Question

Two particles execute SHM of same amplitude of $20cm$ with same period along the same line about the same equilibrium position. The maximum distance between the two is $20cm$. Their phase difference in radians is

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

Answer 1

Answer: (C)

$\frac{\pi }{3}$

Let the equations of motions be
$x=A\sin wt\&x=A\sin (wt+\varphi )$
with usual notations, then,
$\Delta x=A\sin (wt+\varphi )-A\sin \left(wt\right)$
$\therefore \Delta x=2A\sin \left(\varphi /2\right)\cos \left(\right(2wt+\varphi \left)/2\right)$
since $\sin \left(\varphi /2\right)$ is constant, so maximum value of above is when $\cos \left(\right(2wt+\varphi \left)/2\right)=1$
$\therefore$ maximum distance$=2A\sin \left(\varphi /2\right)=20$
since $A=20$,
$sin\left(\varphi /2\right)=1/2$
$\therefore \varphi /2=\pi /6or\varphi /2=5\pi /6$
$\therefore \varphi =\pi /3or\varphi =5\pi /3=2\pi -\pi /3$
$\therefore$ second particle is either lagging or leading by $\pi /3$