Question

The length of a sonometer wire between two fixed ends is $1.10m$. Where should the two bridges be placed to divide the wire into three segments whose fundamental frequencies are in the ratio of $1:2:3$?

• A. $0.5$ and $0.8$ from left
• B. $0.4$ and $0.6$ from left
• C. $0.7$ and $0.5$ from left
• D. $0.6$ and $0.9$ from left

Answer 1

The correct option is D $0.6$ and $0.9$ from left

Given, fundamental frequencies of the three segments,
$f_1:f_2:f_3=1:2:3$
$f_1=f,f_2=2f,f_3=3f$

Length of the sonometer wire,
$L=1.10m$

Let lengths of three segments be $l_1{l}_2$ and $l_3$.
$l_1+l_2+l_3=L=1.10$ ... (i)
For sonometer wire, $f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

For given $T$ and $\mu$,
$f\propto \frac{1}{l}$ or $fl=$ constant
$f_1{l}_1=f_2{l}_2=f_3{l}_3=k$
$f{l}_1=2f{l}_2=3f{l}_3=k$
$l_1=\frac{k}{f},l_2=\frac{k}{2f},l_3=\frac{k}{3f}$

Substitute these values in equation (i)
$\frac{k}{f}+\frac{k}{2f}+\frac{k}{3f}=1.10$
$\frac{k}{f}(1+\frac{1}{2}+\frac{1}{3})=1.1$
$\frac{k}{f}=1.1\left(\frac{6}{11}\right)$
$\frac{k}{f}=0.6$
So,
$l_1=\frac{k}{f}=0.6m,l_2=\frac{k}{2f}=0.3m$

The bridges must be placed at $l_1=0.6m$ and $l_1+l_2=0.6+03=0.9m$ from left



Final answer: (a)