Question

Two particles are in S.H.M. along parallel straight lines with same amplitude and time period. If they cross each other in opposite directions at the midpoint of mean and extreme positions. Phase difference between them is:

• A. ${30}^{\circ }$
• B. ${120}^{\circ }$
• C. ${150}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

Answer: (B)

${120}^{\circ }$

Since $x=A\sin (\omega t+\varphi )$
$\frac{A}{2}=A\sin (\omega t+\varphi )\Rightarrow \sin (\omega t+\varphi )=\frac{1}{2}\Rightarrow \omega t+\varphi ={30}^{\circ }or{150}^{\circ }$
So the one particle has phase of ${30}^{\circ }$ and another has ${150}^{\circ }$
So the phase difference is ${150}^{\circ }-{30}^{\circ }={120}^{\circ }$