the square root of the gravitational acceleration.
435 Medium
435 of 445
Analyzing +4 | -1
Two pendulum oscillate with a constant phase difference of 45∘45^{\circ} and same amplitude. If the maximum velocity of one of them is vv and that of other is v+xv+x, then the value of xx will be
View answer and explanation
Correct option: A
0
The correct answer is 0 because if two pendulums oscillate with the same amplitude and a constant phase difference, their maximum velocities will be the same.
436 Medium
436 of 445
Application +4 | -1
A circular disc rotating with frequency f0=1.3f_0=1.3 rev/sec comes to a stop in 30 seconds. The approximate angular acceleration is:
View answer and explanation
Correct option: B
−0.27 rad/sec2-0.27 \mathrm{~rad} / \mathrm{sec}^2
Option B (-0.27 rad/sec²) is correct because it accurately calculates the angular acceleration using the given initial frequency and the time it takes to stop.
437 Medium
437 of 445
Remembering +4 | -1
Which one of the following is the correct relation between frequency f\mathrm{f} and angular frequency ω\omega ?
View answer and explanation
Correct option: B
ω=2πf\omega=2 \pi \mathrm{f}
Option B (ω=2πf\omega=2 \pi \mathrm{f}) is correct because the angular frequency is related to the frequency by the formula ω=2πf\omega = 2 \pi f.
438 Medium
438 of 445
Application +4 | -1
The displacement of two particles of same mass executing simple harmonic motions are x1=16sin⁡(10t+π2)x_1=16 \sin \left(10 t+\frac{\pi}{2}\right) and x2=25cos⁡(ωt)x_2=25 \cos (\omega t) respectively.
The value of ' ω\omega ' for which the energies of both the particles remain the same is
View answer and explanation
Correct option: D
6.4 unit
Option D (6.4 unit) is correct because it is the value of ω\omega that equates the energies of the two particles, derived by setting E1=E2E_1 = E_2 and solving for ω2\omega_2.
439 Medium
439 of 445
Application +4 | -1
A spring compressed by 0.2 m0.2 \mathrm{~m} develops a restoring force of 25 N25 \mathrm{~N}. A body of mass 5 kg5 \mathrm{~kg} is placed over it. Deduce:
(i) force constant of the spring
(ii) the depression of the spring under the weight of the body and
(iii) the period of oscillation, if the body is disturbed.
Take g=10 N kg−1g=10 \mathrm{~N} \mathrm{~kg}^{-1}.
View answer and explanation
Correct option: A
125 Nm−1;0.4 m;2π5 s125 \mathrm{~Nm}^{-1}; 0.4 \mathrm{~m}; \frac{2 \pi}{5} \mathrm{~s}
Option A (125 Nm−1;0.4 m;2π5 s125 \mathrm{~Nm}^{-1}; 0.4 \mathrm{~m}; \frac{2 \pi}{5} \mathrm{~s}) is correct because it correctly calculates the spring constant, depression, and period of oscillation.
440 Medium
440 of 445
Application +4 | -1
The displacement of a particle from its mean position (in mm ) is given by y=y= 0.4sin⁡(5πt+2.5π)cos⁡(5πt+2.5π)0.4 \sin (5 \pi t+2.5 \pi) \cos (5 \pi t+2.5 \pi). The motion of the particle is
View answer and explanation
Correct option: C
S.H.M. with period 0.2 s
Option C (S.H.M. with period 0.2 s) is correct because the given equation can be simplified to a standard SHM equation with a period of 0.2 s.
441 Medium
441 of 445
Understanding +4 | -1
All oscillatory motions are necessarily periodic motions but:
View answer and explanation
Correct option: A
All periodic motions are not oscillatory
Option A (All periodic motions are not oscillatory) is correct because while oscillatory motions are always periodic, periodic motions do not necessarily involve back-and-forth movement around a mean position.
442 Medium
442 of 445
Application +4 | -1
A particle is subjected to three SHMsSHMs

X1=3sin⁡ωtX_1=3 \sin \omega t

X2=4cos⁡ωtX_2=4 \cos \omega t

X3=6sin⁡(ωt+π)X_3=6 \sin (\omega t+\pi)

Where XX is in metre. The resultant amplitude of particle is
View answer and explanation
Correct option: D
5  m5\, {~m}
Option D (5  m5 \, {\text{~m}}) is correct because it is the resultant amplitude calculated by vector addition of the SHMs.
443 Medium
443 of 445
Application +4 | -1
A mass MM is suspended from a massless spring. An additional mass mm stretches the spring further by a distance xx. The combined mass will oscillate with a period
View answer and explanation
Correct option: A
2π{(M+m)xmg}2 \pi \sqrt{\left\{\frac{(M+m) x}{m g}\right\}}
Option A (2π{(M+m)xmg}2 \pi \sqrt{\{{\frac{{(M + m)x}}{{mg}}\}}}) is correct because it accurately gives the period of oscillation for the combined mass.
444 Medium
444 of 445
Analyzing +4 | -1
Two SHM are represented by equations,
y1=6cos⁡(6πt+π6),y2=3(3sin⁡3πt+cos⁡3πt){y_1} = 6\cos \left( {6\pi t + \frac{\pi }{6}} \right),{y_2} = 3\left( {\sqrt 3 \sin 3\pi t + \cos 3\pi t} \right)
View answer and explanation
Correct option: A
ratio of their amplitudes is 1
Option A (ratio of their amplitudes is 1) is correct because after simplifying the second equation, the amplitudes of both SHMs are found to be equal.
445 Medium
445 of 445
Application +4 | -1
A uniform disc of radius rr is to be suspended through a small hole made in the disc. Find the minimum possible time period of the disc for small oscillations. What should be the distance of the hole from the centre for it to have minimum time period?
View answer and explanation
Correct option: A
2πr2g,r/22 \pi \sqrt{\frac{r \sqrt{2}}{g}}, r / \sqrt{2}
Option A (2πr2g,r/22 \pi \sqrt{\frac{{r \sqrt{{2}}}}{{g}}}, r / \sqrt{{2}}) is correct because it represents the minimum time period calculated using the moment of inertia and the condition for minimum time period.
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