# Open pipe resonance

I read somewhere, a closed pipe has resonance with a sound of wave length 4 times the length of pipe (from open end to closed end), when putting a source above the open end (since it reinforces the vibration). Am I right? I also read that an OPEN pipe (so both ends open) has resonance with a sound of wave length 4 times its pipe length. But why is this?

BTW I don't know if I am even right about my statements here, please correct if you could!

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Have you read this? en.wikipedia.org/wiki/Acoustic_resonance – joshphysics Aug 19 '13 at 18:51
Regarding terminology, when one says "closed pipe," one usually means "closed at both ends." Such a pipe cannot be driven from the end, but it still has resonant modes. Rather you seem to be discussing a "half-open" or "half-closed" pipe. – Chris White Aug 19 '13 at 18:53

So for a closed pipe, as in the picture above, we have a node at one end and an antinode at the other, and the distance between a node and antinode is $(2n + 1)\lambda/4$, where $n = 0$ gives you the fundamental, $n = 1$ is the first overtone and so on (the diagram actually shows the fifth overtone). For the fundamental frequency, $n = 0$, the length of the pipe is equal to $\lambda/4$ so the wavelength is four times the length of the pipe as you say.
If you take an open pipe the air will be moving, i.e. there is an antinode, at both ends. The distance between two antinodes is $n\lambda/2$, where this time $n = 1$ gives us the fundamental, so for the fundamental the length of the pipe is equal to $\lambda/2$ i.e. the wavelength is twice the length of the pipe.