We have $x = a\sin \left( {\omega t + {\pi \over 6}} \right)$

$\therefore$ Velocity, $v = {{dx} \over {dt}} = a\omega \cos \left( {\omega t + {\pi \over 6}} \right)$

Maximum velocity = a$\omega$
According to question,

${{a\omega } \over 2} = a\omega \cos \left( {\omega t + {\pi \over 6}} \right)$

$\Rightarrow \cos \left( {\omega t + {\pi \over 6}} \right) = {1 \over 2} = \cos {60^o}$ or ${\cos {\pi \over 3}}$

$\Rightarrow \omega t = {\pi \over 3} - {\pi \over 6} \Rightarrow \omega t = {\pi \over 6}$

$\Rightarrow {{2\pi } \over T}.t = {\pi \over 6} \Rightarrow t = {T \over {12}}$