Question

Two simple harmonic motions are represented by the equations $y_{1}=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)$ and $y_{2}=0.1 \cos \pi t .$ The phase difference of the velocity of particle $1$ with respect to the velocity of particle $2$ is at $t=0$

• A. $\frac{-\pi}{3}$
• B. $\frac{\pi}{6}$
• C. $\frac{-\pi}{6}$
• D. $\frac{\pi}{3}$

Answer 1

Answer: (C)

$y_{1} =0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)$
$ v_{1} =\frac{d y_{1}}{d t}=0.1 \times 100 \pi \cos \left(100 \pi t+\frac{\pi}{3}\right) $
$=10 \pi \cos \left(100 \pi t+\frac{\pi}{3}\right) $
$=10 \pi \sin \left(\frac{\pi}{2}+100 \pi t+\frac{\pi}{3}\right) $
$ y_{2} =0.1 \cos \pi t $
$v_{2} =\frac{d y_{2}}{d t}=-0.1 \pi \sin \pi t $
$=0.1 \pi \sin (\pi+\pi t)$
At $ t =0$, phase difference of $ v_{1}$ w.r.t. $ v_{2}$
$=\left(\frac{\pi}{2}+\frac{\pi}{3}\right)-\pi=-\frac{\pi}{6}$