Physics · Simple Harmonic Motion

JEE Advanced 2025 — Paper 2 — Question 3

As shown in the figures, a uniform rod $O O^{\prime}$ of length $l$ is hinged at the point $O$ and held in place vertically between two walls using two massless springs of same spring constant. The springs are connected at the midpoint and at the top-end $\left(O^{\prime}\right)$ of the rod, as shown in Fig. 1 and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $f_{1}$ . On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2 and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is $f_{2}$ . Ignoring gravity and assuming motion only in the plane of the diagram, the value of $\frac{f_{1}}{f_{2}}$ is

Option A:
2
Option B:
$\sqrt{2}$
Option C:
$\sqrt{\frac{5}{2}}$
Correct
Option D:
$\sqrt{\frac{2}{5}}$

Answer: C

Step-by-step solution

$\frac{\mathrm{ML}^{2}}{3} \ddot{\theta}+\mathrm{K} \cdot \frac{\ell}{2} \theta \cdot \frac{\ell}{2}+\mathrm{K} \cdot \ell \cdot \theta \cdot \ell=0$

$\ddot{\theta}+\frac{15}{4} \frac{\mathrm{k}}{\mathrm{M}} \theta=0$

$\ddot{\theta}=-\left(\frac{15}{4} \frac{\mathrm{k}}{\mathrm{M}}\right) \theta$

$\omega_{1}=\sqrt{\frac{15 \mathrm{~K}}{4 \mathrm{M}}}$

$\frac{1}{3} \mathrm{ML}^{2} \ddot{\theta}+2 \mathrm{~K} \frac{\mathrm{~L}}{2} \theta \cdot \frac{\mathrm{~L}}{2}=0$

$\ddot{\theta}+\frac{3}{2} \frac{\mathrm{~K}}{\mathrm{M}} \theta=0$

$\ddot{\theta}=-\frac{3}{2} \frac{K}{M} \theta$

$\omega_{2}=\sqrt{\frac{3}{2} \frac{K}{M}}$

$\frac{\omega_{1}}{\omega_{2}}=\sqrt{\frac{15}{4} \times \frac{2}{3}}=\sqrt{\frac{5}{2}}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2025
Paper: Paper 2
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Miscellaneous Problems in SHM