Oscillations
50 JEE Physics previous year questions on Oscillations — options free on every question; 5 include the answer & explanation free, the rest unlock with PYQ Pass.
The measured value of the length of a simple pendulum is \(20 \mathrm{~cm}\) with \(2 \mathrm{~mm}\) accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is \(\mathrm{N} \%\). The value of \(\mathrm{N}\) is:
[JEE Main 2024, 31 Jan (Shift 2)]
6
$$\begin{aligned}& \mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{~g}}} \\& \mathrm{~g}=\frac{4 \pi^2 \ell}{\mathrm{~T}^2} \\& \text { Percentage error, } \frac{\Delta \mathrm{g}}{\mathrm{~g}}=\frac{\Delta \ell}{\ell}+\frac{2 \Delta \mathrm{~T}}{\mathrm{~T}} \\& =\frac{0.2}{20}+2\left(\frac{1}{40}\right)=\frac{1.2}{20}\end{aligned}$$
percentage change = \(\frac{1.2}{20}\times 100=6\%\)
If \(R\) is the radius of the earth and the acceleration due to gravity on the surface of earth is \(g =\pi^2 m / s ^2\), then the length of the second's pendulum at a height \(h =2 R\) from the surface of earth will be, :
\(\frac{1}{9} m\)
At altitude \(h=2R\) above Earth’s surface, the distance from Earth’s center is \(r=R+h=3R\).
Acceleration due to gravity there is
\[g' = g\left(\frac{R}{r}\right)^2 = g\left(\frac{R}{3R}\right)^2 = \frac{g}{9}.\]
For a seconds pendulum \(T=2\,\text{s}\) and
\[T=2\pi\sqrt{\frac{L}{g'}} \;\Rightarrow\; L=\frac{g' T^2}{4\pi^2}.\]
With \(g=\pi^2\,\text{m s}^{-2}\),
\[L=\frac{\left(\dfrac{g}{9}\right)\, (2)^2}{4\pi^2}=\frac{g}{9}\cdot\frac{4}{4\pi^2}=\frac{g}{9\pi^2}=\frac{\pi^2}{9\pi^2}=\frac{1}{9}\ \text{m}.\]
\[\boxed{\dfrac{1}{9}\ \text{m}}\]
The equation of motion of a particle is given by \(x=a\sin (50t+\pi /3)cm\). The particle will come to rest at time \({t}_{1}\) and it will have zero acceleration at time \({t}_{2}\). The \({t}_{1}\) and \({t}_{2}\) respectively are __________.
[02 April, 2026 (Shift-II)]
\(\frac{\pi }{300}s,\frac{\pi }{75}s\)
At rest (v = 0):
\(\cos (50t+\pi /3)=0\Rightarrow 50t+\pi /3=\frac{\pi }{2}\)\({t}_{1}=\frac{\pi /2-\pi /3}{50}=\frac{\pi }{300}\)
Acceleration \(\left(a=-{\omega }^{2}x\right)\) :
\(a=-2500a\sin (50t+\pi /3)\)
For zero acceleration:
\(\sin (50t+\pi /3)=0\Rightarrow 50t+\pi /3=\pi\)
\({t}_{2}=\frac{\pi -\pi /3}{50}=\frac{\pi }{75}\)
A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm . If D and d are the total distance and displacement covered by the particle in 12.5 s , then \(\frac{D}{d}\) is
25
The period T is 2 seconds, so the number of complete cycles in 12.5 seconds is:
\(n=\frac{12.5}{2}=6.25\text{ cycles. }\)
In one full cycle, the particle moves a distance of 4 A (since it goes from +A to -A and back to +A).
So, in 6.25 cycles, the total distance covered is:
\(D=6.25\times 4A=6.25\times 4\times 1=25cm\)
Since after 6 full cycles, the particle returns to its starting position, it only moves A in the last incomplete cycle.
Therefore, the displacement is d =1 cm.
\(\frac{D}{d}=\frac{25}{1}=25\)
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain.
Reason (R): Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa.
