A stretched wire of length 110 cm is divided in to 3 segments whose frequencies are in the ratio 1 : 2 : 3 . Their respective lengths must be

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The fundamental frequency of a stretched wire is given by $f = \frac{1}{2 l} \frac{T}{μ}$ .

Since the tension $T$ and linear mass density $μ$ are constant for the segments of the same wire, the frequency is inversely proportional to the length, i.e., $f \propto \frac{1}{l}$ .

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

The ratio of their lengths will be $l_{1} : l_{2} : l_{3} = \frac{1}{f _{1}} : \frac{1}{f _{2}} : \frac{1}{f _{3}} = 1 : \frac{1}{2} : \frac{1}{3}$ .

Multiplying by $6$ , we get $l_{1} : l_{2} : l_{3} = 6 : 3 : 2$ .

Let the lengths be $6 x$ , $3 x$ , and $2 x$ . The total length of the wire is $110$ cm.

$6 x + 3 x + 2 x = 110$

$11 x = 110 \Rightarrow x = 10$ cm

The respective lengths are:

$l_{1} = 6 \times 10 = 60$ cm

$l_{2} = 3 \times 10 = 30$ cm

$l_{3} = 2 \times 10 = 20$ cm

Answer: $60$ cm, $30$ cm, $20$ cm

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