Question

Two particles execute SHM of same amplitude of $20\text{cm}$ with same period along the same line about the same equilibrium position. The maximum distance between the two is $20\text{cm}$. 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

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


Let two SHMs be
$x_1=A\sin \omega t$
$x_2=A\sin (\omega t+\varphi )$
Distance between them at any instant is
$x_2-x_1=A\sin (\omega t+\varphi )-A\sin \omega t$
$x_2-x_1=2A\sin \left(\varphi /2\right)\cos (\omega t+\varphi /2)$

As $\sin \left(\varphi /2\right)$ is constant, $x_2-x_1$ is maximum when $\cos (\omega t+\varphi /2)=1$
Given,
Maximum value of $x_2-x_1=20\text{cm}$
Amplitude of SHM $A=20\text{cm}$

$1=2\sin \left(\varphi /2\right)$
Phase difference between the particles is $\varphi =\frac{\pi }{3}.$