In the light of the above statements, choose the most appropriate answer from the options given below :
[JEE Main 2025, 29 Jan (Shift 1)]
Both (A) and (R) are true and (R) is the correct explanation of (A)
\(\text{ }T=2\pi \sqrt{\frac{ℓ}{g}}\\ g=\frac{{g}_{0}{R}^{2}}{(R+h{)}^{2}}\)
As h increases, g decreases, and T increases.
A simple pendulum doing small oscillations at a place R height above earth surface has time period of \({T}_{1}=4\mathrm{s}\). \({T}_{2}\) would be it's time period if it is brought to a point which is at a height 2R from earth surface. Choose the correct relation [R = radius of earth]:
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Two particles of same mass are performing SHM vertically with two different springs of spring constants \(K_1\) and \(K_2\). If amplitude of both is same. Find ratio of the maximum speed of two particles.
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A light hollow cube of side length 10 cm and mass 10 g, is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is \(y\pi \times {10}^{-2}\mathrm{s}\), where the value of \(y\) is
(Acceleration due to gravity, \(g=10\mathrm{m}/{\mathrm{s}}^{2}\), density of water \(={10}^{3}\mathrm{kg}/{\mathrm{m}}^{3}\) )
[JEE Main 2025, 23 Jan (Shift 1)]
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A simple pendulum doing small oscillations at a place R height above earth surface has time period of \({T}_{1}=4\mathrm{s}\). \({T}_{2}\) would be it's time period if it is brought to a point which is at a height 2R from earth surface. Choose the correct relation [R = radius of earth]:
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A spring stretches by 2 mm when it is loaded with a mass of 200 g. From equilibrium position the mass is further pulled down by 2 mm and released. The frequency associated with the system and maximum energy in the spring are _____ Hz and ________ J, respectively.
[JEE Main 2026, 6 Apr (Shift 2)]
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A simple pendulum doing small oscillations at a place R height above earth surface has time period of \({T}_{1}=4\mathrm{s}\). \({T}_{2}\) would be it's time period if it is brought to a point which is at a height 2R from earth surface. Choose the correct relation [R = radius of earth]:
[JEE Main 2024, 05 Apr (Shift 1)]
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A simple pendulum doing small oscillations at a place R height above earth surface has time period of \({T}_{1}=4\mathrm{s}\). \({T}_{2}\) would be it's time period if it is brought to a point which is at a height 2R from earth surface. Choose the correct relation [R = radius of earth]:
[JEE Main 2024, 05 Apr (Shift 1)]
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A particle is executing simple harmonic motion. Its amplitude is \(A\) and time period is 5 sec. The time required by it to move from \(x=A\) to \(x=\frac{A}{\sqrt{2}}\) is ______sec.
[JEE Main 2026, 6 Apr (Shift 1)]
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Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet.
Reason (R): The mass of the pendulum remains unchanged at Earth and the other planet. In the light of the above statements, choose the correct answer from the options given below :
[JEE Main 2025, 22 Jan (Shift 2)]
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Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A) : Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain.
Reason (R): Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa.
In the light of the above statements, choose the most appropriate answer from the options given below :
[JEE Main 2025, 29 Jan (Shift 1)]
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The frequency of oscillation of a mass m suspended by a spring is \({v}_{1}\). If the length of the spring is cut to half, the same mass oscillates with frequency \({v}_{2}\). The value of \({v}_{2}/{v}_{1}\) is ______.
[JEE Main 2026, 8 Apr (Shift 2)]
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Two particles of same mass are performing SHM vertically with two different springs of spring constants \(K_1\) and \(K_2\). If amplitude of both is same. Find ratio of the maximum speed of two particles.
