A sonometer wire of length \(114\, cm\) is fixed at both the ends. Where should the two bridges be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio \(1 : 3 : 4\) ?

\(\mathrm{n}_{1}: \mathrm{n}_{2}: \mathrm{n}_{3}=1: 3: 4\)

Let \(L_{1}, L_{2}\) and \(L_{3}\) be the lengths of the three parts

As \(n \propto \frac{1}{L}\)

\(\therefore \quad \mathrm{L}_{1}: \mathrm{L}_{2}: \mathrm{L}_{3}=\frac{1}{1}: \frac{1}{3}: \frac{1}{4}=12: 4: 3\)

\(\therefore \quad \mathrm{L}_{1}=72 \mathrm{cm}\left(\frac{12}{12+4+3} \times 114\right)\)

\(\mathrm{L}_{2}=24 \mathrm{cm}\left(\frac{4}{19} \times 114\right)\)

and \(\mathrm{L}_{3}=18 \mathrm{cm}\left(\frac{3}{19} \times 114\right)\)

Hence the bridges should be placed at

\(72 \mathrm{cm}\) and \(72+24=96 \mathrm{cm}\) from oneend