Question

A particle is executing SHM 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

The correct option is D $\frac{7T}{12}$

Let the whole distance which is covered by the particle $=4A$

$\frac{5}{8}$ oscillations means, it has already completed $\frac{1}{2}$ oscillation

Therefore total distance will be $2A$

also the half way is $\frac{A}{2}$

$\frac{A}{2}=Asin\omega t$

$\omega =\frac{2\pi }{T}$

putting the value of $\omega$

$t=\frac{T}{12}$

Therefore

Total time taken =Time to complete previous one half $+$ Time to complete $\frac{A}{2}$

$=\frac{t}{2}+\frac{t}{12}=\frac{7T}{12}$