A) \[y=-a\cos (\omega \,t+kx)\]
B) \[y=-a\sin (\omega \,t+kx)\]
C) \[y=a\sin (\omega \,t+kx)\]
D) \[y=a\cos (\omega \,t+kx)\]
Correct Answer: B
Solution :
In closed organ pipe. If \[{{y}_{incident}}=a\sin (\omega t-kx)\] then \[{{y}_{reflected}}=a\sin (\omega t+kx+\pi )=-a\sin (\omega t+kx)\] Superimposition of these two waves give the required stationary wave.You need to login to perform this action.
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