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EAMCET Medical EAMCET Medical Solved Paper-2000

  • question_answer
    The fundamental frequency of a closed pipe is 220 Hz. If \[\frac{1}{4}\] th of the pipe is filled with water the frequency of the first overtone of the pipe now is:

    A)  220 Hz                                 

    B)  440 Hz

    C)  880 Hz                                 

    D)  1760 Hz

    Correct Answer: C

    Solution :

                     Fundamental frequency for a closed organ pipe is\[n=\frac{v}{4l}\] When one-fourth of the pipe is filled with water, then New length \[l=l-\frac{l}{4}=\frac{3l}{4}\] \[\therefore \] New frequency \[n=\frac{v}{4l}\] \[=\frac{v}{4\times \frac{3l}{4}}=\frac{v}{3l}\] \[\frac{n}{n}=\frac{v/3l}{v/4l}\] \[n=\frac{4}{3}n\] \[=\frac{4}{3}\times 220=\frac{830}{3}\,Hz\] For closed organ pipe, first overtone \[=3\times \] fundamental frequency = 3n \[=3\times \frac{880}{3}=880\,Hz\]


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