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)

$\sqrt{ 2 gl ( 1 - \cos \, \theta)}$

Answer 2

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