Question

A particle perform oscillatory motion with amplitude $4 cm$. If the time period of particle is $1 s$, then time taken by the particle to reach $2 cm$ 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, $a=4 cm$
Time period, $T=1 s$,
Displacement, $y=2 cm$
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 $
$\sin \frac{\pi}{6} =\sin \omega t$
$\Longrightarrow$
$\omega t =\frac{\pi}{6}, \frac{2 \pi}{T} \cdot t=\frac{\pi}{6} \left(\because \omega=\frac{2 \pi}{T}\right)$
$t =\frac{T}{12}=\frac{1}{12} s$