If n_1, n_2,and n_3 are the fundamental frequ | vedclass

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

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 string of length $L$ is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.Since $T$ and $\mu$ are constant for all segments,we have $L = \frac{1}{2n} \sqrt{\frac{T}{\mu}} = \frac{k}{n}$,where $k = \frac{1}{2} \sqrt{\frac{T}{\mu}}$ is a constant.The string is divided into three segments of lengths $L_1, L_2$,and $L_3$ such that $L = L_1 + L_2 + L_3$.Substituting the expression for length in terms of frequency: $\frac{k}{n} = \frac{k}{n_1} + \frac{k}{n_2} + \frac{k}{n_3}$.Dividing both sides by $k$,we get $\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$.

If $n_1, n_2$,and $n_3$ are the fundamental frequencies of three segments into which a string is divided,then the original fundamental frequency $n$ of the string is given by

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