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

Vedclass · Accepted Answer

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}$

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(C) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.Since $T$ and $\mu$ are constant,we have $f \propto \frac{1}{L}$,which implies $fL = 	ext{constant} = k$.Therefore,$L = \frac{k}{f}$.When the wire of length $L$ is divided into three segments of lengths $L_1, L_2, L_3$,we have $L = L_1 + L_2 + L_3$.Substituting $L = \frac{k}{f}$,$L_1 = \frac{k}{f_1}$,$L_2 = \frac{k}{f_2}$,and $L_3 = \frac{k}{f_3}$ into the equation,we get:$\frac{k}{f} = \frac{k}{f_1} + \frac{k}{f_2} + \frac{k}{f_3}$.Dividing both sides by $k$,we obtain the relation: $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_3}$.

When a stretched wire of fundamental frequency $f$ is divided into three segments,the fundamental frequencies of these three segments are $f_1, f_2$ and $f_3$ respectively. Then the relation among $f_1, f_2, f_3$ and $f$ is (Assume tension is constant).

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