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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 $1 : 3 : 4$ ?
At $36\, cm$ and $84\, cm$ from one end
At $24\, cm$ and $72\, cm$ from one end
At $48\, cm$ and $96\, cm$ from one end
At $72\, cm$ and $96\, cm$ from one end
Solution
Total length of the wire, $L=114 \mathrm{cm}$
$\mathrm{n}_{1}: \mathrm{n}_{2}: \mathrm{n}_{3}=1: 3: 4$
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
