Question

A sonometer wire has a total length of $2\text{m}$ between the fixed ends. Where should the two bridges be placed, so that the three segments of the wire have fundamental frequencies in the ratio $1:3:5$ ?

• A. $\frac{30}{23}\text{m}$ and $\frac{40}{23}\text{m}$ from one end.
• B. $\frac{6}{23}\text{m}$ and $\frac{16}{23}\text{m}$ from other end.
• C. $\frac{30}{23}\text{m}$ from one end and $\frac{6}{23}\text{m}$ from the other end.
• D. $\frac{30}{23}\text{m}$ from one end and $\frac{10}{23}\text{m}$ from the other end.

Answer 1

Answer: (A), (B), (C)

$\frac{30}{23}\text{m}$ from one end and $\frac{6}{23}\text{m}$ from the other end.


The fundamental frequency of the string, fixed at both ends, is given by,

$f=\frac{v}{2l}$

Let $l_1$ be the length between one end and the first bridge,

$l_2$ be the length between the two bridges,

$l_3$ be the length between second bridge and the other end

Given that, $l_1+l_2+l_3=2\text{m}---\left(1\right)$

Also, $\frac{v}{2l_1}:\frac{v}{2l_2}:\frac{v}{2l_3}=1:3:5---\left(2\right)$

From $\left(2\right)$,

$l_2=\frac{l_1}{3}$

$l_3=\frac{l_1}{5}$

Substituting the above values in $\left(1\right)$,

$\Rightarrow l_1+\frac{l_1}{3}+\frac{l_1}{5}=2$

$\Rightarrow l_1\left(\frac{23}{15}\right)=2$

$\Rightarrow l_1=\frac{30}{23}\text{m}$

$\Rightarrow l_2=\frac{10}{23}\text{m}$

$\Rightarrow l_3=\frac{6}{23}\text{m}$

Therefore,

The distance of first bridge from one end $=l_1=\frac{30}{23}\text{m}$

The distance of second bridge from one end $=l_1+l_2=\frac{40}{23}\text{m}$

The distance of first bridge from other end $=l_2+l_3=\frac{16}{23}\text{m}$

The distance of second bridge from other end $=l_3=\frac{6}{23}\text{m}$

Hence, options $A,B\text{and}C$ are correct.