The angular amplitude of a simple pendulum is $\theta_0$. The maximum tension in its string will be

The simple pendulum at angular amplitude $\theta_0$ is shown in the figure. $T_{\max }=m g+\frac{m v^2}{I} \quad \ldots(i)$ When, bob of pendulum comes from $A$ to $B$, it covers a vertical distance $h$. $\therefore \quad \cos \theta_0=\frac{l-h}{l}$ $\Rightarrow \quad h=I(I-\cos \theta)\ldots(ii)$ Also during $A$ to $B$, potential energy of bob converts into kinetic energy $\therefore \quad m g h=\frac{1}{2} m v^2 \therefore \quad v=\sqrt{2 g h} \quad \ldots(iii)$ Thus, using Eqs. (i), (ii) and (iii) we obtain $T_{\max }=m g+\frac{2 m g}{I} I\left(1-\cos \theta_0\right)$ $=m g+2 m g\left[1-1+\frac{\theta_0{ }^2}{2}\right]$ (By using binomial theorem) $=m g\left(1+\theta_0^2\right)$