Question

The simple harmonic motion of a particle is represented by the equation $y_1=0.\sin \left(100\pi t+\frac{\pi }{3}\right)$ and $y_2=0.1\cos \pi t$ The phase difference of the velocity of particle $1$ w.r.t the velocity of the particle $2$ is$:$

• A. $\frac{\pi }{3}s$
• B. $\frac{\pi }{6}s$
• C. $\frac{\pi }{4}s$
• D. $\frac{\pi }{8}s$

Answer 1

The correct option is B $\frac{\pi }{6}s$

Given equations of wave are$:$

$y_1=0.1sin(100\pi t+\pi /3)$

$y_2=0.1cos\pi t=0.1sin(\pi /2+\pi t)=0.1sin(\pi t+\pi /2)$

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

$d{y}_1/dt=v_1=0.1\times 100\pi cos(100\pi t+\pi /3)=10\pi cos(100\pi t+\pi /3)$

similarly for $2nd$ equation$,$

$d{y}_2/dt=v_2=0.1\times \pi cos(\pi t+\pi /2)=0.1\pi cos(\pi t+\pi /2)$

We know $,$

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

Similarly at $t=0$ phase of 1st particle velocity is $\pi /3$

at $t=0$, phase of velocity of $2nd$ particle is $\pi /2$

Now$,$ phase difference $=$ phase of $1st$ particle at $t=0$ phase of $2nd$ particle at $t=0$

$=\pi /3-\pi /2=-\pi /6$

Hence,

option $\left(B\right)$ is correct answer.