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The length of the wire between two ends of sonometer is $100 \mathrm{~cm}$. 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 $1: 3: 5$.
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Verified Answer
The correct answer is:
$\frac{1500}{23} \mathrm{~cm}, \frac{2000}{23} \mathrm{~cm}$
According to question
Length of wire $=L=100 \mathrm{~cm}$
And ratio of fundamental frequencies
$$
=1: 3: 5
$$
$$
\begin{aligned}
L_1: L_2: L_3 & =\frac{1}{1}: \frac{1}{3}: \frac{1}{5} \\
& =15: 5: 3
\end{aligned}
$$
Let $x=$ comman factor then
$$
\begin{aligned}
& 15 x+5 x+3 x=100 \\
& \Rightarrow \quad 2 x=100 \\
& \Rightarrow \quad x=\frac{100}{23} \\
& \therefore \quad L_1=15 \times \frac{100}{23} \\
& =\frac{1500}{23} \mathrm{~cm} \\
& \therefore \quad L_2=5 \times \frac{100}{23} \\
& =\frac{1500}{23} \mathrm{~cm} \\
&
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad L_3 & =3 \times \frac{100}{23} \\
& =\frac{300}{23} \mathrm{~cm}
\end{aligned}
$$
This shows that bridge should be placed from $A$ at $\frac{1500}{23} \mathrm{~cm}$ and $\frac{2000}{23} \mathrm{~cm}$ respectively
Length of wire $=L=100 \mathrm{~cm}$
And ratio of fundamental frequencies
$$
=1: 3: 5
$$
$$
\begin{aligned}
L_1: L_2: L_3 & =\frac{1}{1}: \frac{1}{3}: \frac{1}{5} \\
& =15: 5: 3
\end{aligned}
$$
Let $x=$ comman factor then
$$
\begin{aligned}
& 15 x+5 x+3 x=100 \\
& \Rightarrow \quad 2 x=100 \\
& \Rightarrow \quad x=\frac{100}{23} \\
& \therefore \quad L_1=15 \times \frac{100}{23} \\
& =\frac{1500}{23} \mathrm{~cm} \\
& \therefore \quad L_2=5 \times \frac{100}{23} \\
& =\frac{1500}{23} \mathrm{~cm} \\
&
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad L_3 & =3 \times \frac{100}{23} \\
& =\frac{300}{23} \mathrm{~cm}
\end{aligned}
$$
This shows that bridge should be placed from $A$ at $\frac{1500}{23} \mathrm{~cm}$ and $\frac{2000}{23} \mathrm{~cm}$ respectively
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