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The frequency of the second overtone of the open pipe is equal to the frequency of the first overtone of the closed pipe. The ratio of the lengths of the open pipe and the closed pipe is
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The correct answer is:
$2: 1$
The frequency of first overtone for closed pipe,
$f_{c}=\frac{3 \mathrm{v}}{4 I_{c}}$
Given, i.e.,
$\begin{aligned}
f_{c} &=f_{0} \\
\frac{3 v}{4 I_{c}} &=\frac{3 v}{2 I_{o}} \Rightarrow \frac{I_{o}}{I_{c}}=\frac{2}{1}
\end{aligned}$
$f_{c}=\frac{3 \mathrm{v}}{4 I_{c}}$
Given, i.e.,
$\begin{aligned}
f_{c} &=f_{0} \\
\frac{3 v}{4 I_{c}} &=\frac{3 v}{2 I_{o}} \Rightarrow \frac{I_{o}}{I_{c}}=\frac{2}{1}
\end{aligned}$
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