A particle executes SHM with a period T. The time taken to complete 3/8^th oscillation starting from the mean position is

The time to complete 3/8th oscillation = time to complete 1/4th oscillation + time to complete 1/8th oscillation from the extreme.  The time to complete 1/4th oscillation =  $\frac{T}{4}$
The time taken to complete 1/8th oscillation from extreme position is obtained from the equation,
y = $\frac{a}{2}$  = a cos $\omega t$   

$\therefore \cos \omega \mathrm{t}=\frac{1}{2}\Rightarrow \cos \frac{2\pi }{\mathrm{T}}\mathrm{t}=\frac{1}{2}$

$t=\frac{T}{6}\mathrm{s}$
$\therefore \text{The total time}=\frac{\mathrm{T}}{4}+\frac{\mathrm{T}}{6}=\frac{5\mathrm{T}}{12}$