The time period of simple harmonic motion of mass M in the given figure is π √(α M/5 K) , where the value of α is \ \ \ \ \ \ \ ・

Physics · Simple Harmonic Motion

JEE Main 2024 — 31 January, Shift 2 — Question 56

The time period of simple harmonic motion of mass M in the given figure is $\pi \sqrt{\frac{\alpha M}{5 K}}$ , where the value of $\alpha$ is $\_\_\_\_\_\_\_$ ・

Answer: 12

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{k}_{\mathrm{eq}}=\frac{2 \mathrm{k} \cdot \mathrm{k}}{3 \mathrm{k}}+\mathrm{k}=\frac{5 \mathrm{k}}{3}$

Angular frequency of oscillation

$(\omega)=\sqrt{\frac{\mathrm{k}_{\mathrm{eq}}}{\mathrm{m}}}$

$(\omega)=\sqrt{\frac{5 k}{3 m}}$ Period of oscillation

$(\tau)=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{3 \mathrm{~m}}{5 \mathrm{k}}}$

$$=\pi \sqrt{\frac{12 \mathrm{~m}}{5 \mathrm{k}}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 31 January, Shift 2
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Miscellaneous Problems in SHM