In S.H.M. the ratio of kinetic energy at mean position to the potential energy when the displacement is half of the amplitude is

Options

A. `(4)/(1)`
B. `(2)/(3)`
C. `(4)/(3)`
D. `(1)/(2)`

Answer

A

Explanation

`K.E.=(1)/(2)m omega^(2)(a^(2)-y^(2))` and `P.E.=(1)/(2)m omega^(2)y^(2)`
At mean position, `y=0:.K.E.=(1)/(2)m omega^(2)a^(2)`
At, `y=(a)/(2),P.E.=(1)/(2)m omega^(2)(a^(2))/(4)=(1)/(8)m omega^(2)a^(2)`
`:.` Ratio `=((1//2)m omega^(2)a^(2))/((1//8)m omega^(2)a^(2))=("K.E. at mean position")/("PE at "a//2)=(4)/(1)`