Thank you for reporting, we will resolve it shortly
Q.
A particle is executing S.H.M. with time period $T$. Starting from mean position, time taken by it to complete $\frac{5}{8}$ oscillations, is
Oscillations
Solution:
Total distance covered by particle $=4 A$
For $\frac{5}{8}$ of oscillation means that it has completed $\frac{1}{2}$ the oscillation taking $\frac{T}{2}$ seconds. Now it has to cover $\frac{1}{8}$ oscillation more. The whole path may be divided into 8 parts of $\frac{A}{2}$ hence it has to travel $\frac{A}{2}$ distance from mean position.
$\frac{A}{2}=A \sin \omega t$
$\frac{\pi}{6}=8\, \omega t$
$t=\frac{T}{12}$
$\left[\right.$ Putting $\left.\omega=\frac{2 \pi}{T}\right]$
Total time $=\frac{T}{2}+\frac{T}{12}=\frac{7 T}{12}$