Question

The displacements of two particles executing S.H.M. on the same line are given as $y_{1}=a \sin \left(\frac{\pi}{2} t+\phi\right)$ and $y_{2}=b \sin \left(\frac{2 \pi}{3} t+\phi\right)$. The phase difference between them at $t=1 s$ is

• A. $\pi$
• B. $\frac{\pi}{2}$
• C. $\frac{\pi}{4}$
• D. $\frac{\pi}{6}$

Answer 1

Answer: (D)

Phase difference between them is just difference in the angular values.
Phase difference $=\left(\frac{2 \pi}{3}+\phi\right)-\left(\frac{\pi}{2}+\phi\right)=\frac{\pi}{6}$