Question

The bob of simple pendulum having length $l$, is displaced from mean position to an angular position $q$ with respect to vertical. If it is released, then velocity of bob at equilibrium position

• A. $ \sqrt{ 2 gl ( 1 - \cos \, \theta)} $
• B. $ \sqrt{ 2 gl ( 1 + \cos \, \theta)} $
• C. $ \sqrt{ 2 gl \cos \, \theta} $
• D. $ \sqrt { 2gl}$

Answer 1

Answer: (A)

In $\triangle O A C, \cos \theta=A / l$
or, $O A=l \cos \theta$
$\therefore A B=l(1-\cos \theta)=h$
At point, $C$ the velocity of bob $=0$.
The vertical acceleration $=g$
$\therefore v^{2}=2 g h$
or, $ v=\sqrt{2 g l(1-\cos \theta)}$