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 centre 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:
Answer: (C) $\dfrac{1}{2\pi}\sqrt{\dfrac{6k}{m}}$
For a small rotation $\theta$, each end moves $\dfrac l2\theta$ along the spring, so each spring exerts $k\dfrac l2\theta$ with lever arm $\dfrac l2$. Both torques oppose the rotation:
$$\tau = -2k\left(\frac l2\right)^2\theta = -\frac{kl^2}{2}\theta$$
With $I = \dfrac{ml^2}{12}$:
$$\omega^2 = \frac{kl^2/2}{ml^2/12} = \frac{6k}{m} \Rightarrow f = \frac{1}{2\pi}\sqrt{\frac{6k}{m}}$$
Solution by Sreeraj P, M.Sc Physics