Question

Two particle P and Q are executing simple harmonic motion with equal amplitude and frequency. At a distance of half the amplitude one of them is moving away from the mean position while the other is observed at a distance half the amplitude to be moving towards mean position as shown in figure. The phase difference between them is:

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{4\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is B $\frac{\pi }{3}$

Equation of SHM: $x=a\sin wt$

for $P$ $a\sin (wt+\varphi )=a/2$

$\sin \left(wt\right)1/2$

let $t=0$

$a\sin \varphi =a/2$

$\varphi =\pi /6$

for $Q:a\sin (wt+\delta )=-a/2$

$\sin \left(\delta \right)=-1/2$

$\delta =-\pi /6$

Phase difference $=\varphi -\delta$

$=\frac{\pi }{6}-(-\frac{\pi }{6})$

$\pi /3$