Question

A uniform rod of length $L$ and mass $M$ is pivoted 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 $\theta$ in one direction and released. The frequency of oscillation is

• A. $\frac{1}{2 \pi} \sqrt{ \frac{2k}{M}}$
• B. $\frac{1}{2 \pi} \sqrt{ \frac{k}{M}}$
• C. $\frac{1}{2 \pi} \sqrt{ \frac{6k}{M}}$
• D. $\frac{1}{2 \pi} \sqrt{ \frac{24k}{M}}$

Answer 1

Answer: (C)

$\frac{1}{2 \pi} \sqrt{ \frac{6k}{M}}$

Answer 2

$x = \frac{L}{2} \theta$
Restoring torque = $- (2kx) \frac{L}{2}$
$\propto = - \frac{ kL(L/ 2 \theta )}{I} = - \bigg[ \frac{ k L^2/2}{ ML^2 /12} \bigg] \theta = - \bigg( \frac{6 k}{M} \bigg) \theta$
$\therefore \, \, \, f =\frac{1}{2 \pi} \sqrt{ |\frac{\propto}{ \theta }|} =\frac{ 1}{2\pi} \sqrt{\frac{ 6k}{M}}$