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JEE Main & Advanced Physics Wave Mechanics Question Bank Organ Pipe (Vibration of Air Column)

  • question_answer
    The stationary wave \[y=2a\sin kx\cos \omega \,t\] in a closed organ pipe is the result of the superposition of \[y=a\sin (\omega \,t-kx)\]and                                              [Roorkee 1994]

    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.


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