Does a pipe hit at the top behave like an open-open system or an open-closed system?

Question

I'm making an instrument from an open-ended tube that produces sound when the top (opening) is struck. To find the lengths corresponding to pitches, I used

$$ V = \lambda f \tag{1} \label{1} $$

and combined it with the fundamental frequency of an open-open system. However, I've found that I am one octave below my desired pitch in testing.

If I were to treat it as an open-closed system where the length equals 1/4 of the wavelength, my calculations would work out.

Why is this? Is it a coincidence?

Answer 1

Like with Boomwhackers but striking on the end? Yes, that would cause an octave shift. Strike the tube on its side.

Open ends are nodes for the sound pressure of a standing wave. A closed end is an antinode for the sound pressure.