Question

Two particles execute SHM of same amplitude of 20 cm with same period along the same line about the same equilibrium position.The maximum distance between the two is 20 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 $\frac{\pi }{3}$

lets say
$x_1=20\sin \omega t{x}_2=20\sin (\omega t+\varphi )$
distance between them is given by
$|x_2-x_1|=|20\sin (\omega t+\varphi )-20\sin \omega t|=20\left|(\sin (\omega t+\varphi )-\sin \omega t)\right|\Rightarrow |x_2-x_1|=40\left|\cos (\omega t+\frac{\theta }{2})\sin \left(\frac{\theta }{2}\right)\right|$
The distance between them is maximum when $\cos (\omega t+\frac{\theta }{2})=1$ i.e. $40\sin \left(\frac{\theta }{2}\right)=20$
$\Rightarrow \sin \left(\frac{\theta }{2}\right)=\frac{1}{2}\Rightarrow \frac{\theta }{2}=\frac{\pi }{6}\Rightarrow \theta =\frac{\pi }{3}$