The PE of a simple harmonic oscillator 0.1 s after crossing the mean position is 1/4 of its total energy. Then the period of its oscillation is

Options

A. 0.2 s
B. 0.3 s
C. 0.9 s
D. 1.2 s

Answer

D

Explanation

`PE=(1)/(4)TE" at x"=(A)/(2)`
`x=Asinomegat`
`(A)/(2)=Asinomegat therefore sinomegat=(1)/(2)`
`therefore omegat=sin^(-1)((1)/(2))=(pi)/(6) therefore (2pi)/(T)t=(pi)/(6)`
`therefore T=1.2s`