The time period of oscillation of a particle that executes `SHM` is `1.2s`. The time starting from mean position at which its velocity will be half of its velocity at mean position is

Options

A. 0.1 s
B. 0.2 s
C. 0.4 s
D. 0.6 s

Answer

A

Explanation

from extreme position `y = A cos omega t`
`V = (dy)/(dt) = A omega xx - sin omega t`
`(A omega)/(2)=A omega sin omega t`
`(2pi)/(T) xx t = (pi)/(6) rArr t = (T)/(12) = 0.1 s`