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 velocity of particle 2 at $t=0$ is

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

Answer 1

The correct option is A $-\frac{\pi }{6}$

Velocity $v=\frac{dy}{dt}$

So, $v_1=\frac{d{y}_1}{dt}=10\pi cos(100\pi t+\pi /3)$

and $v_2=\frac{d{y}_2}{dt}=-0.1\pi sin\left(\pi t\right)=0.1\pi cos(\pi /2+\pi t)$

Now, phase difference $\delta \theta =100\pi t+\pi /3-\pi /2-\pi t$

At $t=0$, $\delta \theta =-\pi /6$