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A spring-block system has a time period of 2 seconds. If D is the total distance traveled by the block and \(d\) is the displacement of the block in 12.5 seconds, find the ratio D/d.(Shift I - Memory Based)
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A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm . If D and d are the total distance and displacement covered by the particle in 12.5 s , then \(\frac{D}{d}\) is
[JEE Main 2025, 24 Jan (Shift 1)]
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Match List - I with List - II.
| List - I | List - II | ||
| A. | \({Sin}^{2}\omega t\) | I. | Periodic with time period \(T=\frac{\pi }{\omega }\) but not simple harmonic motion (SHM) |
| B. | \({Sin}^{3}(2\omega t)\) | II. | Periodic with time period \(T=\frac{2\pi }{\omega }\) but Not SHM |
| C. | \(Sin(\omega t)+\cos (\pi \omega t)\) | III. | Periodic with time period \(T=\frac{\pi }{\omega }\) and SHM |
| D. | \(\cos \omega t+\cos 2\omega t\) | IV. | Non-periodic |
Choose the correct answer from the options given below:
[JEE Main 2026, 5 Apr (Shift 2)]
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The measured value of the length of a simple pendulum is \(20 \mathrm{~cm}\) with \(2 \mathrm{~mm}\) accuracy. The time for 50 oscillations was measured to be 40 seconds with 1 second resolution. From these measurements, the accuracy in the measurement of acceleration due to gravity is \(\mathrm{N} \%\). The value of \(\mathrm{N}\) is:
[JEE Main 2024, 31 Jan (Shift 2)]
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A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm . If D and d are the total distance and displacement covered by the particle in 12.5 s , then \(\frac{D}{d}\) is
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Two beads, each with charge 𝑞 and mass 𝑚, are on a horizontal, frictionless, non-conducting, circular hoop of radius 𝑅. One of the beads is glued to the hoop at some point, while the other one performs small oscillations about its equilibrium position along the hoop. The square of the angular frequency of the small oscillations is given by [\({\epsilon }_{0}\) is the permittivity of free space.]
[JEE Advanced 2024]
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For a periodic motion represented by the equation \(y=\sin \omega t+\cos \omega t\). The amplitude of the motion is
[JEE Main 2023, 10 Apr (Shift 2)]
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For a body executing S.H.M. :
(1) Potential energy is always equal to its K.E.
(2) Average potential and kinetic energy over any given time interval are always equal.
(3) Sum of the kinetic and potential energy at any point of time is constant.
(4) Average K.E. in one time period is equal to average potential energy in one time period.
Choose the most appropriate option from the options given below:
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If the time period of a two meter long simple pendulum is \(2 s\), the acceleration due to gravity at the place where pendulum is executing S.H.M. is:
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Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion \( A \): A pendulum clock when taken to Mount Everest becomes fast.
Reason \( R \): The value of \( g \) (acceleration due to gravity) is less at Mount Everest than its value on the surface of earth.
In the light of the above statements, choose the most appropriate answer from the options given below
[JEE Main 2023, 24 Jan (Shift 2)]
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Assertion: At the peak of mountain, time period of pendulum increases.
Reason: Time period of pendulum increases with decrease in g.
(Shift I Memory Based)
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If the time period of a two meter long simple pendulum is \(2 s\), the acceleration due to gravity at the place where pendulum is executing S.H.M. is:
[JEE Main 2021, 25 Feb (Shift 1)]]
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The function of time representing a simple harmonic motion with a period of \(\frac{\pi}{\omega}\) is:
[JEE Main 2021, 18 Mar (Shift 2)]
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For a body executing S.H.M. :
(1) Potential energy is always equal to its K.E.
(2) Average potential and kinetic energy over any given time interval are always equal.
(3) Sum of the kinetic and potential energy at any point of time is constant.
(4) Average K.E. in one time period is equal to average potential energy in one time period.
Choose the most appropriate option from the options given below:
[JEE Main 2021, 31 Aug (Shift 2)]
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Two particles \(A\) and \(B\) of equal masses are suspended from two massless springs of spring constants \(k_{1}\) and \(k_{2}\), respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of \(A\) and \(B\) is
[JEE Main 2021, 17 Mar (Shift 2)]
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A particle starts executing simple harmonic motion (SHM) of amplitude ' \(a\) ' and total energy \(E\). At any instant, its kinetic energy is \(3 E / 4\) then its displacement ' \(y\) ' is given by:
[JEE Main 2021, 27 Jul (Shift 1)]
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A block of mass \(1 \ kg\) attached to a spring is made to oscillate with an initial amplitude of \(12 \ cm\). After 2 minutes the amplitude decreases to \(6 \ cm\). Determine the value of the damping constant for this motion.
