Two identical point masses P and Q , suspended from two separate massless springs of spring constants k 1 and k 2 , respectively, oscillate vertically. If their maximum speeds are the same, the ratio ( A Q / A P ) of the amplitude A Q of mass Q to the amplitude A P of mass P is :

Physics · Simple Harmonic Motion

NEET (UG) 2025 — Question 41

Two identical point masses $P$ and $Q$ , suspended from two separate massless springs of spring constants $\mathrm{k}_{1}$ and $\mathrm{k}_{2}$ , respectively, oscillate vertically. If their maximum speeds are the same, the ratio ( $A_{Q} / A_{P}$ ) of the amplitude $A_{Q}$ of mass $Q$ to the amplitude $A_{P}$ of mass $P$ is :

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

Answer: D

Step-by-step solution

Given $m_{P}=m_{Q}$

Also $\left(V_{\max }\right)_{P}=\left(V_{\max }\right)_{Q}$

$\therefore \mathrm{A}_{P} \omega_{P}=\mathrm{A}_{Q} \omega_{Q}$

$A_{P} \sqrt{\frac{k_{1}}{m_{P}}}=A_{Q} \sqrt{\frac{k_{2}}{m_{Q}}}\left[\because \omega=\sqrt{\frac{k}{m}}\right]$

$\therefore \frac{\mathrm{A}_{\mathrm{Q}}}{\mathrm{A}_{\mathrm{P}}}=\sqrt{\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}}$

Answer key and solution verified before publishing.

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Exam: NEET (UG) 2025
Paper: 2025 paper
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Linear SHM and Spring-Pulley-Block Systems