Blowing Across a Bottle: the Helmholtz Resonance

Q: Why does the note get higher as I pour water in?

Because V in the formula is the volume of air, and pouring water in reduces it. f goes as 1/√V, so halving the air doubles nothing but multiplies the frequency by √2 — about six semitones. Quartering the air raises the note by a full octave. If the note came from the glass, adding water would lower it, which is exactly what happens when you tap the bottle with a spoon instead. The two experiments side by side are the cleanest demonstration that there are two separate resonators in one bottle.

Blow across the top of a bottle and it sings one note. That note is not a pipe resonance and it has almost nothing to do with the bottle’s height: it is a mass of air in the neck bouncing on a spring of air in the body, f = (c/2π)√(A/VL). Give the page the volume, the neck and the temperature and it gives the note — with and without the end correction, which moves the answer by about a sixth.

Last updated 11 Nov 2026

Resonant frequency of a bottle, box or cavity

Neck mass on a cavity spring, end correction optional

Volume of air in the cavityIn millilitres, which is the same as cubic centimetres, so a 500 mL bottle is 500 here. It is the air volume, not the bottle volume. Drink half the bottle and the air volume goes up, not down, and the note goes DOWN — which is the opposite of what most people guess and is the easiest prediction to test in a kitchen. The volume enters under a square root, so quadrupling it halves the frequency: an octave costs a factor of four in volume.

Inside diameter of the neckMeasured across the hole, inside the glass. A typical drinks bottle is around 22 mm. The frequency depends on the neck area, which is πd²/4, and area is under the same square root, so the frequency is proportional to the diameter itself: double the hole and you raise the note by an octave. That is why a bottle with a wide mouth sings higher than a narrow-necked one of the same size, and why covering part of the hole with your lip flattens the note.

Length of the neckThe straight part of the neck, from the lip down to where it flares into the body. This is the slug of air that does the moving, so a longer neck means more mass on the same spring and a lower note — again as a square root, so quadrupling the neck length drops the note an octave. On a real bottle the boundary between neck and body is a judgement call, and that uncertainty is usually bigger than everything else on this page put together.

End correction on the neckThis is the convention choice on this page and it is worth more than one mark. The air that oscillates does not stop at the ends of the neck: it drags a little of the air outside and a little inside along with it, so the effective length is longer than the one you measured. UNSW’s acoustics pages advise about 0.6 radii at the outside end and about one radius at the inside end, so 1.6 radii in total, and describe that as very approximate. Switching it on lowers the frequency, by 16% at the default numbers. Most textbook exercises leave it off and most real bottles need it, which is exactly why this page shows both numbers side by side instead of picking one.

Air temperatureSets the speed of sound, which scales the whole answer: the frequency is directly proportional to c, so a 10-degree rise lifts the note by about 1.7%, or 30 cents — audible to a good ear but not to a tuner’s patience. The speed is computed here from the same exact-SI expression as the speed-of-sound record on this site, so the two pages cannot disagree.

The bottle in cross-section, drawn to your numbers. The body grows and shrinks with the air volume, the neck lengthens and widens with the two neck fields, and the dashed red extension below the neck is the end correction — extra effective length that is not made of glass. The arrow across the top is where you blow. The scale on the right is logarithmic from 20 Hz to 2 kHz, with the dot at the note this bottle sings.

183.20HzExample

A 500 mL bottle with a 22 mm neck 50 mm long, air at 20 °C, end correction on

A mass on a spring, where both the mass and the spring are air

f = (c / 2π) √(A / V L_eff) · L_eff = L + 1.6 r · A = πd²/4 · equivalently f = (1/2π)√(k/m) with k = γP A²/V and m = ρAL_eff

the picture, first: the air in the neck is the mass: a slug of air that slides up and down as a lump. The air in the body is the spring: push the slug in and you compress the body, which pushes back. Nothing else matters — not the bottle’s height, not its shape, not what it is made of. That is why a bottle and a box of the same volume with the same neck sing the same note
A: the neck’s cross-sectional area, πd²/4, in m². It appears on top under the root, so f ∝ d: double the hole, raise the note by an octave. Note that A enters the stiffness as A² and the mass as A, which is why it survives as a single power of A in the ratio
V: the volume of air in the cavity, in m³. 500 mL is 5×10⁻⁴ m³. f ∝ 1/√V, so four times the air is an octave down. This is the air volume, not the container volume, and it is the one you change by pouring water in
L_eff: the effective neck length, longer than the one you measured because air just outside and just inside the neck moves with the slug. UNSW’s guidance is about 0.6 r at the open end and about 1.0 r at the inner end, giving L + 1.6 r, and describes the figures as very approximate. The selector above turns this on and off and the page prints both answers either way
c: the speed of sound in the gas filling the resonator. f ∝ c, so the note rises with temperature by about 0.17% per degree. Fill the same bottle with helium and the note rises by a factor of 2.93 — a useful demonstration that the note belongs to the air and not to the glass
what it is NOT: this is not the quarter-wave pipe resonance f = c/4L. A bottle blown across the top does both things at once, but the Helmholtz mode is far lower and far louder, and it is the note you hear. Use the pipe formula on a bottle and you will be out by a factor of several

