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JEE • PHYSICS • OSCILLATIONS

Oscillations practice · Mixed difficulty

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451 Medium
451 of 483
Understanding +4 | -1
A particle is performing a motion on the xx-axis with its position given by x=A+Bsin⁡ωt.x=A+B \sin \omega t. The amplitude of particle is
View answer and explanation
Correct option: D
BB
Option D is correct because the amplitude of the sinusoidal motion is the coefficient of the sine function, which is B.
452 Easy
452 of 483
Understanding +4 | -1
I. In SHM to find time taken in moving from one point to another point, we cannot apply the relation, Time = Distance  Speed =\frac{\text { Distance }}{\text { Speed }}
II. In SHM, speed is constant.
Which of the following statement(s) is/are correct?
View answer and explanation
Correct option: B
Only I
Option B (Only I) is correct because speed is not constant in SHM, so Time = Distance/Speed cannot be directly applied.
453 Medium
453 of 483
Application +4 | -1
A particle executes simple harmonic motion of amplitude AA. At what distance from the mean position is its kinetic energy equal to its potential energy?
View answer and explanation
Correct option: A
±0.71 A\pm 0.71 \mathrm{~A}
Option A (±0.71 A\pm 0.71 \mathrm{~A}) is correct because at this distance, the kinetic and potential energies in SHM are equal, which is A/2A/\sqrt{2}.
454 Easy
454 of 483
Analyzing +4 | -1
Statement-1: For a particle of mass 1 kg1 \mathrm{~kg}, executing S.H.M., if slope of restoring force vs displacement graph is =−1=-1, then the time period of oscillation will be 6.28 s6.28 \mathrm{~s}.
Statement-2: If 1 kg1 \mathrm{~kg} mass is replaced by 2 kg2 \mathrm{~kg} mass and rest of the information remains same as in statement-1, then the time period of oscillation will remain 6.28 s6.28 \mathrm{~s}.
View answer and explanation
Correct option: C
Statement-1 is true, Statement-2 is false
Option C (Statement-1 is true, Statement-2 is false) is correct because the time period depends on the mass and the slope of the restoring force vs displacement graph, and changing the mass while keeping the slope the same will change the time period.
455 Medium
455 of 483
Understanding +4 | -1
To show that a simple pendulum executes simple harmonic motion, it is necessary to assume that
View answer and explanation
Correct option: C
Amplitude of oscillation is small
The correct answer is Amplitude of oscillation is small because the approximation sin⁡θ≈θ\sin \theta \approx \theta is valid only for small angles, which is necessary for SHM.
456 Medium
456 of 483
Application +4 | -1
A particle having mass 10 g10 \mathrm{~g} oscillates according to the equation x=(2.0 cm)sin⁡[(100 s−1)t+π/6]x=(2.0 \mathrm{~cm}) \sin \left[\left(100 \mathrm{~s}^{-1}\right) t+\pi / 6\right]. Find the position, the velocity and the acceleration at t=0.t=0.
View answer and explanation
Correct option: A
1.0 cm,1.73 m s−1,100 m s−21.0 \mathrm{~cm}, 1.73 \mathrm{~m} \mathrm{~s}^{-1}, 100 \mathrm{~m} \mathrm{~s}^{-2}
Option A (1.0 cm, 1.73 m s⁻¹, 100 m s⁻²) is correct because it correctly calculates the position, velocity, and acceleration at t=0 using the given equation.
457 Medium
457 of 483
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.
458 Medium
458 of 483
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.
459 Easy
459 of 483
Application +4 | -1
The maximum acceleration of a particle in SHM is made two times keeping the maximum speed to be constant. It is possible when
View answer and explanation
Correct option: C
frequency is doubled while amplitude is halved
Option C (frequency is doubled while amplitude is halved) is correct because maximum acceleration is ω2A\omega^2 A and maximum speed is ωA\omega A. To double acceleration while keeping speed constant, ω\omega must double and AA must halve.
460 Medium
460 of 483
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.
461 Medium
461 of 483
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.
462 Medium
462 of 483
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.
463 Medium
463 of 483
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.
464 Medium
464 of 483
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.
465 Medium
465 of 483
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.
466 Medium
466 of 483
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.
467 Medium
467 of 483
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.
468 Medium
468 of 483
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.
469 Medium
469 of 483
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.
470 Medium
470 of 483
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.
471 Medium
471 of 483
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.
472 Medium
472 of 483
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.
473 Medium
473 of 483
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.
474 Hard
474 of 483
Analyzing +4 | -1
I. Average kinetic energy in one oscillation during SHM of a body is 14mω2A2\frac{1}{4} m \omega^2 A^2.

II. Maximum kinetic energy is 12mω2A2\frac{1}{2} m \omega^2 A^2.

Which of the following statement(s) is/are correct?
View answer and explanation
Correct option: C
Both I and II
Option C (Both I and II) is correct because both the average kinetic energy and the maximum kinetic energy formulas are correctly stated for SHM.
475 Medium
475 of 483
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.
476 Medium
476 of 483
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.
477 Medium
477 of 483
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.
478 Medium
478 of 483
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.
479 Medium
479 of 483
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.
480 Medium
480 of 483
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.
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