The length of a sonometer wire AB is 176 cm. Where should the two bridges be placed from end A to divide the wire into three segments whose fundamental frequencies are in the ratio 1 : 2 : 3 ?

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Let the lengths of the three segments be $l_{1}$ , $l_{2}$ , and $l_{3}$ .

The fundamental frequency of a stretched string is given by $f = \frac{v}{2 l}$ . Since the tension and mass per unit length are constant for the wire, $f \propto \frac{1}{l}$ .

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

Therefore, the ratio of their lengths is $l_{1} : l_{2} : l_{3} = 1 : \frac{1}{2} : \frac{1}{3} = 6 : 3 : 2$ .

Let $l_{1} = 6 x$ , $l_{2} = 3 x$ , and $l_{3} = 2 x$ .

The total length of the wire is $l_{1} + l_{2} + l_{3} = 176 cm$ .

$6 x + 3 x + 2 x = 176 \Rightarrow 11 x = 176 \Rightarrow x = 16 cm$ .

The lengths of the individual segments are:
$l_{1} = 6 \times 16 = 96 cm$
$l_{2} = 3 \times 16 = 48 cm$
$l_{3} = 2 \times 16 = 32 cm$

The positions of the two bridges from end A are $l_{1}$ and $l_{1} + l_{2}$ .

First bridge position $= 96 cm$ .

Second bridge position $= 96 + 48 = 144 cm$ .

Answer: 96 cm and 144 cm

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