A) \[a\sin \,(kx-\omega t)\]
B) \[-a\cos (kx-\omega t)\]
C) \[-a\cos (kx+\omega t)\]
D) \[-a\cos (kx+\omega t)\]
Correct Answer: D
Solution :
Since, the point \[x=0\] is a node and reflection is taking place from point \[x=0\] i.e,, reflection must be taking place from the fixed end. Hence,; the reflected ray must suffer an additional phase change of \[\pi \]. So, if \[{{y}_{incident}}=a\,\cos (kx-\omega t)\] \[\therefore \] \[{{y}_{reflected}}=a\cos (-kx-\omega t+\pi )\] \[=-a\cos (\omega t+kx)\]
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