ANSWER:

Spring force acting at the end gives restoring torque

$\tau=r F \sin \theta$

$\Rightarrow \tau_0=-\left(k x \frac{L}{2} \cos \theta\right) \times 2$

Value of displacement of spring,

$x=\frac{L}{2} \sin \theta$

For the value of angular displacement tends to zero,

$\theta \rightarrow 0$

$\Rightarrow \cos \theta \cong 1, \sin \theta \cong \theta$

Then torque,

$\Rightarrow \tau=-k \frac{L}{2} \times \frac{L}{2} \theta=-\frac{k L^2}{2} \theta \ldots$

From equation $(i)$,

$\Rightarrow \tau=1 a=\frac{k L^2}{2} \theta$

$\Rightarrow \frac{M L^2}{12} a=-\frac{k L^2}{2} \theta$

$\Rightarrow a=-\frac{6 k}{M} \theta=-\omega^2 \theta$

$\Rightarrow \omega=\sqrt{\frac{6 k}{M}}$

Time period,

$T =\frac{2 \pi}{\omega}$

$\Rightarrow T=2 \pi \sqrt{\frac{M}{6 k}}$

Frequency,

$f=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{\overline{M k}}{M}}$

Final answer: $(c)$