A particle executed S.H.M. starting from its mean position at `t=0`, If its velocity is `sqrt3bomega`, when it is at a distance b from the mean positoin, when `omega=(2pi)/(T)`, the time taken by the particle to move from b to the extreme position on the same side is

Options

A. `(5pi)/(6omega)`
B. `(pi)/(3omega)`
C. `(pi)/(2omega)`
D. `(pi)/(4omega)`

Answer

B

Explanation

`v^2=omega^2(A^2-b^2)=3omega^2b^2impliesA^2=4b^2`
`b=(A)/(2)=Asinomegatimpliest=(pi)/(6omega)`
Required time `t_1=(T)/(4)-t=(2pi)/(4omega)-(pi)/(6omega)=(pi)/(3omega)`