Oscillations JEE Main previous year questions with solutions

4 solved JEE Main questions on Oscillations, free to read — no sign-in needed. The full chapter has 191 questions; sign in to attempt the remaining 187 in the exam simulator.

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 (29 Jan, Shift 1)Superposition of S.H.M's and Resonance
   Two simple harmonic waves having equal amplitudes of $8cm$ and equal frequency of $10Hz$ are moving along the same direction. The resultant amplitude is also $8cm$. The phase difference between the individual waves is _____ degree.
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   Answer: 120

   If two waves have the same amplitude, frequency, and direction, but are out of phase by $\phi$ degrees, then the amplitude of the resultant waves is given by ${A}_{R}=2A\cos \left(\dfrac{\Delta \phi }{2}\right)$ Here, ${A}_{1}={A}_{2}={A}_{R}=8cm$ So, $\cos \left(\dfrac{\Delta \phi }{2}\right)=\dfrac{1}{2}$ Or phase difference between the individual waves is $\dfrac{\Delta \phi }{2}=60^{\circ}\Rightarrow \Delta \phi =120^{\circ}$.