A) \[\frac{\pi }{3}\]
B) \[\frac{4\pi }{3}\]
C) \[\frac{3\pi }{3}\]
D) \[\frac{5\pi }{6}\]
Correct Answer: D
Solution :
\[{{y}_{1}}=2a\,\sin \,\left( \omega t+\frac{\pi }{6} \right)=-2a\,\cos \,\left( \frac{\pi }{2}+\omega t+\frac{\pi }{6} \right)\] \[\Rightarrow \] \[{{\phi }_{1}}=\frac{\pi }{2}+\omega t+\frac{\pi }{6}\] and \[{{y}_{2}}=-2a\,\cos \,\left( \omega t-\frac{\pi }{6} \right)\] \[\Rightarrow \] \[{{\phi }_{2}}=\omega t-\frac{\pi }{6}\] \[\therefore \] Phase difference \[={{\phi }_{1}}-{{\phi }_{2}}\] \[=\frac{\pi }{2}+\omega t+\frac{\pi }{6}-\left( \omega t-\frac{\pi }{6} \right)\] \[=\frac{\pi }{2}+\frac{\pi }{3}=\frac{5\pi }{6}\]
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