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A simple pendulum with a bob of mass ‘m’...

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

zero

B

2gh

C

mgh

D

`sqrt(2gh)`

Text Solution

Verified by Experts

The correct Answer is:
D
By the law of conservation of energy
P.B. at A = K.E. at B.
`mgh_(A)=(1)/(2)mv_(B)^(2)`
`V_(B)=sqrt(2gh)`
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