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