A tuning fork resonates with a sonometer wire | vedclass

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Question

(C) The frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is length,$T$ is tension,and $\mu$ is linear mass

Vedclass · Accepted Answer

$6$

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(C) The frequency of a sonometer wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is length,$T$ is tension,and $\mu$ is linear mass density.Since $L$ and $\mu$ are constant,$f \propto \sqrt{T}$.Let $f_0$ be the frequency of the tuning fork.For $T_1 = 6 \ N$,the wire resonates with the tuning fork,so $f_1 = f_0 = k\sqrt{6}$,where $k = \frac{1}{2L\sqrt{\mu}}$.For $T_2 = 54 \ N$,the frequency of the wire is $f_2 = k\sqrt{54} = k\sqrt{9 	imes 6} = 3k\sqrt{6} = 3f_0$.The number of beats per second is $|f_2 - f_0| = 12$.Substituting $f_2 = 3f_0$,we get $|3f_0 - f_0| = 12$.$2f_0 = 12$,which gives $f_0 = 6 \ Hz$.

$A$ tuning fork resonates with a sonometer wire of length $1 \ m$ stretched with a tension of $6 \ N$. When the tension in the wire is changed to $54 \ N$,the same tuning fork produces $12$ beats per second with it. The frequency of the tuning fork is $Hz$.

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