A thin wire of length 99 cm is fixed at both | vedclass

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

Vedclass · Accepted Answer

$54, 27, 18$

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(C) The fundamental frequency of a vibrating wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.Since the tension $T$ and mass per unit length $\mu$ are the same for all segments,the frequency $n$ is inversely proportional to the length $l$ of the segment $(n \propto \frac{1}{l})$.Given the ratio of fundamental frequencies is $n_1 : n_2 : n_3 = 1 : 2 : 3$.Therefore,the ratio of their lengths is $l_1 : l_2 : l_3 = \frac{1}{n_1} : \frac{1}{n_2} : \frac{1}{n_3} = \frac{1}{1} : \frac{1}{2} : \frac{1}{3} = 6 : 3 : 2$.The total length of the wire is $L = l_1 + l_2 + l_3 = 99 \ cm$.Sum of the ratio parts $= 6 + 3 + 2 = 11$.Calculating individual lengths:$l_1 = \frac{6}{11} 	imes 99 = 54 \ cm$$l_2 = \frac{3}{11} 	imes 99 = 27 \ cm$$l_3 = \frac{2}{11} 	imes 99 = 18 \ cm$Thus,the lengths are $54 \ cm, 27 \ cm, 18 \ cm$.

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