A) \[a\cos (kx-\omega t+\pi )\]
B) \[a\cos (kx-\omega t+\pi )\]
C) \[a\cos \left( kx+\omega t+\frac{\pi }{2} \right)\]
D) \[a\cos \left( kx-\omega t+\frac{\pi }{2} \right)\]
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 \[\pi \]or a path change of \[\frac{\lambda }{2}.\] So, if\[{{y}_{incident}}=a\cos (kx-\omega t)\] \[\Rightarrow \]\[{{y}_{incident}}=a\cos (-kx-\omega t+\pi )\] \[=-a\cos (\omega t+kx)\] Hence equation for the other wave \[y=a\cos (kx+\omega t+\pi )\]
You need to login to perform this action.
You will be redirected in
3 sec
