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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 391-420 of 483 questions

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391 Medium
391 of 483
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.
392 Medium
392 of 483
Application +4 | -1
A block of mass 1 kg1 \mathrm{~kg} is fastened to a spring of spring constant of 100 Nm−1100 \mathrm{~Nm}^{-1}. The block is pulled to a distance x=10 cmx=10 \mathrm{~cm} from its equilibrium position (x=0 cm)(x=0 \mathrm{~cm}) on a frictionless surface from rest t=0t=0. The kinetic energy and the potential energy of the block when it is 5 cm5 \mathrm{~cm} away from the mean position is
View answer and explanation
Correct option: A
0.375 J,  0.125 J0.375 \mathrm{~J},\,\,0.125 \mathrm{~J}
Option A (0.375 J, 0.125 J) is correct because the kinetic energy and potential energy are calculated using the formulas for SHM at the given displacement.
393 Medium
393 of 483
Application +4 | -1
The displacement of a particle (in meter) from its mean position is given by the equation y=0.2(cos⁡2πt2−sin⁡2πt2)y=0.2\left(\cos ^2 \frac{\pi t}{2}-\sin ^2 \frac{\pi t}{2}\right). The motion of the above particle is
View answer and explanation
Correct option: B
simple harmonic with amplitude 0.2 m0.2 \mathrm{~m}
Option B (simple harmonic with amplitude 0.2 m0.2 \mathrm{~m}) is correct because the given equation can be simplified to a standard SHM equation with an amplitude of 0.2 m.
394 Medium
394 of 483
Application +4 | -1
A pendulum having time period equal to two seconds is called a seconds pendulum. Those used in pendulum clocks are of this type. Find the length of a seconds pendulum at a place where g=π2 m s−2g=\pi^2 \mathrm{~m} \mathrm{~s}^{-2}.
View answer and explanation
Correct option: A
1 cm
The correct answer is 1 cm because the length of a seconds pendulum is calculated using the formula T=2πlgT = 2\pi\sqrt{\frac{l}{g}} with T=2T=2 seconds and g=π2 m s−2g=\pi^2 \mathrm{~m} \mathrm{~s}^{-2}.
395 Medium
395 of 483
Understanding +4 | -1
Identify the wrong statement from the following?
View answer and explanation
Correct option: B
The effective spring constant k\mathrm{k} of springs in parallel is given by 1k=1k1+1k2+…\frac{1}{\mathrm{k}}=\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{k}_2}+\ldots
Option B (The effective spring constant k\mathrm{k} of springs in parallel is given by 1k=1k1+1k2+…\frac{1}{\mathrm{k}}=\frac{1}{\mathrm{k}_1}+\frac{1}{\mathrm{k}_2}+\ldots) is correct because springs in parallel add directly, not reciprocally.
396 Medium
396 of 483
Understanding +4 | -1
Average velocity of a particle executing SHM in one complete vibration is:
View answer and explanation
Correct option: D
Zero
Option D (Zero) is correct because in SHM, the average velocity over one complete cycle is zero as the particle returns to its starting point.
397 Medium
397 of 483
Understanding +4 | -1
When a particle executes SHM, there is always a constant ratio between its displacement and
View answer and explanation
Correct option: B
Acceleration
Option B (Acceleration) is correct because in SHM, acceleration is directly proportional to the displacement and acts in the opposite direction, resulting in a constant ratio.
398 Medium
398 of 483
Understanding +4 | -1
Statement I The total energy of a particle executing simple harmonic motion, can be negative.

Statement II Potential energy of a system can be negative.
View answer and explanation
Correct option: A
Statement I is true, Statement II is true; Statement II is the correct explanation for Statement I
Option D is correct because Statement I is false (total energy of SHM cannot be negative), and Statement II is true (potential energy can be negative).
399 Medium
399 of 483
Analyzing +4 | -1
Assertion: In a SHM, kinetic and potential energies become equal when the displacement is 1/21 / \sqrt{2} times the amplitude.
Reason: In SHM, kinetic energy is zero when potential energy is maximum.
View answer and explanation
Correct option: B
If both assertion and reason are true but reason is not correct explanation of the assertion
Option B (If both assertion and reason are true but reason is not correct explanation of the assertion) is correct because while both statements are true, the reason doesn't explain why KE and PE are equal at the specified displacement.
400 Medium
400 of 483
Understanding +4 | -1
Statement-I: During the oscillation of a simple pendulum the tension in string is greatest at the mean position.
