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

B

Explanation

`omega = (2pi)/(1.2) = (5pi)/(3)`
`x = A sin omega t`
`v = A omega cos omega t, (A omega)/(2) = A omega cos t`
`rArr cos omega t = (1)/(2) rArr omega t = 60^(@)`
`t = ((pi)/(3))/(omega) = ((pi)/(3))/((5pi)/(3)) = 0.2 s`