A) \[y=A\sin (\omega t+0.8kx)\]
B) \[y=-0.8A\sin (\omega t+kx)\]
C) \[y=A\sin (\omega t+kx)\]
D) \[y=0.8A\sin (\omega t+kx)\]
Correct Answer: B
Solution :
On getting reflected from a rigid boundary the wave suffers. Hence, if \[{{y}_{incident}}=A\sin (\omega t-kx)\] Then,\[{{y}_{reflected}}=(0.8A)\sin \{\omega t-k(-x)+\pi \}\] \[=-0.8A\sin (\omega t+kx)\] an additional phase change of \[\pi .\]
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