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A simple pendulum of length 'L' has ...

A simple pendulum of length 'L' has mass 'M' and it oscillates freely with amplitude energy is
(g = acceleration due to gravity)

A

`(MgA^(2))/(2L)`

B

`(MgA)/(2L)`

C

`(MgA^(2))/(L)`

D

`(2MgA^(2))/(L)`

Text Solution

Verified by Experts

The correct Answer is:
A
Potential energy of a simple pendulum is given as,
`= 1/2 Momega^(2) A^(2) = 1/2 M. (g)/(L) .A^(2) ( :' omega = sqrt((g)/(L)))`
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