A particle executing SHM of amplitude `a` has displacement `(a)/(2)` at `t=(T)/(4)` and a negative velocity. The epoch of the particle is

Options

A. `(pi)/(3)`
B. `(2pi)/(3)`
C. `pi`
D. `(5pi)/(3)`

Answer

A

Explanation

Equation of SHM`=ysin(omegat+phi)`
When `y=(a)/(2)`,`t=(T)/(4)=(2pi)/(4omega)=(pi)/(2omega)`
`v=aomegacos(omegat+phi)`, velocity is negative
`(pi)/(2)(omegat+phi)lt(3pi)/(2)`
`(a)/(2)=asin(omegat+phi)impliessin(omegat+phi)=(1)/(2)`
`((pi)/(2)+phi)=(5pi)/(6)`
Substituting in the above equation, we get `phi=(pi)/(3)`