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Simple Pendulum

A pendulum bo...

Question

A pendulum bob of mass m is held in the horizontal position and then released show that the velocity of bob at lowest position is $\sqrt{2 g l}$ .

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Solution

Loss in Gravitational potential energy = Gain in Kinetic energy
$= m g (l) = \frac{1}{2} m v^{2}$
$V = \sqrt{2 g l}$

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