Question

Two particles are executing SHM of same amplitude and frequency along the same straight line path. They pass each other when going in opposite directions, each time their displacement is half of their magnitude. What is the phase difference between them?

• A. $\pi /6$
• B. $4\pi /6$
• C. $\pi /3$
• D. $2\pi /6$

Answer 1

The correct option is B $4\pi /6$

$\begin{array}{c}Wehave\\ x=A\sin \left(\omega t+\delta \right);where\delta isthephasecons\tan t.\\ Forparticle1\\ atX=Aandt=0\\ \frac{A}{2}=A\sin \delta \\ \Rightarrow \frac{1}{2}=\sin \delta \\ \Rightarrow \delta =\frac{\pi }{6},\frac{5\pi }{6}\\ Thevelocityis\\ v=\frac{dx}{dt}=A\omega \cos \left(\omega t+\delta \right)\\ Att=0,v=A\omega t\cos \delta ]Now,\\ \cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}and\cos \left(\frac{5\pi }{6}\right)=-\frac{\sqrt{3}}{2}\\ Asvispositiveforparticle1,\delta mustbeequalto\frac{\pi }{6}\\ for,particle2,Visnegative,\delta mustbeequalto\frac{5\pi }{6}\\ Thus,thephasedifferencebetweenthemis=\frac{5\pi }{6}-\frac{\pi }{6}\\ =\frac{4\pi }{6}\\ Hence,theoptionBisthecorrectanswer.\end{array}$