Length of a son meter wire AB is 110 cm . Where should two bridges be placed from A to divide wire i

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14.Waves and Sound

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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 $1:2:3$?

A

$40\; cm$ and $80\; cm$

B

$60\; cm$ and $90\; cm$

C

$30\; cm$ and $60\; cm$

D

$30\; cm$ and $90\; cm$

(AIPMT-1995)

Solution

$f=\frac{V}{2 L} \Rightarrow f \propto \frac{1}{L}$

$f_1: f_2: f_3=1: 2: 3$

$\Rightarrow \frac{1}{L_1}: \frac{1}{L_2}: \frac{1}{L_3}=1: 2: 3$

$\Rightarrow L_1: L_2: L_3=\frac{1}{1}: \frac{1}{2}: \frac{1}{3}$

$\Rightarrow L_1: L_2: L_3=6: 3: 2$

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 $A$.

Standard 11

Physics

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