Physics · Simple Harmonic Motion

JEE Main 2025 — 29 January, Evening Shift — Question 26

Two bodies A and B of equal mass are suspended from two massless springs of spring constant

$\mathrm{k}_{1}$ and $\mathrm{k}_{2}$ , 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

Option A:
$\sqrt{\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}}$
Correct
Option B:
$\frac{k_{1}}{k_{2}}$
Option C:
$\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}$
Option D:
$\sqrt{\frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}}$

Answer: A

Step-by-step solution

$\mathrm{V}_{1}=\mathrm{A}_{1} \omega_{1}$

$\mathrm{V}_{2}=\mathrm{A}_{2} \omega_{2}$

$\mathrm{A}_{1}=\mathrm{A}_{2}$

$\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{\omega_{1}}{\omega_{2}}=\frac{\sqrt{\frac{\mathrm{K}_{1}}{\mathrm{~m}}}}{\sqrt{\frac{\mathrm{~K}_{2}}{\mathrm{~m}}}}$

$\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\sqrt{\frac{\mathrm{K}_{1}}{\mathrm{~K}_{2}}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Evening Shift
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Linear SHM and Spring-Pulley-Block Systems