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The fundamental frequency of a closed organ pipe of length is equal to the second overtone of an organ pipe open at both the ends. The length of the organ pipe open at both the ends is
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Verified Answer
The correct answer is:
For closed organ pipe,
\(n_c=\frac{V}{4 l}\)
For open organ pipe,
\(n_0=\frac{V}{2 l^{\prime}}\)
For the second overtone of an open organ pipe,
\(\begin{aligned}
& n^{\prime}=3 n_0=\frac{3 V}{2 l^{\prime \prime}} \\
& n_c=n^{\prime} \\
& \frac{V}{4 l}=\frac{3 V}{2 l^{\prime}} \\
& l^{\prime}=6 l=6 \times 20 \\
& l^{\prime}=120 \mathrm{~cm}
\end{aligned}\)
\(n_c=\frac{V}{4 l}\)
For open organ pipe,
\(n_0=\frac{V}{2 l^{\prime}}\)
For the second overtone of an open organ pipe,
\(\begin{aligned}
& n^{\prime}=3 n_0=\frac{3 V}{2 l^{\prime \prime}} \\
& n_c=n^{\prime} \\
& \frac{V}{4 l}=\frac{3 V}{2 l^{\prime}} \\
& l^{\prime}=6 l=6 \times 20 \\
& l^{\prime}=120 \mathrm{~cm}
\end{aligned}\)
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