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Two uniform stretched strings $A$ and $B$, made of steel, are vibrating under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$, the ratio of the lengths of the strings is
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Verified Answer
The correct answer is:
1 : 3
Given, 1st overtone of $A=2$ nd overtone of $B$
$\begin{aligned}
\Rightarrow & \frac{1}{l_1} \sqrt{\frac{T}{\pi r_1^2 d}} =\frac{3}{2 l_2} \sqrt{\frac{T}{\pi r_2^2 d}} \\
\Rightarrow & \frac{r_2}{r_1} =\frac{3}{2}\left(\frac{l_1}{l_2}\right)
\end{aligned}$
But $r_1=2 r_2$
$\begin{aligned}
& \therefore \quad \frac{r_2}{2 r_2}=\frac{3}{2}\left(\frac{l_1}{l_2}\right) \\
& \text { or } \quad \frac{l_1}{l_2}=\frac{1}{3} \Rightarrow l_1: l_2=1: 3 \\
&
\end{aligned}$
$\begin{aligned}
\Rightarrow & \frac{1}{l_1} \sqrt{\frac{T}{\pi r_1^2 d}} =\frac{3}{2 l_2} \sqrt{\frac{T}{\pi r_2^2 d}} \\
\Rightarrow & \frac{r_2}{r_1} =\frac{3}{2}\left(\frac{l_1}{l_2}\right)
\end{aligned}$
But $r_1=2 r_2$
$\begin{aligned}
& \therefore \quad \frac{r_2}{2 r_2}=\frac{3}{2}\left(\frac{l_1}{l_2}\right) \\
& \text { or } \quad \frac{l_1}{l_2}=\frac{1}{3} \Rightarrow l_1: l_2=1: 3 \\
&
\end{aligned}$
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