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A simple pendulum with a bob of mass m o...

A simple pendulum with a bob of mass m oscillates from A to C and back to A, such that PB=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

`sqrt(2gh)`

D

mgh

Text Solution

Verified by Experts

The correct Answer is:
C
P.E. at A = K.E. at B
`therefore Mgh =(1)/(2) mv_(B)^(2) therefore v_(B)=sqrt(2gh)`
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