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JEE Main & Advanced Physics Wave Mechanics Question Bank Beats

  • question_answer
    The frequency of tuning forks A and B are respectively 3% more and 2% less than the frequency of tuning fork C. When A and B are simultaneously excited, 5 beats per second are produced. Then the frequency of the tuning fork 'A' (in Hz) is                                                                   [EAMCET 2001]

    A)            98  

    B)            100

    C)            103

    D)            105 

    Correct Answer: C

    Solution :

               Let n be the frequency of fork C then                     \[{{n}_{A}}=n+\frac{3n}{100}=\frac{103n}{100}\] and  \[{{n}_{B}}=n-\frac{2n}{100}=\frac{98}{100}\] but \[{{n}_{A}}-{{n}_{B}}=5\]Þ \[\frac{5n}{100}=5\]Þ \[n=100\,Hz\] \ \[{{n}_{A}}=\frac{(103)(100)}{100}=103\,Hz\]


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