Helmholtz resonator revisited

We finished a bottle of wine this evening, and I blew across the top as I often do. (Don’t worry: I only do this at home. If we’re ever in a restaurant together, I won’t embarrass you by blowing across the neck of an empty bottle.)

The pitch sounded lower than I expected, so I revisited some calculations I did last year.

As I wrote about here, a wine bottle is approximately a Helmhotz resonator. The geometric approximation is not very good, but the pitch prediction usually is. An ideal Helmholtz resonator is a cylinder attached to a sphere, and a typical wine bottle is more like a cylinder attached to a larger cylinder. But the formula predicting pitch is robust to departures from ideal assumptions.

As noted before, the formula for the fundamental frequency of a Helmholtz resonator is

f = \frac{v}{2\pi} \sqrt{\frac{A}{LV}}

where the variables are as follows:

  • f, frequency in Hz
  • v, velocity of sound
  • A, area of the opening
  • L, length of the neck
  • V, volume

The opening diameter was 2 cm, the neck length 9 cm, and the volume 750 cm³. All these are typical. The predicted frequency is f = 118 Hz. The measured frequency was 106 Hz, measured by the Sonic Tools phone app.

The actual frequency was about 10% lower than predicted. This is about a whole step lower in musical terms. I could certainly hear an interval that large if I heard the two pitches sequentially. But I don’t have perfect pitch, and so I’m skeptical whether I could actually notice a pitch difference of that size from memory.

One thought on “Helmholtz resonator revisited

  1. Diethard Michaelis

    German Wikipedia
    gives a resonance frequency formula with end correction
    (End-Korrektur der Resonatorlänge)
    which results in a lower frequency:
    L is replaced by (L + diameter*pi/4) i.e.
    9 is replaced by (9+2*pi/4) = 10.57 and sqrt(9/10.57) = 0.92
    which is quite close to the observed 106/118 = 0.90

    Typo: 2nd para, 3rd sentence: An idea Helmholtz … -> An ideal … [missing l]

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