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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • question_answer
    A metal wire of linear mass density of 9.8 g/m is stretched with a tension of 10 kg-wt between two rigid supports 1 m apart. The wire passes at its middle point between the poles of a permanent magnet and it vibrates in resonance when carrying an alternating current of frequency n. The frequency n of the alternating source is

    A)  50 Hz                                   

    B)  100 Hz

    C)  200 Hz                 

    D)         25 Hz

    Correct Answer: A

    Solution :

    The wire will vibrate with the same frequency as that of source. This can be considered as an example of forced vibration.                 \[T=10\times 9.8N=98N\]                 \[m=9.8\times {{10}^{-3}}kg/m\] Frequency of wire\[f=\frac{1}{2L}\sqrt{\left( \frac{T}{m} \right)}\]                                       \[=\frac{1}{2\times 1}\sqrt{\left( \frac{98}{9.8\times {{10}^{-3}}} \right)}\]                                      \[=50\,\,Hz\]


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