A uniform stick of mass M and length L is pivoted at its centre. Its ends are tied to two springs each of force constant K. In the position shown in figure, the strings are in their natural length. When the string is displaced through a small angle $\mathrm{\theta }$ and released. The stick:

 If stick is rotated through a small angle $\mathrm{\theta }$, the spring is stretched by a distance $\frac{\mathrm{L}}{2}\mathrm{\theta }$. Therefore restoring force is spring = K($\frac{\mathrm{L}}{2}\mathrm{\theta }$). Each spring causes a torque $\mathrm{\tau }$(= Fd) =($\frac{\mathrm{KL\theta }}{2}$)$\frac{\mathrm{L}}{2}$ in the same direction. 
Therefore equation of rotational motion is
$\mathrm{\tau }$ = I$\mathrm{\alpha }$

$-2\left(\frac{\mathrm{KL\theta }}{2}\right)\frac{\mathrm{L}}{2} = \mathrm{I\alpha } \mathrm{with} \mathrm{I} = \frac{{\mathrm{ML}}^2}{12}$

$\mathrm{\alpha } = -\left(\frac{6\mathrm{K}}{\mathrm{M}}\right)\mathrm{\theta }$
Standard equation of angular S.H.M. is $\mathrm{\alpha } = -{\mathrm{\omega }}^2\mathrm{\theta }$
${\mathrm{\omega }}^2 = \frac{6\mathrm{K}}{\mathrm{M}}$
frequency f = $\frac{\mathrm{\omega }}{2\mathrm{\pi }} =\frac{1}{2\mathrm{\pi }}\sqrt{\frac{6\mathrm{K}}{\mathrm{M}}}$