A) 0
B) \[\frac{\pi }{2}\]
C) p
D) \[\frac{\pi }{4}\]
Correct Answer: B
Solution :
\[{{y}_{1}}=a\sin \omega t,\]and \[{{y}_{2}}=b\cos \omega t=b\sin \left( \omega \,t+\frac{\pi }{2} \right)\] So phase difference \[\varphi =\pi /2\]
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