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(C) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is tension and $\mu$ is mass per uni

Vedclass · Accepted Answer

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

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(C) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is tension and $\mu$ is mass per unit length.Since $T$ and $\mu$ are constant for the segments,we have $n \propto \frac{1}{l}$,which implies $nl = k$ (constant).Given that the total length $l = l_1 + l_2 + l_3$,we can express each segment length as $l_1 = \frac{k}{n_1}$,$l_2 = \frac{k}{n_2}$,and $l_3 = \frac{k}{n_3}$.The original length is $l = \frac{k}{n}$.Substituting these into the length equation: $\frac{k}{n} = \frac{k}{n_1} + \frac{k}{n_2} + \frac{k}{n_3}$.Dividing both sides by $k$,we get $\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$.

When a string of length $l$ is divided into three segments of length $l_1, l_2$ and $l_3$,the fundamental frequencies of the three segments are $n_1, n_2$ and $n_3$ respectively. The original fundamental frequency $n$ of the string is:

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