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

Answer: (A)

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

$y_1=0.1sin(100\pi t+\frac{\pi }{3})$

$y_2=0.1cos\pi t=0.1sin(\frac{\pi }{2}+\pi t)=0.1sin(\pi t+\frac{\pi }{2})$

Now, for finding velocity of particle , differentiate both equations with respect to time .

$\frac{d{y}_1}{dt}=v_1=0.1\times 100\pi cos(100\pi t+\frac{\pi }{3})$

similarly for 2nd equation,

$\frac{d{y}_1}{dt}=v_2=0.1\times \pi cos(\pi t+\frac{\pi }{2})=0.1\pi cos(\pi t+\frac{\pi }{2})$

if equation $x=Asin(\omega t+\varphi )$ is given then, at $t=0$ phase of motion is $\varphi$

similarly at $t=0$ phas of 1st particle velocity is $\frac{\pi }{3}$

at $t=0$ phase of velocity of 2nd particle is $\frac{\pi }{2}$

now thw phase difference = phase of 1st particle at $t=0$ - phase of 2nd particle at$t=0$

$=\frac{\pi }{3}-\frac{\pi }{2}=-\frac{\pi }{6}$