A) \[\frac{\pi }{4}\]
B) p
C) \[\frac{\pi }{8}\]
D) \[\frac{\pi }{2}\]
Correct Answer: D
Solution :
\[{{y}_{1}}=a\sin (\omega t-kx)\] and \[{{y}_{2}}=a\cos (\omega t-kx)=a\sin \,\left( \omega \,t-kx+\frac{\pi }{2} \right)\] Hence phase difference between these two is \[\frac{\pi }{2}.\]
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