Displacement from the mean position

$y = a\sin \left( {{{2\pi } \over T}} \right)t$

According to problem y = a/2

$a/2 = a\sin \left( {{{2\pi } \over T}} \right)t$

$\Rightarrow {\pi \over 6} = \left( {{{2\pi } \over T}} \right)t \Rightarrow t = T/12$

This is the minimum time taken by the particle to travel half of the amplitude from the equilibrium position.