Oscillations JEE Main previous year questions with solutions

5 solved JEE Main questions on Oscillations, free to read — no sign-in needed. The full chapter has 190 questions; sign in to attempt the remaining 185 in the exam simulator.

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1. Q1JEE Main 2026 (08 Apr, Shift 2)Applications of SHM
   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 ________.
   1. A.$1$
   2. B.$2$
   3. C.$\sqrt{2}$
   4. D.$\sqrt{3}$
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   Answer: (C)

   The frequency of oscillation of a spring-mass system is given by $v = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}$, where $k$ is the spring constant and $m$ is the mass. The spring constant $k$ is inversely proportional to the length of the spring $L$, so $k \propto \dfrac{1}{L}$. When the length of the spring is cut to half, the new length is $L' = \dfrac{L}{2}$. The new spring constant becomes $k' = 2k$. The new frequency of oscillation is $v_2 = \dfrac{1}{2\pi} \sqrt{\dfrac{k'}{m}} = \dfrac{1}{2\pi} \sqrt{\dfrac{2k}{m}} = \sqrt{2} v_1$. Therefore, the ratio $\dfrac{v_2}{v_1} = \sqrt{2}$. Answer: $\sqrt{2}$
2. Q2JEE Main 2025 (29 Jan, Shift 2)Simple Harmonic Motion
   Two bodies A and B of equal mass are suspended from two massless springs of spring constant $k_1$ and $k_2$, respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of $A$ to the maximum velocity of $B$ is
   1. A.$\frac{\mathrm{k}_1}{\mathrm{k}_2}$
   2. B.$\sqrt{\frac{\mathrm{k}_1}{\mathrm{k}_2}}$
   3. C.$\sqrt{\frac{\mathrm{k}_2}{\mathrm{k}_1}}$
   4. D.$\frac{\mathrm{k}_2}{\mathrm{k}_1}$
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   Answer: (B)

   $\begin{aligned} & \mathrm{V}_1=\mathrm{A}_1 \omega_1 \\ & \mathrm{~V}_2=\mathrm{A}_2 \omega_2 \\ & \mathrm{~A}_1=\mathrm{A}_2 \\ & \frac{\mathrm{~V}_1}{\mathrm{~V}_2}=\frac{\omega_1}{\omega_2}=\frac{\sqrt{\frac{\mathrm{K}_1}{\mathrm{~m}}}}{\sqrt{\frac{\mathrm{~K}_2}{\mathrm{~m}}}} \\ & \frac{\mathrm{~V}_1}{\mathrm{~V}_2}=\sqrt{\frac{\mathrm{K}_1}{\mathrm{~K}_2}}\end{aligned}$
3. Q3JEE Main 2024 (29 Jan, Shift 1)Energy of Simple Harmonic Motion
   When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is $\dfrac{x}{8}$, where $x=$_________.
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   Answer: 9

   The total energy of the harmonic oscillator is given by the formula $E=\dfrac{1}{2}k{A}^{2}...\left(1\right)$ where, $k$ is the force constant and $A$ is the amplitude. The potential energy of the oscillator at the given displacement can be written as $\begin{matrix}U & = & \dfrac{1}{2}k{\left(\dfrac{A}{3}\right)}^{2} \\ & = & \dfrac{k{A}^{2}}{2\times 9} \\ & = & \dfrac{E}{9}\end{matrix}$ Hence, its kinetic energy is given by $\begin{matrix}KE & = & E-\dfrac{E}{9} \\ & = & \dfrac{8E}{9}\end{matrix}$ Thus, the required ratio is $\begin{matrix}\dfrac{E}{KE} & = & \dfrac{E}{\dfrac{8E}{9}} \\ & = & \dfrac{9}{8}\end{matrix}$ Hence, $x=9$.
4. Q4JEE Main 2023 (30 Jan, Shift 2)Time Period and Frequency
   The velocity of a particle executing SHM varies with displacement ($x$) as $4{v}^{2}=50–{x}^{2}$. The time period of oscillations is $\dfrac{x}{7}s$. The value of $x$ is ______. [Take $\pi =\dfrac{22}{7}$]
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   Answer: 88

   Given: $4{v}^{2}=50-{x}^{2}$ $\Rightarrow v=\dfrac{1}{2}\sqrt{50-{x}^{2}}$ Comparing with standard equation of SHM, $v=\omega \sqrt{{A}^{2}-{x}^{2}}$, we get $\omega =\dfrac{1}{2}$ Now, $T=\dfrac{2\pi }{\omega }=4\pi =\dfrac{88}{7}$ Hence, required $x=88$.
5. Q5JEE Main 2021 (26 Aug, Shift 2)Errors of Measurement
   If the length of the pendulum in pendulum clock increases by $0.1$, then the error in time per day is:
   1. A.$43.2s$
   2. B.$8.64s$
   3. C.$86.4s$
   4. D.$4.32s$
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   Answer: (A)

   $T=2\pi \sqrt{\dfrac{l}{g}}$ $T\propto {l}^{\dfrac{1}{2}}$ $\dfrac{\Delta T}{T}=\dfrac{1}{2}\dfrac{\Delta l}{l}$ $\dfrac{\Delta T}{24\times 3600}=\dfrac{1}{2}\times \dfrac{0.1}{100}$ $\Delta T=\dfrac{1}{2}\times \dfrac{0.1}{100}\times 24\times 3600$ $\Delta T=43.2\text{s}$