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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 :
A
$\sqrt{2gl(1 - \cos\theta)}$
B
$\sqrt{2gl(1 + \cos\theta)}$
C
$\sqrt{2gl\cos\theta}$
D
$\sqrt{2gl}$
Detailed Solution
At angle $\theta$, the bob is at height $h = l - l\cos\theta = l(1 - \cos\theta)$ above the lowest point.
[add image: pendulum of length l displaced by angle $\theta$; vertical drop $l(1 - \cos\theta)$]
Potential energy at the extreme position = kinetic energy at the mean position:
$mgl(1 - \cos\theta) = \dfrac{1}{2}mv^2$
$v = \sqrt{2gl(1 - \cos\theta)}$
[add image: pendulum of length l displaced by angle $\theta$; vertical drop $l(1 - \cos\theta)$]
Potential energy at the extreme position = kinetic energy at the mean position:
$mgl(1 - \cos\theta) = \dfrac{1}{2}mv^2$
$v = \sqrt{2gl(1 - \cos\theta)}$