Question

Two particles execute SHM of the same amplitude and frequency along the same straight line. They pass one another when going in opposite directions each time their displacement is half their amplitude. What is the phase difference between them?

• A. ${60}^{\circ }$
• B. ${30}^{\circ }$
• C. ${90}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is D ${120}^{\circ }$

For a particle executing SHM, we can say that
$x=A\sin (\omega t+\varphi )$
Given two particles are executing SHM, we can write that
$x_1=A\sin \left({\delta }_1\right)$ and $x_2=A\sin \left({\delta }_2\right)$
Where ${\delta }_1$ and ${\delta }_2$ are the phase of two particles.
From the data given in the question, two particles pass one another going in opposite directions when their displacements are half of their amplitudes.
$\therefore$ $x_1=x_2=\frac{A}{2}$, we get
For particle 1:
Assuming particle is moving away from mean position.
$\Rightarrow \frac{A}{2}=A\sin \left({\delta }_1\right)\Rightarrow \sin \left({\delta }_1\right)=\frac{1}{2}\Rightarrow {\delta }_1=\frac{\pi }{6}$
Similarly, For particle 2:
Assuming particle is moving towards mean position.
$\Rightarrow \frac{A}{2}=A\sin \left({\delta }_2\right)\Rightarrow \sin \left({\delta }_2\right)=\frac{1}{2}\Rightarrow {\delta }_2=\frac{5\pi }{6}$

Phase difference between the two particles is $\delta =|{\delta }_1-{\delta }_2|=\frac{5\pi }{6}-\frac{\pi }{6}=\frac{2\pi }{3}$ or ${120}^{\circ }$
Thus, option (d) is the correct answer.

Alternate solution :
Phasor diagram of the particles executing SHM is shown below.


From the diagram we can deduce that,$\cos \varphi =\frac{x}{A}$
When the particles are crossing each other at $x=\frac{A}{2}$ we get,
$\cos \varphi =\frac{\frac{A}{2}}{A}\Rightarrow \varphi ={\cos }^{-1}\frac{1}{2}\Rightarrow \varphi =\frac{\pi }{3}$
so, phase difference between the two particles is given by,
$\delta =2\varphi =\frac{2\pi }{3}$ or ${120}^{\circ }$
Thus, option (d) is the correct answer.