The bob of a simple pendulum having length l | vedclass

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Question

(A) Let $O$ be the point of suspension and $B$ be the equilibrium position. The bob is displaced to point $C$ at an angle $	heta$ with the vertical.I

Vedclass · Accepted Answer

$\sqrt{2 g l(1-\cos 	heta)}$

Vedclass · Answer

(A) Let $O$ be the point of suspension and $B$ be the equilibrium position. The bob is displaced to point $C$ at an angle $	heta$ with the vertical.In $	riangle OAC$,the vertical distance $OA = l \cos 	heta$.The vertical height $h$ through which the bob falls from $C$ to $B$ is $h = OB - OA = l - l \cos 	heta = l(1 - \cos 	heta)$.By the law of conservation of energy,the potential energy lost equals the kinetic energy gained at the equilibrium position $B$:$mgh = \frac{1}{2}mv^2$$v^2 = 2gh$Substituting $h = l(1 - \cos 	heta)$:$v^2 = 2gl(1 - \cos 	heta)$$v = \sqrt{2gl(1 - \cos 	heta)}$

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

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