
Analysing the Helmholtz resonanceNow let's get quantitative: First of all, we'll assume that the wavelength of the sound produced is much longer than the dimensions of the resonator. For the bottles in the animation at the top of this page, the wavelengths are 180 and 74 cm respectively, so this approximation is pretty good, but it is worth checking whenever you start to describe something as a Helmholtz oscillator. The consequence of this approximation is that we can neglect pressure variations inside the volume of the container: the pressure oscillation will have the same phase everywhere inside the container.Let the air in the neck have an effective length L and cross sectional area S. Its mass is then SL times the density of air ρ. (Some complications about the effective length are discussed at the end of this page.) If this 'plug' of air descends a small distance x into the bottle, it compresses the air in the container so that the air that previously occupied volume V now has volume V − Sx. Consequently, the pressure of that air rises from atmospheric pressure P_{A} to a higher value P_{A} + p.
substituting for F and m gives: So the restoring force is proportional to the displacement. This is the condition for Simple Harmonic Motion, and it has a frequency which is 1/2π times the square root of the constant of proportionality, so
Resonance, impedance, phase and frequency dependenceThis section can be read on its own, but if you want more detailed background, see Oscillations, Forced Oscillations and Acoustic compliance, inertance and impedance. Let's return to the mechanical representation and look at the Helmholtz resonator from the outside, as shown in the first schematic: we are pushing with an oscillatory force F, with frequency f ,the mass m (the air in the neck of the resonator), which is supported on the spring (the enclosed air) with spring constant k, whose other end is fixed (the air in the resonator can't escape). Unrealistically, we'll neglect gravity and friction (for now).
The force required to accelerate the mass is proportional to the acceleration and so proportional to f^{2}. At sufficiently low frequency, the force required to accelerate the mass is negligible, so F only has to compress and extend the spring. So in that limit, applied force F = +kx. (The spring force is –kx.) The mechanical impedance of the system at this low frequency is compliant or springlike. Acoustically, it looks like an acoustic compliance: we push the mass of air in the neck and compress the air in the resonator. The applied force F to the right is in phase with x measured to the right, so it is 90° behind the velocity and the applied pressure is 90° behind the acoustic flow into the resonator. Remember that f^{2} dependence of the acceleration. So, at sufficiently high frequency, the spring force is negligible in comparison with that accelerating the mass. So F is in phase with the acceleration, which puts it 90° ahead of the velocity. Acoustically, the applied pressure is 90° ahead of the acoustic flow into the resonator. It is a mechanical or acoustic inertance. At the resonant frequency (still talking about the sketch above), the amplitude can be large for very small force. So at resonance, the impedance is very low, when viewed from force applied to the mass from outside (the diagram above). At resonance, there is maximum flow into and out of the resonator. Now consider what happens if we look at the pressure inside the resonator (away from the neck). Here we are looking at the force on the spring, so our mechanical analogue looks like this schematic.
This time, low frequency means that the force can be small for a given amplitude: the spring and mass move together as a mass, and the system this time looks inertive at low frequency. At high frequency, the mass hardly moves, and the system is springlike or compliant. So the phase relations are opposite to what we had before: at low f, p leads U by 90°; at high f, p lags U by 90°. Complications involving the effective lengthThe first diagram on this page draws the 'plug' of air as though it were a cylinder that terminates neatly at either end of the neck of the bottle. This is oversimplified. In practice, an extra volume both inside and outside moves with the air in the neck – as suggested in the animation above. The extra length that should be added to the geometrical length of the neck is typically (and very approximately) of 0.6 times the radius at the outside end, and one radius at the inside end).
Helmholtz resonances and guitars* I said above that the air in the body of a guitar acts almost like a Helmholtz oscillator. This case is complicated because the body can swell a little when the air pressure rises inside – and also because the air 'in' the sound hole of the guitar has a geometry that is less easily visualised than that in the neck of a bottle. Indeed, in the case of the guitar body, the length of the plug of the air is approximately equal to the two 'end effects' at the end of a 'pipe' which is only a couple of mm thick. The end effects, however, are related to and of similar size to the radius of the hole, so the mass of air is substantial. The length of the end effect of a cylindrical pipe that opens onto an infinite, plane baffle is 0.85 times the radius of the pipe. Although the soundboard of a guitar is not infinite, one would expect a similar end effect, and so the effective length of the 'plug' of air would be about 1.7 times the radius of the hole. (Some makers increase this by fixing a short tube below the soundhole, with equal radius.)A couple of people have written asking how big the sound hole should be for a given instrument. Well, we can use the equation above to start to answer that question. However, the swelling of the body is important. This makes the 'spring' of the air rather softer, and so lowers the frequency. The purely Helmholtz resonance can be investigated by keeping the body volume constant. When measuring this, a common practice is to bury the guitar in sand, to impede the swelling or 'breathing' of the body. However, guitars are not usually played in this situation. So the Helmholtz calculation will give an overestimate of the frequency of resonance for a real, flexible body. Let's assume a circular sound hole with radius r, so S = πr^{2}, and L = 1.7r as explained above. When we substitute into the equation for the Helmholtz frequency, using c = 340 m/s, we get: It is more complicated when the tone holes are not circular, because the end effect is not equal to that of a circle with the same area. PhD student and luthier John McLennan is writing up a report of some measurements about this, which we'll post here soon. On guitar and violin family instruments, the Helmholtz (plus body) resonance is often near or a little below the frequency of the second lowest string, around D on a violin or GA on a guitar. You can reduce or shift the Helmholtz frequency substantially by covering all or part of the hole with a suitably shaped pieced of stiff cardboard. If you then play a note near the resonance and then slide the card so it alternately covers and reveals the hole, you'll clearly hear the effect of the resonance. Is the 0.85r effect reasonable? Ra Inta, who did a PhD on guitar
acoustics in our lab, suggests an interesting demonstration:
Tuning the Helmholtz resonanceAmong the publications of John McLennan, a PhD student in this lab, is an article in which he varies the Helmholtz resonance by varying the speed of sound.

Happy birthday, theory of relativity!As of June 2005, relativity is 100 years old. Our contribution is Einstein Light: relativity in brief... or in detail. It explains the key ideas in a short multimedia presentation, which is supported by links to broader and deeper explanations. 