Q: A uniform rod of length L and mass M is provided at the centre. Its two ends are attached to two springs of equal spring constants k . The springs are fixed to rigid supports as shown in the figure. And rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle θ in one direction and released. The frequency of oscillation is

(a) $\large \frac{1}{2\pi} \sqrt{\frac{2 k}{M}}$

(b) $\large \frac{1}{2\pi} \sqrt{\frac{k}{M}}$

(c) $\large \frac{1}{2\pi} \sqrt{\frac{6 k}{M}}$

(d) $\large \frac{1}{2\pi} \sqrt{\frac{24 k}{M}}$

Ans: (c)

Sol: The rod is gently pushed through a small angle θ

$\large \theta = \frac{x}{L/2} = \frac{2 x}{L}$

Restoring Torque $\large \tau = – (2 k x)\frac{L}{2}$

$\large I \alpha = – \frac{k}{L} \frac{L \theta}{2}$

$\large \alpha = – \frac{k L^2}{2 I} \theta$

$\large \alpha = – \frac{k L^2}{2 (M L^2/12)} \theta$

$\large \alpha = – \frac{6 k}{M} \theta $

Frequency $\large f = \frac{1}{2 \pi} \sqrt{\frac{6 k}{M}}$