A thin uniform vertical rod of mass m and length l pivoted at point O is shown in the given figure. The combined stiffness of the springs is equal to k. The mass of the spring is negligible. The angular frequency of small oscillation is

Let the bar be rotated through a small angle $\mathrm{\theta }$.
The restoring torque of the forces mg, k₁x and k₂x about O can be given as

  $\mathrm{\tau }=-\left[\mathrm{mg}\left(\frac{\mathrm{l}}{2}\right)\sin \mathrm{\theta }+{\mathrm{k}}_1\mathrm{x}(\mathrm{lcos}\mathrm{\theta })+{\mathrm{k}}_2\mathrm{x}(\mathrm{lcos}\mathrm{\theta })\right]$

Since $\mathrm{\theta }$ is small, $\sin \mathrm{\theta }\cong \mathrm{\theta },\mathrm{x}=\mathrm{l\theta }\text{ and }\cos \mathrm{\theta }\cong 1$

Putting ${\mathrm{k}}_1+{\mathrm{k}}_2=\mathrm{k}$, we obtain

$\begin{array}{l} \mathrm{\tau }=-\left[{\mathrm{kl}}^2+\mathrm{mg}\left(\frac{\mathrm{l}}{2}\right)\right]\mathrm{\theta }\\ \text{ or }\mathrm{I\alpha }=-\left[{\mathrm{kl}}^2+\mathrm{mg}\left(\frac{\mathrm{l}}{2}\right)\right]\mathrm{\theta }\\ \Rightarrow {\mathrm{\omega }}_{\text{osc. }}=\sqrt{\left(\frac{{\mathrm{kl}}^2+\mathrm{mg}(\mathrm{l}/2)}{\left({\mathrm{ml}}^2/3\right.}\right)}\\ =\sqrt{\frac{3\mathrm{k}}{\mathrm{m}}+\frac{3\mathrm{g}}{2\mathrm{l}}}\end{array}$