Question
The bob of simple pendulum having length $l$, is displaced from mean position to an angular position $\theta$ with respect to vertical. If it is released, then velocity of bob at lowest position
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Answer
(c)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 mgh = \frac{1}{2}mv_{\max }^2$
$ \Rightarrow mgh = \frac{1}{2}mv_{\max }^2$
$ \Rightarrow {v_{\max }} = \sqrt {2gh} $
Also, from figure $\cos \theta = \frac{{l - h}}{l}$
$ \Rightarrow h = l(1 - \cos \theta )$
So, ${v_{\max }} = \sqrt {2gl(1 - \cos \theta )} $
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