Question

The length of the wire between two ends of a sonometer is $100cm$. What should be the positions of two bridges below the wire so that the three segments of the wire have their fundamental frequencies in the ratio $1:3:5$.

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

Answer 1

Answer: (D)

Let $L(=100cm)$ 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 , $\upsilon \propto \frac{1}{L}$
As the fundamental frequencies are in the ratio of $1:3:5$,
$\therefore L_1:L_2:L_3=\frac{1}{1}:\frac{1}{3}:\frac{1}{5}=15:5:3$
Let x be the common factor. Then
$15x+5x+3x=L=100$
$23x=100$ or $x=\frac{100}{23}$
$\therefore L_1=15\times \frac{100}{23}=\frac{1500}{23}cm$
$L_2=5\times \frac{100}{23}=\frac{500}{23}cm$
$L_2=3\times \frac{100}{23}=\frac{300}{23}cm$
$\therefore$ The bridges should be placed from $A$ at $\frac{1500}{23}cm$ and $\frac{2000}{23}cm$respectively.