1.

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 isA. `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» Correct Answer - B
`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)`


Discussion

No Comment Found

Related InterviewSolutions