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JEE Main & Advanced Physics Wave Mechanics Question Bank Self Evaluation Test - Waves

  • question_answer
    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)  0.02 

    B)  0.03  

    C) 0.04 

    D) 0.01

    Correct Answer: A

    Solution :

    [a] For fundamental mode, \[f=\frac{1}{2\ell }\sqrt{\frac{T}{\mu }}\] Taking logarithm on both sides, we get \[\log \,f=\log \left( \frac{1}{2\ell } \right)+\log \left( \sqrt{\frac{T}{\mu }} \right)\] \[=\log \left( \frac{1}{2\ell } \right)+\frac{1}{2}\log \left( \frac{T}{\mu } \right)\] or \[\log \,f=\log \left( \frac{1}{2\ell } \right)+\frac{1}{2}[\log \,T-\log \mu ]\] Differentiating both sides, we get \[\frac{df}{f}=\frac{1}{2}\frac{dT}{T}\] (as \[\ell \] and \[\mu \] are constants) \[\Rightarrow \frac{dT}{T}=2\times \frac{df}{f}\] Here \[df=6\] \[f=600Hz\] \[\therefore \,\,\frac{dT}{T}=\frac{2\times 6}{600}=0.02\]


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