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  4. Two waves are given by y1= cos (4t-2x) and y2= sin ( 4t-2x+(π /4) ) . The phase difference between the two waves is

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Q. Two waves are given by $ {{y}_{1}}=\cos \,(4t-2x) $ and $ {{y}_{2}}=\sin \left( 4t-2x+\frac{\pi }{4} \right) $ . The phase difference between the two waves is

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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} $