Worked example

A 500 mL bottle with a 22 mm neck 50 mm long, air at 20 °C, end correction on
EVERYTHING INTO SI FIRST. V = 500 mL = 5.000×10⁻⁴ m³. Neck radius r = 11 mm = 0.011 m, so A = πr² = π × 1.21×10⁻⁴ = 3.8013×10⁻⁴ m². Geometric neck L = 50 mm = 0.050 m. Millilitres to cubic metres is a factor of a million, and it is the step that eats whole answers.
THE END CORRECTION. 1.6 × 0.011 = 0.0176 m, so L_eff = 0.050 + 0.0176 = 0.0676 m. That is 35% more neck than you measured, which should tell you immediately that this is not a detail.
THE SPEED OF SOUND. At 20 °C, c = √(1.4RT/M) = 343.24 m/s, and c/2π = 54.628. That prefactor is the same for every bottle at this temperature, so compute it once.
THE GEOMETRY FACTOR. A/(V L_eff) = 3.8013×10⁻⁴ / (5.000×10⁻⁴ × 0.0676) = 3.8013×10⁻⁴ / 3.380×10⁻⁵ = 11.247 m⁻², and its square root is 3.3537 m⁻¹. Check the units: m² over m³×m is m⁻², so the root is per metre and c × per metre is per second. If your units do not come out as a frequency, you have put a millilitre somewhere.
MULTIPLY. f = 54.628 × 3.3537 = 183.20 Hz. That is close to F#₃, about 1517 cents — a little over an octave and a fourth — below concert A.
AND THE COMPARISON THAT MATTERS. Without the end correction the same bottle gives 54.628 × √(3.8013×10⁻⁴/2.500×10⁻⁵) = 213.02 Hz, which is 16.3% higher — nearly three semitones. That gap is the single biggest reason a bottle you measure does not match a bottle you calculate, and it is bigger than any temperature or measurement error you are likely to make.
CHECK THE ASSUMPTION BEFORE BELIEVING THE ANSWER. λ = 343.24/183.20 = 1.874 m, and the cube root of the volume is 0.079 m. The wavelength is 24 times the size of the bottle, so the pressure really is near enough uniform inside and the lumped model is sound.

Three resonators, checked against published worked examples

The first two rows are not illustrations, they are **the check**. UNSW’s music acoustics pages give both examples with their own answers, and this page reproduces them: 89.9 Hz against “about 90”, and 133.8 Hz against “about 130”. The small differences are the speed of sound — UNSW uses 343 m/s flat, this page computes 343.236 at 20 °C — and rounding in the quoted answers. Note that the first row feeds in the *effective* length with the correction switched off, because UNSW had already added the correction by hand; feeding an effective length into a calculator that is also applying a correction double-counts it, and that is a real and easy mistake.
Resonator	V	Neck area	Neck length used	This page	Source says
UNSW spherical resonator	2920 mL	8.300 cm²	105 mm (effective, correction off)	89.88 Hz	about 90 Hz
UNSW 1-litre bottle	1000 mL	3.000 cm²	50 mm (geometric, correction off)	133.81 Hz	about 130 Hz
This page’s default bottle	500 mL	3.801 cm²	67.6 mm (50 mm + correction)	183.20 Hz	—

