A light inextensible string carrying a spring is passed over two smooth fixed pulleys as shown. If the rod is slightly displaced about the hinge from is equilibrium position, then time period is.

Let the rod is displaced by $\mathrm{\theta }$ towards left:

$\therefore$ net torque on rod is

$-k\left(\frac{l}{2}\right)\sin \theta \cdot \frac{l}{2}\cos \theta +mg\sin \theta \frac{l}{2}+k\frac{l}{2}\sin \theta l\cos \theta$

$\text{ as }\theta \text{ is small. }\sin \theta \to \theta ;\cos \theta \to 1$

$\begin{array}{l}\Rightarrow \tau =\frac{k{l}^2}{2}\theta +\frac{mgl}{2}\cdot \theta -\frac{k{l}^2}{4}\theta \\ =\left[\frac{k{l}^2}{4}+\frac{mgl}{2}\right]\theta =\left(\frac{m{l}^2}{3}\right)\alpha \\ \therefore t_1=\frac{T}{2}=\pi \sqrt{\frac{4ml}{3(kl+2mg)}}. . . . . \left(1\right)\end{array}$

When the rod is displaced right by an angle $\mathrm{\theta }$ ,then string becomes slack i.e., rod becomes physical pendulum.

$\therefore t_{22}=T=\frac{1}{2}=2\pi \sqrt{\frac{I}{mgd}}$

$\text{ where }I=\frac{m{l}^2}{3},d=l/2\text{, time period }=t_1+t_2$