Starting from the origin a body oscillates simple harmonically with a period of 2s . After time `(1)/(x)` second willthe kinetic energy be `75%` of its total energy , then value of x is

Options

A. `(1)/(4)s`
B. `(1)/(2)s`
C. `(1)/(3)s`
D. `(1)/(6)s`

Answer

C

Explanation

"sin"ce the body starts from the mean position
`x=A "sin" omega t therefore v=(dx)/(dt)=A omega "cos" omega t`
`therefore (1)/(2)mv^(2)=(1)/(2)m A^(2)omega^(2)"cos"^(2)omega t`
It is given that `K.E.=(3)/(4)` (total energy)
`therefore (1)mA^(2)omega^(2)"cos"^(2)omega t=(3)/(4)((1)/(2)m omega^(2)A^(2))`
`therefore "cos"^(2) omega t=(3)/(4)`
`therefore "cos" omega t =(sqrt(3))/(2) therefore omega t =30^(@)=(pi)/(6)`
`therefore (2pi t)/(T)=(pi)/(6) therefore (2pi t)/(4)=(pi)/(6)`
`therefore t =(1)/(3) s`