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
$\sqrt {2gl\left( {1 - \cos \theta } \right)} $
$\sqrt {2gl\left( {1 + \cos \theta } \right)} $
$\sqrt {2gl\cos \theta } $
$\sqrt {2gl} $
Solution
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