The displacement of a particle of mass 1 kg in S.H.M. is x = 2 sin `(pit+phi)`m. Then variation of its PE in joule is

Options

A. `U=4pi^(2)sin^(2)(pit+phi)`
B. `U=2pi^(2)sin^(2)(pit+phi)`
C. `U=2pi^(2)cos^(2)(pit+phi)`
D. `U=4pi^(2)cos^(2)(pit+phi)`

Answer

B

Explanation

`P.E.=(1)/(2)kx^(2)=(1)/(2)momega^(2)4sin^(2)(pit+phi)`
`=pi^(2)2sin^(2)(pit+phi)=2pi^(2)sin^(2)(pit+phi)`