Waves

JEE Main level, numerical answer, medium

63 of 142

Question 63 of 142

A tuning fork resonates with a sonometer wire of length 1m1m stretched with a tension of 6N.6N. When the tension in the wire is changed to 54N,54N, the same tuning fork produces 1212 beats per second with it. The frequency of the tuning fork is _______ Hz.Hz.

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Answer: 6

Frequency of sonometer wire is given by, f=12LTμf=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu }}
Therefore, f1=126μ{f}_{1}=\dfrac{1}{2}\sqrt{\dfrac{6}{\mu }} f2=1254μ{f}_{2}=\dfrac{1}{2}\sqrt{\dfrac{54}{\mu }}
Hence, f1f2=13\dfrac{{f}_{1}}{{f}_{2}}=\dfrac{1}{3}
Also it is given that, f2−f1=12{f}_{2}-{f}_{1}=12
So, f1=6Hz{f}_{1}=6Hz.

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