Three similar wires of frequencies n_1, n_2 a | vedclass

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(B) The fundamental frequency of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $l$ is the length,$T$ is the tension,and $

Vedclass · Accepted Answer

$\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$

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(B) The fundamental frequency of a stretched wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $l$ is the length,$T$ is the tension,and $\mu$ is the mass per unit length.Since the wires are similar,the mass per unit length $\mu$ and tension $T$ are the same for all segments.Thus,$n \propto \frac{1}{l}$,which implies $nl = k$ (a constant).For the three wires,we have $n_1 l_1 = n_2 l_2 = n_3 l_3 = k$.This gives $l_1 = \frac{k}{n_1}$,$l_2 = \frac{k}{n_2}$,and $l_3 = \frac{k}{n_3}$.The total length of the combined wire is $l = l_1 + l_2 + l_3$.Substituting the values,we get $\frac{k}{n} = \frac{k}{n_1} + \frac{k}{n_2} + \frac{k}{n_3}$.Dividing by $k$,we obtain $\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$.

Three similar wires of frequencies $n_1, n_2$ and $n_3$ are joined to form a single wire. What will be its resulting frequency $n$?

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