A) \[-\pi /3\]
B) \[\pi /6\]
C) \[-\pi /6\]
D) \[\pi /3\]
Correct Answer: C
Solution :
\[{{V}_{1}}\,=\frac{d{{y}_{1}}}{dt}=0.1\times 100\pi \cos \,(100\pi t+\pi /3)\] \[=10\pi \sin \,\left( 100\pi t+\frac{\pi }{2}\,+\frac{\pi }{3} \right)\] \[{{V}_{2}}=\frac{d{{y}_{2}}}{dt}\] \[=-0.1\times 100\pi \sin (100\pi t)\,=10\pi \sin (100\pi t+\pi )\] Argument of sign function is called phase. Required phase difference = \[\left( 100\pi t+\frac{\pi }{2}\,+\frac{\pi }{3} \right)\,-(100\pi t+\pi )\,=-\frac{\pi }{6}\]
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