A simple pendulum of mass m executes S H M with total energy E. If at an instant it is at one of extreme positions, then its linear momentum after a phase shift of $\frac{\pi }{3}$ rad will be

$$\begin{gathered} \frac{1}{2}m{v}_{\max }^2=E \text{ or } \frac{1}{2}m(\omega A{)}^2=E \\ \text{or } \omega A=\sqrt{\frac{2E}{m}} \end{gathered}$$

Since, the body starts from extreme position where v=0. Hence, we can write

$v=v_{\max }\sin \omega t=\omega A\sin \omega t$

When $\omega t=\frac{\pi }{3},v=\frac{\sqrt{3}\omega A}{2}$

Momentum $p=mv=\frac{m\omega A}{2}\left(\sqrt{3}\right)$

$$\begin{gathered} =\frac{m\sqrt{2E/m}}{2}\left(\sqrt{3}\right) \\ =\sqrt{\frac{mE}{2}}\left(\sqrt{3}\right)=\sqrt{\frac{3mE}{2}} \end{gathered}$$