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

Options

A. `(MgA^(2))/(2L)`
B. `(MgA)/(2L)`
C. `(MgA^(2))/(L)`
D. `(2MgA^(2))/(L)`

Answer

A

Explanation

`P.E.=(1)/(2) M omega^(2)A^(2)`
For the simple pendulum, `T=2pi sqrt((L)/(g))`
`therefore (T)/(2pi)=sqrt((L)/(g))`
`therefore omega =(2pi)/(T)=sqrt((2)/(L))`
`therefore P.E. =(1)/(2) Momega^(2)xxA^(2)=(1)/(2)M. ((g)/(L)). A^(2)=(MgA^(2))/(2L)`