A particle executing SHM of amplitude 'a' has a displacement a/2 at t = T/4 and a negative velocity. The epoch of the particle is

$\text{ Equation of SHM }=y\sin (\omega t+\varphi )$

$\text{ When }y=\frac{a}{2},t=\frac{T}{4}=\frac{2\pi }{4\omega }=\frac{\pi }{2\omega }$

$v=a\omega \cos (\omega t+\varphi ) \mathrm{velocity} \mathrm{is} \mathrm{negative}$

$\begin{array}{l}\frac{\pi }{2}<(\omega t+\varphi )<\frac{3\pi }{2}\\ \frac{a}{2}=a\sin (\omega t+\varphi )\Rightarrow \sin (\omega t+\varphi )=\frac{1}{2}\\ \left(\frac{\pi }{2}+\varphi \right)=\frac{5\pi }{6}\end{array}$

$\text{ Substituting in the above equation, we get }\varphi =\pi /3$