Question

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

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

Answer 1

Answer: (B)

$v_{1}=\frac{d y_{1}}{d t}=0.1 \times 100 \pi \cos \left(100 \pi t+\frac{\pi}{3}\right)$
$v_{2}=\frac{d y_{2}}{d t}=-0.1 \pi \sin \pi t=0.1 \pi \cos \left(\pi t+\frac{\pi}{2}\right)$
$\therefore $ Phase diff. $=\varphi_{1}-\varphi_{2}$
$=\frac{\pi}{3}-\frac{\pi}{2}=\frac{2 \pi-3 \pi}{6}=-\frac{\pi}{6}$