What each input does to the note, from the default bottle

Every row is a square root, so nothing responds as fast as you expect except the diameter, which responds twice as fast because area goes as d². **The row to try at home is the second one**: adding water reduces the air volume and raises the note, so a bottle gets higher as you fill it, which is the opposite of what happens when you tap the glass with a spoon. Tapping excites the glass; blowing excites the air. Notice too that switching off the end correction moves the answer more than warming the room by twenty degrees — the convention matters more than the weather. The diameter row is not quite a clean octave because widening the neck also doubles the end correction.
Change	New value	Frequency	Interval
nothing (default)	—	183.20 Hz	—
drink half, air volume doubles	1000 mL	129.54 Hz	down a fifth (−6.0 st)
air volume quartered	125 mL	366.40 Hz	up an octave
neck diameter doubled	44 mm	317.57 Hz	up 9.5 semitones
neck length quadrupled	200 mm	105.12 Hz	down 9.6 semitones
warm it from 20 °C to 40 °C	40 °C	189.35 Hz	up 57 cents
end correction switched off	—	213.02 Hz	up 2.6 semitones

A spring made of air, and why the bottle’s height is irrelevant

Blow across the top of an empty bottle and you get one clear low note. Tap the same bottle with a spoon and you get a different, higher one. The two have nothing to do with each other: the tap rings the glass, and the blow rings the air. What the air does is the interesting part, and it is not a pipe. A pipe resonance needs a standing wave along the tube, with nodes and antinodes and a wavelength comparable to the length. Here the wavelength is twenty times the size of the whole bottle. Nothing is standing. Instead the bottle is a mass on a spring.

Which part is which. The air sitting in the neck is a slug with real mass, and it can slide up and down the neck like a piston. The air in the body is a gas that resists compression, so when the slug moves inwards the body pushes back. Mass plus restoring force is a harmonic oscillator, and the standard result f = (1/2π)√(k/m) applies. Work out the stiffness of the trapped air, k = γPA²/V, and the mass of the slug, m = ρAL, divide, and the pressure and density collapse into the speed of sound: f = (c/2π)√(A/VL). Every bottle dimension except the neck and the volume has vanished, which is why a tall thin bottle and a squat round one of the same capacity and the same neck sing the same note.

The three experiments that prove it in a kitchen. Pour water in: the air volume falls, V falls, the note rises — and if the note came from the glass it would do nothing. Cover half the mouth with your lip: the effective area falls and the note drops. Put a drinking straw into the neck to lengthen it: the note drops again. All three are predicted by the one formula and none of them is predicted by a pipe model.

The end correction, which is where the arguments start. The slug of air does not have sharp ends. Air just outside the mouth and just inside the body gets dragged along with it, so the moving mass is larger than ρAL and the effective neck is longer than the one your ruler says. UNSW quotes about 0.6 radii at the outside end and about one radius at the inside end, and is careful to call the figures very approximate. For a typical bottle neck that is a third again on the length and nearly three semitones on the note, so this is not a refinement — it is most of the difference between a calculation that matches a real bottle and one that does not. This page implements L + 1.6r and lets you switch it off, and it shows you both answers whichever you choose, because textbook exercises and real bottles genuinely want different ones.

Where you have met one without knowing. The port on a bass-reflex loudspeaker is a Helmholtz resonator: the tube is the neck and the cabinet is the volume, tuned deliberately to reinforce the lowest octave. The sound hole of a guitar and the f-holes of a violin form one with the body, and it is the lowest resonance either instrument has. An ocarina is nothing but a Helmholtz resonator with finger holes that change A. A car with one rear window open at speed booms at a few hertz for the same reason. And the perforated acoustic panels used to tame a noisy room are arrays of tiny resonators, each absorbing hardest at its own frequency — which connects this page directly to the reverberation time of a hall.

Where it stops working. The derivation needs the wavelength to be much bigger than the resonator, so that the pressure inside is essentially the same everywhere at any instant. The page prints that ratio for you. It also assumes rigid walls, which is why UNSW warns that a guitar body — which swells as the pressure rises, softening the spring — resonates lower than the Helmholtz calculation predicts. And it ignores losses entirely, so it tells you the frequency but nothing about how loud or how sustained the note is.

Frequently asked questions

Why does the note get higher as I pour water in?

Because V in the formula is the volume of air, and pouring water in reduces it. f goes as 1/√V, so halving the air doubles nothing but multiplies the frequency by √2 — about six semitones. Quartering the air raises the note by a full octave. If the note came from the glass, adding water would lower it, which is exactly what happens when you tap the bottle with a spoon instead. The two experiments side by side are the cleanest demonstration that there are two separate resonators in one bottle.

Is this the same as a quarter-wave pipe, f = c/4L?

No, and treating a bottle as a closed pipe is the commonest error here. The pipe formula needs a standing wave along the tube and gives, for a 25 cm bottle, 343/(4 × 0.25) = 343 Hz. The real note is nearer 180 Hz. The Helmholtz mode is lower because it does not care about the length of the bottle at all — it is a lumped oscillator, not a wave. A bottle does have pipe modes too, much higher up, but the Helmholtz mode is the loud one you hear.

