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JCECE Medical JCECE Medical Solved Paper-2005

  • question_answer
    Two instruments having stretched strings are being played in unison. When the tension of one of the instruments is increased by 1%, 3 beats are produced in 2s. The initial frequency of vibration of each wire is:

    A)  300 Hz       

    B)  500 Hz

    C)  1000 Hz      

    D)  400 Hz

    Correct Answer: A

    Solution :

     Key Idea: Beats = Difference in frequencies. The frequency of vibration \[n=\frac{1}{2l}\sqrt{\frac{T}{m}}\] ?(i) Given, \[n'=n+\frac{3}{2},T'=T+\frac{T}{100}=\frac{101\,T}{100}\] \[\therefore \] \[n+\frac{3}{2}=\frac{1}{2l}\sqrt{\frac{101T}{100\,m}}\] \[\Rightarrow \] \[n+\frac{3}{2}=1.005\times \frac{1}{2l}\sqrt{\frac{T}{m}}\] ?(ii) From Eqs. (i) and (ii),we get \[n+\frac{3}{2}=1.005\times n\] \[\Rightarrow \] \[n=300\,Hz\]


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