A particle is undergoing simple harmonic motion in a straight line with time period of 3 seconds .starting from the mean position the 3/8 oscillation will be completed in how much time ?
Ans is 1.25 sec how?

Dear Student
Let us say time period of oscillation is T.
That is Time taken by 1 oscillation is T.
3/8th oscillation means 1/4th oscillation +1/8th oscillation
So
Time to complete 1/4th oscillation is T/4s.
Time to complete 1/8​th vibration from extreme position is obtained from 
$$\begin{gathered} y = a \cos \omega t \\ Here y = \frac{a}{2} \\ As in time \frac{T}{4}, particle completes dis\tan ce equal to amplitude a \\ In time T total dis\tan ce covered by particle is 4a \\ So \\ \frac{a}{2} =a \cos \omega t = a \cos \frac{2\mathrm{\pi }}{T}t \\ \frac{1}{2}= \cos \frac{2\mathrm{\pi }}{T}t \\ \cos \frac{\mathrm{\pi }}{3} =\cos \frac{2\mathrm{\pi }}{T}t \\ \frac{\mathrm{\pi }}{3} =\frac{2\mathrm{\pi }}{T}t \\ t =\frac{T}{6} \end{gathered}$$
So time to complete 3/8th oscillation =T/4 +T/6 = 5T/12
Here T =3 sec
So
So time to complete 3/8th oscillation =5T/12 = 15/12  = 1.25 s

Regards