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JEE Main & Advanced Physics Wave Mechanics Question Bank Vibration of String

  • question_answer
    In an experiment with sonometer a tuning fork of frequency 256 Hz resonates with a length of 25 cm and another tuning fork resonates with a length of 16 cm. Tension of the string remaining  constant  the frequency  of the second tuning fork is                                                                     [KCET 2005]

    A)            163.84 Hz                               

    B)            400 Hz

    C)            320 Hz                                     

    D)            204.8 Hz

    Correct Answer: D

    Solution :

                         In case of sonometer frequency is given by \[n=\frac{p}{2l}\sqrt{\frac{T}{m}}\]Þ \[\frac{{{n}_{2}}}{{{n}_{1}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\Rightarrow {{n}_{2}}=\frac{25}{16}\times 256=400\,\,\,Hz\]


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