The initial phase of a simple harmonic oscillator is zero. At what fraction of the period, the velocity is half of its maximum value?

Options

A. 1
B. `1//2`
C. `(2//3)`
D. `1//6`

Answer

D

Explanation

`v=Aomegacosomegat`
`therefore (1)/(2)Aomega=Aomegacosomegat`
`therefore omegat=cos^(-1)((1)/(2))`
`(2pi)/(T)t=(pi)/(3)`
`therefore (t)/(T)=(1)/(6)`