An organ pipe of length $L$ is open at one end and closed at other end. The wavelengths of the three lowest resonating frequencies that can be produced by this pipe are
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$\mathrm{L}=(2 \mathrm{n}-1) \frac{\lambda}{4}$

$\lambda=\frac{4 \mathrm{L}}{(2 \mathrm{n}-1)} \Rightarrow \lambda_{1}=\frac{4 \mathrm{L}}{2(1)-1}=4 \mathrm{L}$

$\lambda_{2}=\frac{4 \mathrm{L}}{4-1}=\frac{4 \mathrm{L}}{3}$ and $\lambda_{3}=\frac{4 \mathrm{L}}{6-1}=\frac{4 \mathrm{L}}{5}$

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