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 :

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

Answer 1

Answer: (A)

$\frac{-\pi}{6}$

Answer 2

Given : $y_{1}=0.1\,sin\left(100\,\pi t+\frac{\pi}{3}\right)$ $\therefore \frac{dy_{1}}{dt}=v_{1}=0.1\times100\,\pi\,cos\left(100\,\pi t+\frac{\pi}{3}\right)$ or $v_{1}=10\pi\,sin\left(100\,\pi t+\frac{\pi}{3}+\frac{\pi}{2}\right)$ or $v_{1}=10\pi \,sin\left(100\,\pi t+\frac{5\pi }{6}\right)$ and $y=0.1\,cos\,\pi t$ $\therefore \frac{dy_{2}}{dt}=v_{2}=-0.1\,sin\,\pi t=0.1\,sin \left(\pi t+\pi\right)$ Hence, phase difference $\Delta\phi=\phi_{1}-\phi_{2}$ $=\left(100\,\pi t+\frac{5\pi}{6}\right)-\left(\pi t+\pi\right)$ $=\frac{5\pi}{6}-\pi$ $=\frac{5\pi }{6}-\pi \left(at\,t=0\right)$ $=-\frac{\pi}{6}$