MGIMS WARDHA MGIMS WARDHA Solved Paper-2011

  • question_answer
    The length of two open organ pipes are \[l\] and \[(l+\Delta l)\]   respectively.   Neglecting   end corrections the beats frequency between them will be approximately

    A) \[\frac{v}{2l}\]                                  

    B) \[\frac{v}{4l}\]

    C)  \[\frac{v\Delta l}{2{{l}^{2}}}\]                                   

    D)  \[\frac{v\Delta l}{l}\]

    Correct Answer: C

    Solution :

                     Frequency of open organ pipe \[n=\frac{v}{2l}\] For first pipe\[{{n}_{1}}=\frac{v}{2l}\] For second pipe\[{{n}_{2}}=\frac{v}{2(l+\Delta l)}\] Number of beats per sec in beats frequency \[=|{{n}_{1}}-{{n}_{2}}|=\left| \frac{v}{2l}-\frac{v}{2(l+\Delta l)} \right|=\frac{v\delta l}{2{{l}^{2}}}\]


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