b
since $V_{0}$ is the maximum speed of the ball in its equilibrium position. $\therefore V_{0}=\omega_{0} A$
$A=\frac{V_{0}}{\omega_{0}}$
So equation of motion becomes $\frac{V_{0}}{\omega_{0}}=A \sin \omega_{0} t$ as there was no initial phase in the $SHM.$
There was no initial phase in the $SHM.$
Hence $B$ is correct.