What is the end correction and do I have to use it?

It is the extra effective length of the neck caused by air outside and inside moving along with the slug. UNSW’s guidance is about 0.6 r at the open end and about 1.0 r at the inner end, hence 1.6 r in total, and that source explicitly calls the numbers very approximate. Use it when you are comparing against a real bottle; leave it off when you are reproducing a textbook exercise that didn’t. At the default numbers it is the difference between 183 Hz and 213 Hz, so you cannot simply ignore the question.

Does the shape of the bottle matter?

Barely, as long as there is a clear neck and a clear body. The formula contains the neck area, the neck length and the cavity volume, and nothing else: no height, no diameter, no curvature. A cube, a sphere and a bottle of the same air volume with the same neck all sing the same note to within the accuracy of the end correction. Where shape starts to matter is when the neck stops being distinguishable from the body — a wide-mouthed jar, for instance — because then there is no slug to separate from the spring.

How do I get a lower note out of the same bottle?

Three ways, and all of them are visible in the formula. Increase the air volume, which usually means find a bigger bottle. Lengthen the neck — pushing a short length of straw or tube into the mouth works and is worth trying. Or narrow the opening, by covering part of it. Each of those is a square-root effect, so you need a factor of four in volume or neck length for one octave, but only a factor of two in diameter.

Does the gas inside change the note?

Strongly, and it is the best demonstration that the note belongs to the air. The frequency is directly proportional to the speed of sound, so filling the bottle with helium multiplies the note by 2.93, and filling it with carbon dioxide lowers it. Temperature does the same thing more gently: this page scales the answer with the speed of sound at the temperature you give, which is about 0.17% per degree.

Related calculators

References

School of Physics, University of New South Wales, Helmholtz Resonance (J. Wolfe, music acoustics pages, read 11 November 2026). Gives the resonance as c/2π times the square root of the geometry factor, advises an effective neck length longer than the drilled one by “typically (and very approximately) 0.6 times the radius at the outside end” plus about one radius at the inside end, and works two examples: a spherical resonator of volume 0.002 92 m³, neck area 0.000 83 m² and geometric neck 0.080 m taken to an effective 0.105 m, giving about 90 Hz at c = 343 m/s; and a 1-litre bottle with a 3 cm² neck 5 cm long giving about 130 Hz. This page reproduces both, entering each neck area as the equivalent diameter: 89.45 Hz and 133.81 Hz. The small differences from the source’s own “about 90” and “about 130” are its rounded 0.105 m effective neck against this page’s 1.6r, and its c = 343 m/s against the 343.24 m/s this page computes at 20 °C. That is the check that the formula implemented here is the formula that source states. The page also warns that a flexible body (a guitar) makes the Helmholtz figure an overestimate, and that the end correction is itself only a rough estimate.
NCERT, Physics Textbook for Class XI, Part II, Chapter 14, Waves. The chapter number was checked by an earlier batch in this project against NCERT’s own answers appendix for Part II, which runs Chapters 8 to 14 and puts Waves at 14; older printings numbered it 15 and much of the web still says 15. Cited for the syllabus treatment of the speed of sound in a gas as the Newton formula v = √(P/ρ) with Laplace’s correction supplying γ, giving v = √(γP/ρ) = √(γRT/M). Citing the chapter is not a claim that any board has approved this page.
National Physical Laboratory (UK), Speed of Sound measurement resource (PDF read 11 November 2026). NPL gives the room-temperature working formula c ≈ 331.3 + 0.606 θ m/s with θ in °C, the fuller c = √(γRT/M) which it approximates as 20.047 √(θ + 273.15), and the expected value 343 m/s at 20 °C. This page reproduces all three: the square-root form returns 343.236 m/s at 20 °C and the linear form 343.42, and the 0.606 coefficient is shown below to be c₀/2T₀ at the freezing point, 0.6065, which is where it comes from.
The International System of Units (SI), 9th edition, Bureau International des Poids et Mesures, 2019. The metre is fixed by defining the speed of light in vacuum as c = 299 792 458 m/s exactly, and the Boltzmann and Avogadro constants are fixed exactly too, which makes the molar gas constant R = 8.314 462 618 153 24 J mol⁻¹ K⁻¹ exact. Every speed of sound on these pages is derived from that exact R rather than copied out of a table.