A) Wave C lags behind in phase by \[\frac{\pi }{2}\] from A and. B leads by \[\frac{\pi }{2}\]
B) Wave C leads in phase by \[\pi \] from A and B lags behind by \[\pi \]
C) Wave C leads in phase by \[\frac{\pi }{2}\] from A and lags behind by \[\frac{\pi }{2}\]
D) Wave C lags behind in phase by \[\pi \] from A and B leads by \[\pi \]
Correct Answer: A
Solution :
According to the given figure, A attains mean position T/4 time earlier than C. So, C lags behind A by \[\frac{\pi }{2}\]. Also B attains the mean position \[\frac{T}{2}\] time earlier. Hence, B leads A by \[\frac{\pi }{2}\].
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