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Question: A simple pendulum with a bob of mass ‘m’ oscillates from A to C and back to A such that PB is H. If ...

A simple pendulum with a bob of mass ‘m’ oscillates from A to C and back to A such 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

mgHmgH

B

2gH\sqrt{2gH}

C

Tp−1pT\frac{p - 1}{p}

D

Zero

Answer

2gH\sqrt{2gH}

Explanation

Solution

At B, the velocity is maximum using conservation of mechanical energy

ΔPE=ΔKE\Delta PE = \Delta KE ⇒ mgH=12mv2mgH = \frac{1}{2}mv^{2} ⇒ v=2gHv = \sqrt{2gH}