Question

Two SHMs are represented by $y_1=0.1\sin (100\pi t+\frac{\pi }{3})$ and $y_2=0.1\cos \pi t$. The phase difference of the velocity of particle 1 with respect to particle 2 is

• A. $\frac{-\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{-\pi }{3}$
• D. $\frac{+\pi }{6}$

Answer 1

Answer: (A)

$\frac{-\pi }{6}$

$y_1=0.1\sin (100\pi t+\frac{\pi }{3})\Rightarrow v_1=10\pi \cos (100\pi t+\frac{\pi }{3})y_2=0.1\cos \pi t=0.1\sin (\frac{\pi }{2}+\pi t)\Rightarrow v_2=0.1\pi \cos (\pi t+\frac{\pi }{2})\Rightarrow \Delta \varphi =\frac{\pi }{3}-\frac{\pi }{2}=\frac{-\pi }{6}$