Statement-II: At the mean position the centripetal acceleration of the bob is maximum.
View answer and explanation
Correct option: A
Statement-I &\& Statement-II are true
Option A (Statement-I &Statement−IIaretrue)iscorrectbecausethetensionismaximumatthemeanpositionduetothecombinedeffectofweightandcentripetalforce,andthecentripetalaccelerationisalsomaximumatthemeanpositionduetomaximumvelocity.\& Statement-II are true) is correct because the tension is maximum at the mean position due to the combined effect of weight and centripetal force, and the centripetal acceleration is also maximum at the mean position due to maximum velocity.
401 Medium
401 of 483
Application +4 | -1
The instantaneous displacement of a simple pendulum oscillator is given by x=Acos⁡(ωt+π4)x=A \cos \left(\omega t+\frac{\pi}{4}\right). Its speed will be maximum at time :-
View answer and explanation
Correct option: A
π4ω\frac{\pi}{4 \omega}
Option A (π4ω\frac{\pi}{4 \omega}) is correct because it represents the time at which the velocity function reaches its maximum value.
402 Medium
402 of 483
Understanding +4 | -1
A particle executes simple harmonic motion between xx =−A  and  x=+A= - A\,\,\mathrm{and}\,\,x = + A The time taken for it to go from 0to  A/2  is  T10\mathrm{ }to\;A/2\,\,is\,\,{T_1} and to go from A/2A/2 to A  is  T2A\,\,\mathrm{is}\,\,{T_2} Then
View answer and explanation
Correct option: A
T1<T2{T_1} < {T_2}
Option A (T1<T2{T_1} < {T_2}) is correct because in SHM, the particle's speed is higher near the equilibrium position (x=0) and lower near the extreme positions (x=A).
403 Medium
403 of 483
Analyzing +4 | -1
Assertion (A): The time period of a simple pendulum of infinite length is finite value.
Reason (R): The time period of a simple pendulum is directly proportional to the square root of length.
View answer and explanation
Correct option: B
If Both A &\& R are true but R is not correct explanation of A
Option B is correct because while the time period of a simple pendulum with infinite length does have a finite value, the reason provided (direct proportionality to the square root of length) is not the correct explanation.
404 Medium
404 of 483
Application +4 | -1
Motion of an oscillating liquid column in a U\mathrm{U}-tube is
View answer and explanation
Correct option: C
Simple harmonic and time period is independent of the density of the liquid
The correct answer is that the motion is simple harmonic and the time period is independent of the density of the liquid because the time period depends only on the height of the liquid column and gravity.
405 Medium
405 of 483
Understanding +4 | -1
Which of the following quantity does NOT
change due to damping of oscillations?
View answer and explanation
Correct option: C
Initial phase
Option C (Initial phase) is correct because damping affects amplitude, angular frequency, and time period, but the initial phase remains unchanged.
406 Easy
406 of 483
Understanding +4 | -1
A particle executing simple harmonic motion along YY-axis has its motion described by the equation y=Asin⁡(ωt)+By=A \sin (\omega t)+B. The amplitude of the simple harmonic motion is
View answer and explanation
Correct option: A
AA
Option A (AA) is correct because in the given equation y=Asin⁡(ωt)+By=A \sin (\omega t)+B, AA represents the amplitude of the simple harmonic motion, while BB is just a vertical shift.
407 Medium
407 of 483
Application +4 | -1
A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of the oscillation is
View answer and explanation
Correct option: D
1 Hz
Option D (1 Hz) is correct because the frequency is calculated from the maximum speed and amplitude of the SHM.
408 Medium
408 of 483
Application +4 | -1
Starting from mean position a body oscillates simple harmonically with a period of 6 s6 \mathrm{~s}. After what time, the kinetic energy will be one third of potential energy?
View answer and explanation
Correct option: A
1 s1 \mathrm{~s}
Option A (1 s1 \mathrm{~s}) is correct because the time is calculated using the relationship between kinetic and potential energy in SHM.
409 Medium
409 of 483
Application +4 | -1
At what displacement from mean position is the K.E. of a particle performing SHM of amplitude 30 cm30 \mathrm{~cm}, eight times its P.E.
View answer and explanation
Correct option: A
10 cm
Option A (10 cm) is correct because it's derived from the relationship between kinetic and potential energy in SHM.
