1.

The initial phase of a simple harmonic oscillator is zero. At what fraction of the period, the velocity is half of its maximum value?A. 1B. `1//2`C. `(2//3)`D. `1//6`

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


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