Question

The bob of a simple pendulum is pulled to a side till the string makes an angle ${60}^{\circ }$ to the horizontal and released. While crossing the mean position, find the velocity of the bob.

Answer 1

using Workenergy theorem

work done by gravity = change in kinetic enery

$mg(l-lsin{60}^o)=\frac{m{v}^2}{2}$

$2gl(1-\frac{\sqrt{3}}{2})=v^2$

$v=\sqrt{0.27gl}$