The K.E. of a particle in S.H.M. 0.2 s after passing the mean position is half of its total energy. Then its period of oscillationis

Options

A. 0.8 s
B. 1.6 s
C. 0.4 s
D. 1.2 s

Answer

B

Explanation

`KE=(1)/(2)T.E.atx=(A)/(sqrt(2))`
`x=Asinomegat therefore(A)/(sqrt(2))=Asinomegat`
`sinomegat=(1)/(sqrt(2))therefore omegat=sin^(-1)((1)/(sqrt(2)))=(pi)/(4)`
`(2pi)/(T)t=(pi)/(4)`
`therefore T=8xx0.2=1.6s`