A) \[\frac{-\pi }{3}\]
B) \[\frac{\pi }{6}\]
C) \[\frac{-\pi }{6}\]
D) \[\frac{\pi }{3}\]
Correct Answer: C
Solution :
\[{{v}_{1}}=\frac{d{{y}_{1}}}{dt}=0.1\times 100\pi \cos \left( 100\pi t+\frac{\pi }{3} \right)\] \[{{v}_{2}}=\frac{d{{y}_{2}}}{dt}=-0.1\pi \sin \pi t=0.1\pi \cos \left( \pi t+\frac{\pi }{2} \right)\] Phase difference of velocity of first particle with respect to the velocity of 2nd particle at t = 0 is \[\Delta \varphi ={{\varphi }_{1}}-{{\varphi }_{2}}=\frac{\pi }{3}-\frac{\pi }{2}=-\frac{\pi }{6}.\]
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