If n_{1}, n_{2} and n_{3} are the fundamental | vedclass

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(C) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.Since the tension $T$ and mass per unit length

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$\frac{1}{n} = \frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}}$

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(C) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.Since the tension $T$ and mass per unit length $\mu$ are constant for the string,we have $n \propto \frac{1}{l}$,which implies $l = \frac{k}{n}$,where $k$ is a constant.When the string is divided into three segments of lengths $l_{1}, l_{2}, l_{3}$ with fundamental frequencies $n_{1}, n_{2}, n_{3}$ respectively,we have $l_{1} = \frac{k}{n_{1}}$,$l_{2} = \frac{k}{n_{2}}$,and $l_{3} = \frac{k}{n_{3}}$.The total length of the string is $l = l_{1} + l_{2} + l_{3}$.Substituting the expressions for length in terms of frequency,we get $\frac{k}{n} = \frac{k}{n_{1}} + \frac{k}{n_{2}} + \frac{k}{n_{3}}$.Dividing by $k$,we obtain the relation $\frac{1}{n} = \frac{1}{n_{1}} + \frac{1}{n_{2}} + \frac{1}{n_{3}}$.

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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