Question

A particle is executing SHM with time period $T$, starting from mean position. Time taken by it to complete $\frac{5}{8}\text{th}$ part of oscillation will be

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

Answer 1

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

Since, the particle is starting its motion from mean position at $t=0$ in the given SHM. Its equation is,
$x=A\sin \omega t$ ...............$\left(1\right)$
Now, number of oscillations completed $=\frac{5}{8}=\frac{1}{2}+\frac{1}{8}$
Thus, particle has completed $\frac{1}{2}$ oscillation in time $\frac{T}{2}$.
Let the $\frac{1}{8}$ oscillation be completed in $t$ time interval.


During $1$ complete oscillation, total distance covered $=4A$

So, In $\frac{1}{8}$ oscillation, distance travelled
$=\frac{4A}{8}=\frac{A}{2}$
Substituting, displacement $x=\frac{A}{2}$ in equation $\left(1\right)$,
$\frac{A}{2}=A\sin \omega t$
$\Rightarrow \sin \omega t=\frac{1}{2}$
$\Rightarrow \omega t=\frac{\pi }{6}$
$\Rightarrow t=\frac{\pi }{6\omega }=\frac{\pi }{6\left(\frac{2\pi }{T}\right)}=\frac{T}{12}$
Thus, the total time taken to complete $\frac{5}{8}$ oscillation is,
$t_{total}=\frac{T}{2}+t$
$\Rightarrow t_{total}=\frac{T}{2}+\frac{T}{12}=\frac{7T}{12}$

Why this question? It challenges your understanding of the time period and distance covered in a complete oscillation. Tips: Always try to break the given oscillation such that you can correlate it with the time period of oscillation.