Let the rod be depressed by a small amount $\mathrm{x}$ (in the figure). Both the springs are compressed by $\mathrm{x}$. When the rod is released, the restoring torque is given by

$\mathrm{\tau }=\left(\mathrm{kx}\right)\times \frac{\mathrm{l}}{2}+\left(\mathrm{kx}\right)\times \frac{1}{2}=\left(\mathrm{kx}\right)\mathrm{l}$

Now $\mathrm{tan\theta }=\frac{\mathrm{x}}{\mathrm{l}/2}=\frac{2\mathrm{x}}{\mathrm{l}}$.

Since $\mathrm{\theta }$ is small, $\mathrm{tan\theta }\simeq \mathrm{\theta }$, where $\mathrm{\theta }$ is expressed in radian.

Thus $\mathrm{\theta }=\frac{2\mathrm{x}}{\mathrm{l}}$ or $\mathrm{x}=\frac{\mathrm{\theta l}}{2}$

$\mathrm{\tau }=\mathrm{k}\left(\frac{\mathrm{\theta l}}{2}\right)\times \mathrm{l}=\frac{{\mathrm{k\theta l}}^2}{2}$



If $\mathrm{I}$ is the moment of inertia of the rod about $\mathrm{O}$, then

$\mathrm{I}\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}=-\left(\frac{{\mathrm{kl}}^2}{2}\right)\mathrm{\theta }$

Or $\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}=-\left(\frac{{\mathrm{kl}}^2}{2\mathrm{I}}\right)\mathrm{\theta }$

Since $\frac{{\mathrm{d}}^2\mathrm{\theta }}{{\mathrm{dt}}^2}\propto \left(-\mathrm{\theta }\right)$, the motion is simple harmonic whose angular frequency is given by

$\mathrm{\omega }=\sqrt{\frac{{\mathrm{kl}}^2}{2\mathrm{I}}}$

Now, $\mathrm{\omega }=\frac{2\mathrm{\pi }}{\mathrm{T}}$ and $\mathrm{I}=\frac{{\mathrm{ml}}^2}{12}$. Therefore, we have

$\frac{2\mathrm{\pi }}{\mathrm{T}}=\sqrt{\frac{{\mathrm{kl}}^2}{2}\times \frac{12}{{\mathrm{ml}}^2}}=\sqrt{\frac{6\mathrm{k}}{\mathrm{m}}}$

Or $\mathrm{T}=\mathrm{\pi }\sqrt{\frac{2\mathrm{m}}{3\mathrm{k}}},\mathrm{ }$ which is choice $\mathrm{\pi }\sqrt{\frac{2\mathrm{m}}{3\mathrm{k}}}$