The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle $\mathrm{\theta }$ and then released. The bob will then pass through its equilibrium position with a speed v, where v equals

If suppose bob rises up to a height h as shown then after releasing potential energy at extreme position becomes kinetic energy of mean position

$\Rightarrow \mathrm{mgh}=\frac{1}{2}{\mathrm{mv}}_{\max }^2\Rightarrow {\mathrm{v}}_{\max }=\sqrt{2\mathrm{gh}}$

Also, from figure $\mathrm{cos\theta }=\frac{\mathrm{l}-\mathrm{h}}{\mathrm{l}}$

$\Rightarrow \mathrm{h}=\mathrm{l}(1-\mathrm{cos\theta })$

So, ${\mathrm{v}}_{\max }=\sqrt{2\mathrm{gl}(1-\mathrm{cos\theta })}$