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

  • question_answer
    A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). 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 2004]

    A)            4    

    B)            6

    C)            8    

    D)            12

    Correct Answer: C

    Solution :

                         \[n\propto \frac{1}{l}\] Þ \[\frac{\Delta n}{n}=-\frac{\Delta l}{l}\] If length is decreased by 2% then frequency increases by 2% i.e., \[\frac{{{n}_{2}}-{{n}_{1}}}{{{n}_{1}}}=\frac{2}{100}\] Þ \[{{n}_{2}}-{{n}_{1}}=\frac{2}{100}\times {{n}_{1}}=\frac{2}{100}\times 392=7.8\approx 8.\]


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