1. Set up the small rotation

Let the rod rotate by a small angle $\theta$ about its centre $O$ in the horizontal plane.

• Length of rod $= \ell$
• Mass of rod $= m$
• Each spring has spring constant $k$

For a small angular displacement $\theta$, each end of the rod moves approximately by $x = \frac{\ell}{2}\theta$ perpendicular to the original position of the rod.

Because of the geometry, one spring gets compressed and the other gets stretched by the same amount $x$.

So extension/compression of each spring is $x = \frac{\ell}{2}\theta$

2. Force due to each spring

The restoring force by each spring is $F = kx = k\left(\frac{\ell}{2}\theta\right)$

Both springs contribute restoring torque in the same rotational sense.

3. Torque due to one spring

Torque of one spring about the centre is $\tau_1 = F\cdot \frac{\ell}{2} = k\left(\frac{\ell}{2}\theta\right)\cdot \frac{\ell}{2} = \frac{k\ell^2}{4}\theta$

Since there are two springs, total restoring torque is $\tau = 2\tau_1 = 2\cdot \frac{k\ell^2}{4}\theta = \frac{k\ell^2}{2}\theta$

Restoring torque is opposite to displacement, so $\tau = -\frac{k\ell^2}{2}\theta$

Thus the effective torsional constant is $C = \frac{k\ell^2}{2}$

4. Moment of inertia of the rod about its centre

For a uniform rod of length $\ell$ about an axis through its centre and perpendicular to the rod (here vertical through $O$), $I = \frac{m\ell^2}{12}$

5. Equation of angular SHM

For rotational SHM, $I\ddot\theta + C\theta = 0$

So $\omega^2 = \frac{C}{I} = \frac{\frac{k\ell^2}{2}}{\frac{m\ell^2}{12}} = \frac{k}{2}\cdot \frac{12}{m} = \frac{6k}{m}$

Hence, $\omega = \sqrt{\frac{6k}{m}}$

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

6. Match with the options

This corresponds to:

$\boxed{\text{B: } \frac{1}{2\pi}\sqrt{\frac{6k}{m}}}$

7. Comparison with stored correct answer

Stored correct answer is B, which matches the derived answer.