A) \[{{y}_{1}}=\frac{1}{2}\sin \,\omega t+\frac{\sqrt{3}}{2}\cos \,\omega t\] \[{{y}_{2}}=\sin \,\omega t+\,\omega t,\,\text{is}\] \[\frac{7\pi }{12}\]
B) \[\frac{\pi }{12}\]
C) \[\frac{-\pi }{6}\]
D) \[\frac{\pi }{6}\]
Correct Answer: B
Solution :
\[{{y}_{1}}=\frac{1}{2}\sin \omega t+\frac{\sqrt{3}}{2}\cos \omega t\] \[=\cos \frac{\pi }{3}\sin \omega t+\sin \frac{\pi }{3}\cos \,\omega t\] \[\therefore \] \[{{y}_{1}}=\sin (\omega t+\pi /3)\] \[{{y}_{2}}=\sqrt{2}\left( \frac{1}{\sqrt{2}}\sin \omega t+\frac{1}{\sqrt{2}}\cos \omega t \right)\] Similarly, \[{{y}_{2}}=\sqrt{2}\sin (\omega t+\pi /4)\] Phase difference\[=\Delta \phi \text{=}\frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}\]
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