Question

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

• A. $\frac{T}{12}$
• B. $\frac{T}{6}$
• C. $\frac{5T}{12}$
• D. $\frac{7T}{12}$

Answer 1

Answer: (D)

Total distance covered by particle $=4A$
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.
$\frac{A}{2}=A\sin \omega t$
$\frac{\pi }{6}=8\omega t$
$t=\frac{T}{12}$
$[$ Putting $\omega =\frac{2\pi }{T}]$
Total time $=\frac{T}{2}+\frac{T}{12}=\frac{7T}{12}$