The length of a sonometer wire ' AB ' is 110 cm . Where should the two bridges be placed from end ' A ' to divide the wire in three segments whose fundamental frequencies are in th

Question

The length of a sonometer wire ' AB ' is 110 cm . Where should the two bridges be placed from end ' A ' to divide the wire in three segments whose fundamental frequencies are in the ratio 1 : 2 : 3 ?

TestHub · Accepted Answer

The fundamental frequency of a sonometer wire is given by $f = \frac{v}{2 l}$ , which implies $f \propto \frac{1}{l}$ .

Given the ratio of fundamental frequencies $f_{1} : f_{2} : f_{3} = 1 : 2 : 3$ .

The ratio of the lengths of the segments will be $l_{1} : l_{2} : l_{3} = \frac{1}{1} : \frac{1}{2} : \frac{1}{3} = 6 : 3 : 2$ .

Let the lengths of the three segments be $6 x$ , $3 x$ , and $2 x$ .

The total length of the wire is $L = 6 x + 3 x + 2 x = 11 x$ .

Given $L = 110$ cm, we have $11 x = 110 \Rightarrow x = 10$ cm.

The lengths of the individual segments are $l_{1} = 60$ cm, $l_{2} = 30$ cm, and $l_{3} = 20$ cm.

The positions of the bridges from end A are at $l_{1} = 60$ cm and $l_{1} + l_{2} = 60 + 30 = 90$ cm.

Answer: 60 cm and 90 cm

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