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BHU PMT BHU PMT Solved Paper-2004

  • question_answer
    A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension\[\left( T \right)\]. If length of the string is decreased, by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is:                                                                                   [BHU PMT-2004]

    A)  4                                            

    B)  6            

    C)  8                                            

    D)  12

    Correct Answer: C

    Solution :

                     Key Idea: According to law of length\[n\propto \frac{1}{l}\]. The frequency of tuning fork is given by                \[n=\frac{1}{2l}\sqrt{\frac{T}{m}}\] Where \[l\] is length of string \[T\] is tension and \[m\] is mass per unit length of string. Given, \[{{l}_{ & 1}}=50\,\,cm,\,\,{{l}_{2}}=\left( 1-\frac{2}{100} \right)\times 50=49\,\,cm\].                                                 \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{49}{50}\] \[\Rightarrow \]                               \[{{n}_{2}}=\frac{50}{49}\times 392=400\,Hz\] Also number of beats per second (beat frequency) \[={{n}_{2}}-{{n}_{1}}=\]difference of the frequencies of sound-sources. \[=400-392=8\]


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