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Question

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

Vedclass · Accepted Answer

$\frac{1}{v} = \frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}$

Vedclass · Answer

(A) The fundamental frequency of a stretched string is given by $v = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is tension and $\mu$ is mass per unit length.Since $T$ and $\mu$ are constant for the string,$v \propto \frac{1}{l}$,which implies $vl = k$ (a constant).Therefore,for the three segments,we have $v_1 l_1 = v_2 l_2 = v_3 l_3 = k$.This gives $l_1 = \frac{k}{v_1}$,$l_2 = \frac{k}{v_2}$,and $l_3 = \frac{k}{v_3}$.The original length $l$ of the string is the sum of the lengths of the segments: $l = l_1 + l_2 + l_3$.Substituting the expressions for lengths in terms of frequencies: $\frac{k}{v} = \frac{k}{v_1} + \frac{k}{v_2} + \frac{k}{v_3}$.Dividing both sides by $k$,we get $\frac{1}{v} = \frac{1}{v_1} + \frac{1}{v_2} + \frac{1}{v_3}$.

When a string is divided into three segments of length $l_1, l_2$ and $l_3,$ the fundamental frequencies of these three segments are $v_1, v_2$ and $v_3$ respectively. The original fundamental frequency $(v)$ of the string is

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