Question

Two particles executing SHM of the same amplitude and frequency on parallel lines side by side. They cross one another when moving in opposite directions each time their displacement is $\frac{\sqrt{3}}{2}$ times their amplitude. The phase difference between them is:

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

since $x=A\sin (\omega t+\varphi )$
$\frac{\sqrt{3}A}{2}=A\sin (\omega t+\varphi )\Rightarrow \sin (\omega t+\varphi )=\frac{\sqrt{3}}{2}\Rightarrow \omega t+\varphi ={60}^{\circ }or{120}^{\circ }$
so the one particle has phase of ${60}^{\circ }$ and another has ${120}^{\circ }$
So the phase difference is ${120}^{\circ }-{60}^{\circ }={60}^{\circ }$