A body of mass m, is attached to a vertical rod of mass M and length L, hung from a pivoted support. A spring of constant K fixed to a support on the left as shown and is attached to the rod at a distance from the pivot. The frequency of the oscillation is:

As rod is displaced by ‘y’ towards the spring will get compressed by ‘y’ and angular shift

$\tan \theta =\frac{y}{x},\frac{1}{2}I{\omega }^2+\frac{1}{2}\text{k}{y}^2+\frac{1}{2}\text{m}{\upsilon }^2$  = constant

$\frac{1}{2}\left[\frac{{\text{ML}}^{\text{2}}}{3}+{\text{mL}}^{\text{2}}\right]{\omega }^2+\frac{1}{2}\text{k}{y}^2+\frac{1}{2}\text{m}{\upsilon }^2$  = constant

$\left(\because \text{v}=\text{L}\omega \right)$

$\frac{1}{2}\left[\frac{\text{M}}{3}+\text{m}\right]{\upsilon }^2+\frac{1}{2}\text{k}{y}^2+\frac{1}{2}\text{m}{\upsilon }^2$  = constant

Differentiating w.r.t time we get

$\frac{1}{2}\left[\frac{\text{M}}{3}+\text{m}\right]2\upsilon \text{a}+\frac{1}{2}\text{k2}y\upsilon +\frac{1}{2}\text{m2}\upsilon \text{a}=\text{0}$

$\text{a}=\frac{-\text{k}y}{\left(\frac{\text{M}}{\text{3}}+2\text{m}\right)}\left(\because \text{a}=-{\omega }^{2}y\right)$

$\omega =\sqrt{\frac{\text{k}}{\left(\frac{\text{M}}{\text{3}}+\text{2m}\right)}}\left(\because \omega =2\pi \text{f}\right)$

$\text{f}=\frac{1}{2\pi }\sqrt{\frac{\text{k}}{\left(\frac{\text{M}}{\text{3}}+\text{2m}\right)}}$