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VIT Engineering VIT Engineering Solved Paper-2009

  • question_answer
    Two identical piano wires have a fundamental frequency of 600 cycle per second when kept under the same tension. What fractional increase in the tension of one wires will lead to the occurrence of 6 beats per second when both wires vibrate simultaneously?

    A)  0.01             

    B)  0.02

    C)  0.03             

    D)  0.04  

    Correct Answer: B

    Solution :

    Beats per second when both the wires vibrate simultaneously. \[{{n}_{1}}\pm {{n}_{2}}=6\] or \[\frac{1}{2l}\sqrt{\frac{T}{m}}\pm \frac{1}{2l}\sqrt{\frac{T}{m}}=6\] or \[\frac{1}{2l}\sqrt{\frac{T}{m}}-\frac{1}{2l}\sqrt{\frac{T}{m}}=6\] or \[\frac{1}{2l}\sqrt{\frac{T}{m}}-600=6\] \[\frac{1}{2l}\sqrt{\frac{T}{m}}-606\]            ??(i) Given that fundamental frequency                      \[\frac{1}{2l}\sqrt{\frac{T}{m}}=600\]         ?..(ii) Dividing Eq. (i) by Eq. (ii), we get        \[\frac{\frac{1}{2l}\sqrt{\frac{T}{m}}}{\frac{1}{2l}\sqrt{\frac{T}{m}}}=\frac{606}{600}\] or \[\sqrt{\frac{T}{T}}=(1.01)\] or \[\frac{T}{T}=(1.02)%\] or \[T=T(1.02)\] Increase in tension                      \[\Delta T=T\times 1.02-T\]        \[=(0.02T)\]    Hence,        \[\Delta T=0.02\]


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