The length of a son meter wire $AB$ is $110\; cm$. Where should the two bridges be placed from $A$ to divide the wire in $3$ segments whose fundamental frequencies are in the ratio of $1:2:3$?
AIPMT 1995, Medium
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$f=\frac{V}{2 L} \Rightarrow f \propto \frac{1}{L}$
$f_1: f_2: f_3=1: 2: 3$
$\Rightarrow \frac{1}{L_1}: \frac{1}{L_2}: \frac{1}{L_3}=1: 2: 3$
$\Rightarrow L_1: L_2: L_3=\frac{1}{1}: \frac{1}{2}: \frac{1}{3}$
$\Rightarrow L_1: L_2: L_3=6: 3: 2$
Let $L_1=6 x, L_2=3 x, L_3=2 x$
Total length of sonometer wire $=110 cm$
$L_1+L_2+L_3=110$
$6 x+3 x+2 x=110$ $x=10$
$\therefore L _1=60 cm , L _2=30 cm , L _3=20 cm$
Hence, bridges should be placed at distance of $60 cm$ and $90 cm$ from $A$.
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