410 Medium
410 of 483
Understanding +4 | -1
A seconds pendulum has a hollow spherical bob of mass 25×10−3kg.25 \times {10^{ - 3}}kg. It is replaced by another small solid bob of same radius but of mass 50×10−3kg50 \times {10^{ - 3}}kg. The new time period will be.
View answer and explanation
Correct option: D
2s
Option D (2s) is correct because the time period of a simple pendulum is independent of the mass of the bob.
411 Medium
411 of 483
Understanding +4 | -1
Two spring of spring constant k1k_1 and k2k_2 are joined in series. The effective spring constant of the combination is given by
View answer and explanation
Correct option: D
k1k2/(k1+k2)k_1 k_2 /\left(k_1+k_2\right)
Option D (k1k2/(k1+k2)k_1 k_2 / (k_1+k_2)) is correct because when springs are connected in series, the reciprocal of the effective spring constant is the sum of the reciprocals of the individual spring constants.
412 Medium
412 of 483
Application +4 | -1
The amplitude of a damped oscillator varies with time as A(t)=A0exp⁡(−bt/2m)A(t)={A}_0 \exp (-b t / 2 m) where b=b= 70 g/s70 \mathrm{~g} / \mathrm{s} and m=200 gm=200 \mathrm{~g}. How long does it take for the mechanical energy to drop to one - fourth of its initial value? [Take ln 2 = 0.7]
View answer and explanation
Correct option: B
4.0 s4.0 \mathrm{~s}
Option B (4.0 s4.0 \mathrm{~s}) is correct because it's calculated by finding the time it takes for the mechanical energy of a damped oscillator to drop to one-fourth of its initial value using the given parameters.
413 Medium
413 of 483
Application +4 | -1
A pendulum has time period TT in air. When
it is made to oscillate in water, it acquired
a time period T′=2TT' = \sqrt 2 T. The density of the
material pendulum bob is equal to
View answer and explanation
Correct option: B
2 gm/cc
Option B (2 gm/cc) is correct because the density is calculated using the time period relationship in air and water.
414 Medium
414 of 483
Understanding +4 | -1
A particle moving along the xx axis executes simple harmonic motion, then the force acting on it is given by
View answer and explanation
Correct option: A
−AKx- A\mathrm{ }Kx
Option A ( −AKx\ - A\mathrm{ }Kx) is correct because the force in SHM is proportional to the displacement and directed towards the equilibrium position.
415 Medium
415 of 483
Application +4 | -1
A pendulum is swinging in an elevator. Its
period will be greatest when the elevator is
View answer and explanation
Correct option: D
accelerating downwards
Option D (accelerating downwards) is correct because the effective gravitational acceleration decreases, leading to a longer period.
416 Medium
416 of 483
Understanding +4 | -1
A body having natural frequency ν\nu' is executing forced oscillations under a driving force of frequency. The system will vibrate
View answer and explanation
Correct option: A
with frequency of driving force ν\nu
Option A (with frequency of driving force ν\nu) is correct because in forced oscillations, the system vibrates with the frequency of the external driving force.
417 Medium
417 of 483
Understanding +4 | -1
If a body is executing simple harmonic motion then
View answer and explanation
Correct option: C
At equilibrium position, the total energy is in the form of kinetic energy
Option C (At equilibrium position, the total energy is in the form of kinetic energy) is correct because at the equilibrium position in SHM, the potential energy is zero, and all the energy is kinetic.
418 Medium
418 of 483
Application +4 | -1
The function ( sin⁡ωt−cos⁡ωt)\sin \omega t-\cos \omega t) represents the S.H.M having a time period TT
View answer and explanation
Correct option: B
2π/ω2 \pi / \omega
Option B (2π/ω2 \pi / \omega) is correct because the time period of a simple harmonic motion is given by T=2π/ωT = 2 \pi / \omega.
419 Medium
419 of 483
Understanding +4 | -1
A particle moving about its equilibrium position with equation y=−ax−bty=-a x-b t. Interpret the condition
View answer and explanation
Correct option: B
It can never perform the SHM
Option B (It can never perform the SHM) is correct because the given equation is not in the form of SHM, which requires acceleration to be proportional to displacement and directed towards the equilibrium position.
420 Medium
420 of 483
Understanding +4 | -1
The acceleration of particle in SHM:
View answer and explanation
Correct option: C
Acts in opposite direction of displacement
The correct answer is C (Acts in opposite direction of displacement) because in SHM, acceleration is always directed towards the mean position, opposite to the displacement.
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