Why does the first resonance occur at λ/4 in a tube closed at one end?

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Vandalus
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A small loudspeaker connected to a signal generator emits a sound of frequency 425 Hz. It is fixed above a long glass tube that is filled with water and has a drain at the bottom so that the water can be slowly released from it. When the level has fallen 20cm from the top of the tube, a standing wave is formed and the air column resonates.

Calculate the wavelength of the sound.

v = fλ

In my book's worked example it says that:

λ/4 = 20cm therefore the λ must be 80cm.

What I don't understand is this:

How do you know that at 20 cm, the standing wave generated will definitely be the first harmonic/fundamental frequency since λ/4 only applies if the standing wave generated is the fundamental frequency?

Isn't 20cm just an arbitrary number?

Or is it that no matter how much the water level falls (no matter the length of the closed tube) the first input of sound will definitely be the fundamental frequency?

Any help is greatly appreciated :D
 
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Vandalus said:
Or is it that no matter how much the water level falls (no matter the length of the closed tube) the first input of sound will definitely be the fundamental frequency?

Yes. It's the (implied) fact that it did not resonate at any shorter length. Had 20cm been, say, the third harmonic, it would have resonated at 4cm.
 
Ah ok. Looks like I should learn to read questions more thoroughly aha.

Thanks :D