Question

A particle performs $SHM$ on x-axis with amplitude $A$ and time period $T$. The time taken by the particle to a distance $A/5$ starting from rest is:

• A. $\frac{T}{20}$
• B. $\frac{T}{2\pi }{\cos }^{-1}\left(\frac{4}{5}\right)$
• C. $\frac{T}{2\pi }{\cos }^{-1}\left(\frac{1}{5}\right)$
• D. $\frac{T}{2\pi }{\sin }^{-1}\left(\frac{4}{5}\right)$

Answer 1

Answer: (D)

$\frac{T}{2\pi }{\sin }^{-1}\left(\frac{4}{5}\right)$

Particle is starting from rest, i.e, from one of its extreme position.
As particle moves a distance $A/5$, we can represent it on a circle as shown.
$\cos \theta =\frac{4A/5}{A}=\frac{4}{5}\Rightarrow \theta ={\cos }^{-1}\left(\frac{4}{5}\right)$
$\omega t={\cos }^{-1}\left(\frac{4}{5}\right)\Rightarrow t=\frac{1}{\omega }{\cos }^{-1}\left(\frac{4}{5}\right)=\frac{T}{2\pi }{\cos }^{-1}\left(\frac{4}{5}\right)$