A sonometer wire has a length of 114 cm, bet | vedclass

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

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$l_1, l_2, l_3=72, 18, 24$

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(C) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.Since $T$ (tension) and $m$ (mass per unit length) are constant for the same wire, $n \propto \frac{1}{l}$.Given the ratio of frequencies $n_1 : n_2 : n_3 = 1 : 3 : 4$, the lengths of the segments must be in the inverse ratio:$l_1 : l_2 : l_3 = \frac{1}{1} : \frac{1}{3} : \frac{1}{4}$.To simplify this ratio, multiply by the least common multiple of $1, 3, 4$, which is $12$:$l_1 : l_2 : l_3 = 12 : 4 : 3$.The sum of the parts is $12 + 4 + 3 = 19$.The total length of the wire is $114 \ cm$.Therefore, the lengths are:$l_1 = \frac{12}{19} 	imes 114 = 12 	imes 6 = 72 \ cm$$l_2 = \frac{4}{19} 	imes 114 = 4 	imes 6 = 24 \ cm$$l_3 = \frac{3}{19} 	imes 114 = 3 	imes 6 = 18 \ cm$.Thus, the lengths are $72 \ cm, 24 \ cm, 18 \ cm$.

$A$ sonometer wire has a length of $114 \ cm$, between two fixed ends. Where should two bridges be placed so as to divide the wire into three segments (in $cm$) whose fundamental frequencies are in the ratio $1:3:4$?

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