A) \[\frac{\pi }{4}\]
B) \[-\frac{\pi }{4}\]
C) \[\frac{3\pi }{4}\]
D) \[\frac{\pi }{2}\]
E) \[\frac{3\pi }{2}\]
Correct Answer: B
Solution :
Equations of waves \[{{y}_{1}}=\cos \,(4t-2x)\] \[=\sin \,\left( 4t-2x+\frac{\pi }{2} \right)\] and \[{{y}_{2}}=\sin \,\left( 4t-2x+\frac{\pi }{4} \right)\] Therefore, phase difference between the two waves is \[\Delta \phi =\left( 4t-2x+\frac{\pi }{4} \right)-\left( 4t-2x+\frac{\pi }{2} \right)\] \[=\frac{\pi }{4}-\frac{\pi }{2}\] \[=-\frac{\pi }{4}\]
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