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An organ pipe with both ends open has a length $L=25 \mathrm{~cm}$. An extra hole is created at position $\frac{L}{2}$. The lowest frequency of sound produced is (assume, speed of sound $=340 \mathrm{~m} / \mathrm{s}$ )
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The correct answer is:
$1360 \mathrm{~Hz}$
We know that, there must be an antinode at the open end. Since, a hole has been made at the position $\frac{L}{2}$, so there must be another antinode at $x=\frac{L}{2}$.
For lowest frequency, $\lambda=L=25 \mathrm{~cm}$
Frequency of sound produced,
$f=\frac{v}{\lambda}=\frac{340}{0.25}=1360 \mathrm{~Hz}$
For lowest frequency, $\lambda=L=25 \mathrm{~cm}$
Frequency of sound produced,
$f=\frac{v}{\lambda}=\frac{340}{0.25}=1360 \mathrm{~Hz}$
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