Physics · Simple Harmonic Motion

JEE Main 2024 — 1 February, Shift 2 — Question 51

A mass $m$ is suspended from a spring of negligible mass and the system oscillates with a frequency $f_{1}$ . The frequency of oscillations if a mass 9 m is suspended from the same spring is $f_{2}$ . The value of $\frac{f_{1}}{f_{2}}$ is $\_\_\_\_\_\_\_$ .

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

$f_{1}=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$

$\mathrm{f}_{2}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{9 \mathrm{~m}}}$

$\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\sqrt{\frac{9}{1}}=\frac{3}{1}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 2
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Linear SHM and Spring-Pulley-Block Systems