A particle undergoes simple harmonic motion having time-period T ( Starting from the mean position). The time taken for 3/8th oscillation is-

$$\begin{gathered} \frac{3}{8}\mathrm{th} \mathrm{oscillation} \mathrm{means} \frac{2}{8}+\frac{1}{8}=\frac{1}{4}+\frac{1}{8} \\ \mathrm{starting} \mathrm{from} \mathrm{mean} \mathrm{position} \frac{1}{4} \mathrm{oscillation} \mathrm{means} \mathrm{it} \mathrm{will} \mathrm{go} \mathrm{from} \mathrm{mean} \mathrm{position} \\ \mathrm{to} \mathrm{extreme} \mathrm{position} \mathrm{and} \mathrm{from} \mathrm{extreme} \mathrm{position} \mathrm{it} \mathrm{has} \mathrm{travelled} \mathrm{half} \mathrm{the} \mathrm{amplitude} \\ \mathrm{i}.\mathrm{e} \frac{1}{8}\mathrm{th} \mathrm{oscillation}. \\ \mathrm{from} \mathrm{extreme} \mathrm{position} \mathrm{to} \mathrm{half} \mathrm{the} \mathrm{amplitude}\frac{ \mathrm{a}}{2}=\mathrm{acos\omega t} \\ \mathrm{t}=\frac{\mathrm{T}}{6} \mathrm{so} \mathrm{total} \mathrm{time} \frac{\mathrm{T}}{4}+\frac{\mathrm{T}}{6}=\frac{5\mathrm{T}}{12} \end{gathered}$$