Correct Answer - C
Let `T` be the time period , then `(T)/(2)=(5pi)/(64)-(pi)/(64)=(4pi)/(64)`
`T=(pi)/(8)s`
`omega=(2pi)/(T)=(2pi)/(((pi)/(8)))=16(rad)/(s)`
Also, `A=10cm` (from the graph) The equation of the sinusoidal wave can be written as `y=10sin(16t+phi)` cm where `phi` is the initial phase. From the graph, corresponding to the crest `=10cm`, when
`t=(3pi)/(64)`.
`=10cm=10sin[16((3pi)/(64))+phi]cm`
`sin[(3pi)/(4)+phi]=1` or `(3pi)/(4)+phi=(pi)/(2)impliesphi=-(pi)/(4)`
`y=10sin(16t-(pi)/(4))`