A) \[y=A\sin (\omega t+kx)\]
B) \[y=-0.8A\sin (\omega t+kx)\]
C) \[y=0.8A\sin (\omega t+kx)\]
D) \[y=A\sin (\omega t+0.8\,kx)\]
Correct Answer: B
Solution :
On getting reflected from a rigid boundary the wave suffers an additional phase change of\[\pi \]. Thus, equation of reflected wave is given by \[y=0.8\text{ }A\text{ }sin(\omega t+kx+\pi )\] \[\therefore \] \[y=-0.8\text{ }A\text{ }sin(\omega t+kx)\]
You need to login to perform this action.
You will be redirected in
3 sec
