Question

Two bodies performing $\text{S.H.M}$. have same amplitude and frequency. Their phases at a certain instant are as shown in the figure. The phase difference between them is

• A. $\frac{11}{6}\pi$
• B. $\pi$
• C. $\frac{\pi }{3}$
• D. $\frac{3}{5}\pi$

Answer 1

Answer: (C)

$\frac{\pi }{3}$

For first body:

$sin\left({\varphi }_1\right)=\frac{0.5A}{A}=\frac{1}{2}$

$\Rightarrow {\varphi }_1=\frac{\pi }{6}$.

For second body:

$sin\left({\varphi }_2\right)=\frac{-0.5A}{A}=-\frac{1}{2}$

$\Rightarrow {\varphi }_1=-\frac{\pi }{6}$.

Therefore, the phase difference is given by,

$△\varphi ={\varphi }_1-{\varphi }_2=\frac{\pi }{6}-(-\frac{\pi }{6})=\frac{\pi }{3}$
Hence, $\left(B\right)$ is the correct answer.