A) \[y=a\sin (kx+\omega \,t)\]
B) \[y=-a\cos (kx+\omega \,t)\]
C) \[y=-a\cos (kx-\omega \,t)\]
D) \[y=-a\sin (kx-\omega \,t)\]
Correct Answer: B
Solution :
Since the point \[x=0\] is a node and reflection is taking place from point \[x=0.\] This means that reflection must be taking place from the fixed end and hence the reflected ray must suffer an additional phase change of p or a path change of \[\frac{\lambda }{2}\]. So, if \[{{y}_{\text{incident}}}=a\cos (kx-\omega \,t)\] Þ \[{{y}_{\text{reflected}}}=a\cos (-kx-\omega \,t+\pi )\]\[=-a\cos (\omega \,t+kx)\]
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