The length of the wire between two ends of a sonometer is 100 cm. What should be the positions of two bridges below the wire so that the three segments of the wire have their fundamental frequencies in the ratio 1: 3: 5.

$\text{ Let }L(=100 \mathrm{cm})\text{ be the length of the wire and }{L}_1,L_2\text{ and }{L}_3\text{ are the lengths of the segments as shown in the figure. }$

$$\begin{gathered} \text{ Fundamental frequency, }\nu \propto \frac{1}{L} \\ \text{ As the fundamental frequencies are in the ratio of }1:3\text{ : 5,} \\ \therefore L_1:L_2:L_3=\frac{1}{1}:\frac{1}{3}:\frac{1}{5}=15:5:3 \end{gathered}$$

$$\begin{gathered} \text{ Let }x\text{ be the common factor.} \\ \text{Then, 15 x+5 x+3 x=L=100} \\ \Rightarrow 23x=100 \text{ or, } x=\frac{100}{23}; \end{gathered}$$

$$\begin{gathered} \therefore L_1=15\times \frac{100}{23}=\frac{1500}{23} \mathrm{cm} ; \\ {\mathrm{L}}_2=5\times \frac{100}{23}=\frac{500}{23} \mathrm{cm} ; \\ {\mathrm{L}}_3=3\times \frac{100}{23}=\frac{300}{23} \mathrm{cm} ; \end{gathered}$$

$$\begin{gathered} \therefore \text{ The bridges should be placed from }\mathrm{A} \\ \text{ at }\frac{1500}{23} \mathrm{cm}\text{ and }\frac{2000}{23} \mathrm{cm}\text{ respectively. } \end{gathered}$$