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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))/(l)/(2L)`

B

`(MgA)/(2L)`

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
Potential energy `=1/2Momega^(2)A^(2)`
`=1/2M.(g)/(L).A^2 (therefore omega=sqrt((g)/(l)))`
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