## A uniform rod of length $L$ and mass $M$ is pivoted at its center.It's 2 ends are attached to 2 springs of equal spring constants $k$.The springs are fixed to rigid supports as shown and the rod is free to oscillate in the horizontal plane.It's gently pushed through a small angle $\theta$ in one direction and released.What is the frequen cy of oscillation?

Apr 15, 2018

$\omega = \sqrt{\frac{1}{12} \frac{M}{k}}$

#### Explanation:

Assuming a small $\theta$ such that $\sin \theta \approx \theta$

$I \ddot{\theta} = - {L}^{2} k \theta$

here

$I = \frac{1}{12} M {L}^{2}$ then

$\frac{1}{12} M {L}^{2} \ddot{\theta} = {L}^{2} k \theta$

or

$\frac{1}{12} \frac{M}{k} \ddot{\theta} + \theta = 0 \Rightarrow \omega = \sqrt{\frac{1}{12} \frac{M}{k}}$