MCQ
Two identical piano wires, kept under the same tension $T$ have a fundamental frequency of $600\,\, Hz.$ The fractional increase in the tension of one of the wires which will lead to occurrence of $6\,\, beats/s$ when both the wires oscillate together would be
  • A
    $0.01$
  • $0.02$
  • C
    $0.03$
  • D
    $0.04$

Answer

Correct option: B.
$0.02$
b
$\text { As } v=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}  \therefore \frac{\Delta v}{v}=\frac{1}{2} \frac{\Delta T}{T}$

 $\frac{\Delta T}{T}=2 \frac{\Delta v}{v}=2 \times \frac{6}{600}=0.02$

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