Question

The length of a sonometer wire $\text{AB}$ is $110\text{cm}$. Where should the two bridges be placed (from $\text{A}$) to divide the wire into $3$ segments, whose fundamental frequencies are in the ratio of $1:2:3$ ?

• A. $30\text{cm}$ and $90\text{cm}$
• B. $40\text{cm}$ and $80\text{cm}$
• C. $60\text{cm}$ and $90\text{cm}$
• D. $30\text{cm}$ and $60\text{cm}$

Answer 1

The correct option is C $60\text{cm}$ and $90\text{cm}$

Fundamental frequency,
$f=\frac{V}{2L}\Rightarrow f\propto \frac{1}{L}$
Given :
$f_1:f_2:f_3=1:2:3$
$\Rightarrow \frac{1}{L_1}:\frac{1}{L_2}:\frac{1}{L_3}=1:2:3$

$\Rightarrow L_1:L_2:L_3=\frac{1}{1}:\frac{1}{2}:\frac{1}{3}$
$\Rightarrow L_1:L_2:L_3=6:3:2$

Let $L_1=6x,L_2=3x,L_3=2x$

Total length of sonometer wire $=110\text{cm}$
$L_1+L_2+L_3=110$
$6x+3x+2x=110$
$x=10$
$\therefore L_1=60\text{cm}$, $L_2=30\text{cm}$, $L_3=20\text{cm}$
Hence, bridges should be placed at distance of $60\text{cm}$ and $90\text{cm}$ from $\text{A}$.