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

  • question_answer
    Stationary waves are set up in air column. Velocity of sound in air is 330 m/s and frequency is 165 Hz. Then distance between the nodes is   [EAMCET (Engg.) 1995; CPMT 1999]

    A)            2 m

    B)            1 m

    C)            0.5 m                                       

    D)            4 m

    Correct Answer: B

    Solution :

               \[v=330\]\[m/s\]; \[n=165\]\[Hz\]. Distance between two successive nodes = \[\frac{\lambda }{2}\] \[=\frac{v}{2n}=\frac{330}{2\times 165}=1m\]


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