Other SHM Systems question from JEE Main 2019 (12 Jan, Shift 1)

Answer: (b)

1. Turn the rod by a small $\theta$: each end moves $\frac{\ell}{2}\theta$, each spring gives force $k\frac{\ell}{2}\theta$ at arm $\frac{\ell}{2}$.
2. Restoring torque $\tau = 2k\left(\frac{\ell}{2}\right)^2\theta = \frac{k\ell^2}{2}\theta$; $I = \frac{m\ell^2}{12}$.
3. $\omega^2 = \frac{k\ell^2/2}{m\ell^2/12} = \frac{6k}{m}$
4. $f = \frac{1}{2\pi}\sqrt{\frac{6k}{m}}$, answer (b).