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 

Total length of thewire, L = 100 cm

${\text{v}}_1:{\text{v}}_2:{\text{v}}_3=1:3:5$

If  $l_1,l_2,l_3$ are the lengths of these parts, then as$\text{V}\propto \frac{1}{l}$

$\therefore l_1:l_2:l_3=\frac{1}{1}:\frac{1}{3}:\frac{1}{5}=15:3:3$

Sum of the ratios = 15 + 5 + 3 = 23

$\therefore l_1=\frac{\text{L}}{\text{23}}\times 15=\frac{100}{23}\times 15=\frac{1500}{23}\text{cm}$

$l_2=\frac{100}{23}\times 5=\frac{500}{23}\text{cm}$

$l_3=\frac{100}{23}\times 3=\frac{300}{23}\text{cm}$

Hence, the bridges should be placed at $\frac{1500}{23}\text{cm}$ and $\frac{2000}{23}\text{cm}$ from the zero end of the wire