Question

The length of the wire between two ends of a sonometer is $100\text{cm}$. What should be the positions of two bridges (from left end of sonometer) below the wire so that the three segments of the wire have their fundamental frequencies in the ratio $1:3:5$

• A. $\frac{1500}{23}\text{cm},\frac{500}{23}\text{cm}$
• B. $\frac{1500}{23}\text{cm},\frac{300}{23}\text{cm}$
• C. $\frac{300}{23}\text{cm},\frac{800}{23}\text{cm}$
• D. $\frac{1500}{23}\text{cm},\frac{2000}{23}\text{cm}$

Answer 1

The correct option is D $\frac{1500}{23}\text{cm},\frac{2000}{23}\text{cm}$

Let $L(=100\text{cm})$ be the length of the wire and
$L_1,L_2$ and $L_3$ are the lengths of the segments as shown in the figure.


Fundamental frequency, $f\propto \frac{1}{L}$
As the fundamental frequencies are in the ratio of $1:3:5$,
$\therefore L_1:L_2:L_3=\frac{1}{1}:\frac{1}{3}:\frac{1}{5}=15:5:3$

As $L=L_1+L_2+L_3$
Let $x$ be the common factor. Then
$15x+5x+3x=100$
$23x=100$
$x=\frac{100}{23}$

$\therefore L_1=15\times \frac{100}{23}=\frac{1500}{23}\text{cm}$
$L_2=5\times \frac{100}{23}=\frac{500}{23}\text{cm};L_3=3\times \frac{100}{23}=\frac{300}{23}\text{cm}$

$\therefore$ The bridges should be placed from $A$ at $\frac{1500}{23}\text{cm}$ and $\frac{2000}{23}\text{cm}$ respectively.