(Take \(\ln 2=0.693\) )
[JEE Main 2021, 17 Mar (Shift 2)]
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A particle executes simple harmonic motion between x = −A and x = +A. If the time taken by the particle to go from x = 0 to A/2 is 2 s, then the time taken by the particle in going from x = A/2 to A is:
[JEE Main 2023, 25 Jan (Shift 2)]
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In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position?
[JEE Main 2021, 25 Jul (Shift 2)]
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When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is:
[JEE Main 2021, 24 Feb (Shift 2)]
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Two simple harmonic motions are represented by the equations
\({x}_{1}=5\sin \left(2\mathrm{πt}+\frac{\pi }{4}\right)\mathrm{and}{x}_{2}=5\sqrt{2}\left(\sin 2\mathrm{πt}+\cos 2\mathrm{πt}\right)\)
The ratio of the amplitudes of \(x_{1}\) and \(x_{2}\) is
[JEE Main 2021, 26 Aug (Shift 2)]
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When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is:
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Given below are two statements :
Statement-I: A second's pendulum has a time period of 1 second.
Statement-II: It takes precisely one second to move between the two extreme positions.
In the light of the above statements, choose the correct answer from the options given below:
[JEE Main 2021, 26 Feb (Shift 2)]
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A block of mass \(1 \ kg\) attached to a spring is made to oscillate with an initial amplitude of \(12 \ cm\). After 2 minutes the amplitude decreases to \(6 \ cm\). Determine the value of the damping constant for this motion.
(Take \(\ln 2=0.693\) )
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A bob of mass ' \(m\) ' suspended by a thread of length \(\text{l}\) undergoes simple harmonic oscillations with time period \(T\). If the bob is immersed in a liquid that has density \(\frac{1}{4}\) times that of the bob and the length of the thread is increased by \({\left(\frac{1}{3}\right)}^{rd}\) of the original length, then the time period of the simple harmonic oscillations will be
[JEE Main 2021, 31 Aug (Shift 2)]
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A particle executes SHM of amplitude \(A\). The distance from the mean position when its's kinetic energy becomes equal to its potential energy is :
[JEE Main 2023, 13 Apr (Shift 2)]
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A particle executes S.H.M. of amplitude A along \(x\)-axis. At \(t=0\), the position of the particle is \(x=\frac{A}{2}\) and it moves along positive \(x\)-direction. If the displacement of particle is \(x=A \sin (\omega t+\delta)\), then the value of \(\delta\) will be :
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The function of time representing a simple harmonic motion with a period of \(\frac{\pi}{\omega}\) is:
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An object of mass \(0.5 \ kg\) executing simple harmonic motion. Its amplitude is \(5 \ cm\) and time period \((T)\) is \(0.2 \ s\). What will be the potential energy of the object at an instant \(t=\frac{T}{4} s\) starting from mean position. Assume that the initial phase of the oscillation is zero.
[JEE Main 2021, 27 Jul (Shift 2)]
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If two similar spring each of spring constant \(K_1\) are joined in series, the new spring constant and time period would be changed by a factor :
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A particle executes S.H.M. of amplitude A along \(x\)-axis. At \(t=0\), the position of the particle is \(x=\frac{A}{2}\) and it moves along positive \(x\)-direction. If the displacement of particle is \(x=A \sin (\omega t+\delta)\), then the value of \(\delta\) will be :
[JEE Main 2023, 10 Apr (Shift 1)]
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The maximum potential energy of a block executing simple harmonic motion is \(25 J\). A is amplitude of oscillation. At \(A / 2\), the kinetic energy of the block is;
[JEE Main 2023, 31 Jan (Shift 1)]
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Two simple harmonic motions are represented by the equations
\({x}_{1}=5\sin \left(2\mathrm{πt}+\frac{\pi }{4}\right)\)
\({x}_{2}=5\sqrt{10}\left(\sin 2\mathrm{πt}+\cos 2\mathrm{πt}\right)\)
The ratio of the amplitude of \(x_{1}\) and \(x_{2}\) is
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