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A sonometer wire of length $114 \mathrm{~cm}$ 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 $1: 3: 4$ ?
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At $72 \mathrm{~cm}$ and $96 \mathrm{~cm}$ from one end
At $72 \mathrm{~cm}$ and $96 \mathrm{~cm}$ from one end
Total length of the wire, $\mathrm{L}=114 \mathrm{~cm}$ $\mathrm{n}_1: \mathrm{n}_2: \mathrm{n}_3=1: 3: 4$
Let $\mathrm{L}_1, \mathrm{~L}_2$ and $\mathrm{L}_3$ be the lengths of the three parts
As $\mathrm{n} \propto \frac{1}{\mathrm{~L}}$
$$
\begin{aligned}
& \therefore \quad \mathrm{L}_1: \mathrm{L}_2: \mathrm{L}_3=\frac{1}{1}: \frac{1}{3}: \frac{1}{4}=12: 4: 3 \\
& \therefore \quad \mathrm{L}_1=72 \mathrm{~cm}\left(\frac{12}{12+4+3} \times 114\right) \\
& \mathrm{L}_2=24 \mathrm{~cm}\left(\frac{4}{19} \times 114\right) \\
& \text { and } \quad \mathrm{L}_3=18 \mathrm{~cm}\left(\frac{3}{19} \times 114\right)
\end{aligned}
$$
Hence the bridges should be placed at $72 \mathrm{~cm}$ and $72+24=96 \mathrm{~cm}$ from one end.
Let $\mathrm{L}_1, \mathrm{~L}_2$ and $\mathrm{L}_3$ be the lengths of the three parts
As $\mathrm{n} \propto \frac{1}{\mathrm{~L}}$
$$
\begin{aligned}
& \therefore \quad \mathrm{L}_1: \mathrm{L}_2: \mathrm{L}_3=\frac{1}{1}: \frac{1}{3}: \frac{1}{4}=12: 4: 3 \\
& \therefore \quad \mathrm{L}_1=72 \mathrm{~cm}\left(\frac{12}{12+4+3} \times 114\right) \\
& \mathrm{L}_2=24 \mathrm{~cm}\left(\frac{4}{19} \times 114\right) \\
& \text { and } \quad \mathrm{L}_3=18 \mathrm{~cm}\left(\frac{3}{19} \times 114\right)
\end{aligned}
$$
Hence the bridges should be placed at $72 \mathrm{~cm}$ and $72+24=96 \mathrm{~cm}$ from one end.
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