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JCECE Engineering JCECE Engineering 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 \[2\,\,s\]. 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,   \[n'=n+\frac{3}{2},\,\,T'=T'+\frac{T}{100}=\frac{101T}{100}\] \[\therefore \]  \[n+\frac{3}{2}=\frac{1}{2l}\sqrt{\frac{101T}{100m}}\] \[\Rightarrow \]               \[n+\frac{3}{2}=1.005\times \frac{1}{2l}\sqrt{\frac{T}{m}}\]                         ? (ii) From Eqs. (i) and (ii), we get                 \[n+\frac{3}{2}=1.005\times n\] \[\Rightarrow \]                     \[n=300\,\,Hz\]


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