Question

Two simple harmonic motions are represented by the equations
$y_1=0.1\sin (100\pi t+\frac{\pi }{3})$ 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)

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

On differentiating the equations given, we know that there will be the phase change in respective equations of $-90$.Now in first equation phase is -$90+60=-30$ and in other case phase will be 0, therefore phase difference is $-30$,which is nothing but option C.

Phase difference ($\varphi$) $=99\pi t+\pi /3-\pi /2$

at $t=0$ ; $\varphi$$=-\pi /6$