Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length l and mass m. The rod is pivoted at its center 'O' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is:

For moment of inertia of uniform rod about its center O: $$I = \frac{m l^2}{12}$$

For restoring torque after small angular displacement $$\theta$$:

$$\text{Elongation/compression of each spring at the ends: } x = \frac{l}{2}\theta$$

$$\text{Restoring force from each spring: } F = kx = k\frac{l}{2}\theta$$

$$\text{Net restoring torque about O: } \tau = -2 \times \left(F \times \frac{l}{2}\right) = -2 \times \left(k\frac{l}{2}\theta \times \frac{l}{2}\right) = -\frac{k l^2}{2}\theta$$

For frequency of oscillation: $$\omega = \sqrt{\frac{C}{I}} = \sqrt{\frac{\frac{k l^2}{2}}{\frac{m l^2}{12}}} = \sqrt{\frac{6k}{m}}$$

$$f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{6k}{m}}$$