Question

A particle perform oscillatory motion with amplitude $4cm$. If the time period of particle is $1s$, then time taken by the particle to reach $2cm$ from the mean position is given by

• A. $(1/2)s$
• B. $(1/4)s$
• C. $(1/12)s$
• D. $(1/6)s$

Answer 1

Answer: (C)

Given, amplitude, $a=4cm$
Time period, $T=1s$,
Displacement, $y=2cm$
Since, particle perform oscillatory motion, hence its displacement equation is
$y=a\sin \omega t$
Substituting the given values in the above equation, we get
$2=4\sin \omega t$
$\frac{1}{2}=\sin \omega t$
$\sin \frac{\pi }{6}=\sin \omega t$
$⟹$
$\omega t=\frac{\pi }{6},\frac{2\pi }{T}\cdot t=\frac{\pi }{6}\left(\because \omega =\frac{2\pi }{T}\right)$
$t=\frac{T}{12}=\frac{1}{12}s$