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Physics Question on Oscillations

A simple pendulum with a bob of mass m oscillates from A to C and back to A such B that PB is H. If the acceleration due to gravity is g, then the velocity of the bob as it passes through B is

A

mgH

B

2gH\sqrt{2gH}

C

zero

D

2gH

Answer

2gH\sqrt{2gH}

Explanation

Solution

Potential energy at A (or C) = Kinetic energy at B. Thus 12mvB2=mgH \frac{1}{2} mv_B^2 = mg H or vb=2gHv_b = \sqrt{ 2 g H}