- Introduction
- Lumped Parameters
- The Cylindrical Pipe
- Branching Pipes
- The Conical Pipe
- Combined Systems
- References

A close analogy can be established between the propagation of sound in pipes and chambers and electrical circuits, which is a great aid in acoustical problems of many kinds, since all the electrical circuit theorems may be applied. For example, the design of loudspeakers makes good use of the circuit analogy. When the dimensions of the region in which the sound propagates are much smaller than the wavelength, the flow is almost incompressible, and a lumped-parameter model is appropriate. On the other hand, a long pipe is the acoustical analogue of an electrical transmission line, specified with distributed parameters.

We'll consider only propagation in air, but the results apply to water and other media as well, with only a change in the properties of the medium. Air has a density of about ρ = 1.293 kg/m^{3} at STP (0°C, 1 atm), and the speed of sound is c = 331.5 m/s under those conditions. At room temperature, air is a little lighter, and sound moves a little more rapidly. Sound consists of very small departures from the atmospheric pressure of 1.013 x 10^{5} Pa, less than about 100 Pa at the most. We will use p to represent this *overpressure*, or departure from the static pressure. The change in density resulting from p may be expressed as p = ρc^{2}s, where s, the *condensation*, is the relative change in density, Δρ/ρ. Since ρc^{2} = 1.42 x 10^{5} Pa, we see that density changes are also very small.

For these small overpressures and condensations, the dynamical equations are very close to linear, so acoustic pressure obeys the principle of superposition. This permits us to take advantage of the properties of harmonic variations that depend on time through e^{jωt}. Fourier's theorem assures us that we can build up any solution as a superposition of harmonic solutions. Also, we can employ frequency-dependent parameters that sidestep the difficult problem of time variations. This, of course, is also done in electrical circuits.

The important acoustic variables will be taken to be the pressure p, and the *volume velocity* Q, which is the particle velocity times the area of flow, or Q = vA. It is clear that Q is a conserved quantity, from the conservation of mass combined with the approximate constancy of the density. It would be more rigorous to use the mass rate of flow, but the volume rate of flow is equivalent in our problems. The product of p and Q has dimensions (N/m^{2})(m^{3}/s) = watt, and is a power, just like the product of V and I.

For the electrical analogy, we take p ↔ V and Q ↔ I. Since the product VI is power, as well as pQ, this is consistent. If p and Q are *phasor* quantities, the ratio p/Q is analogous to the impedance Z = V/I, and is called the *acoustic impedance* Z' (we'll use primes to distinguish acoustical quantities from electrical ones). Then, Z' = R' + j(ωL' - 1/ωC'), which defines acoustical resistance, inductance and capacitance. Distinctive names have been suggested for these quantities, but use of the electrical ones aids clarity. The phasors p and Q may be either peak or rms quantities, which differ only by a factor of √2. In terms of peak quantities, the average power P = Re[pQ*/2] = Re[p*Q/2] = |p||Q|cos θ. As in electrical circuits, cos θ is the power factor, where θ = tan^{-1}(X'/R'), and X' = ωL' - 1/ωC' is the *acoustic reactance*. There is no definite relation between V and p, but once a correspondence has been chosen, all other phasor and energy relations are determined. Just as in the electrical picture, the circuit parameters are usually independent of frequency, which makes analysis straightforward.

I use MKS units in this article, but it would be just as easy to use cgs units, since only 1 m = 100 cm and 1 kg = 1000 g are involved in any conversion. The μbar, or dyne/cm^{2}, has been the traditional pressure unit in acoustics, though the pascal has made some headway. 1 Pa = 10 μbar, so even this conversion is easy. The confusion that arises in electromagnetism does not appear here. It is only necessary to be aware of the units at all times, and to make whatever conversions are required. All of the equations are the same in consistent MKS and cgs systems.

Let us first consider a system of passageways and volumes whose maximum dimensions are much less than a wavelength. Our lumped parameter results will be approximately valid if this dimension is taken as less than λ/4. At 1000 Hz, a quarter-wavelength is 8.25 cm, or 3-1/4". The essential thing here is that the phase is roughly constant throughout the system. Exactly the same restrictions apply to electrical circuits, but here the phase velocity is much higher, and the wavelengths correspondingly longer, so the lumped-parameter approach is valid for most circuits at all reasonable frequencies.

Consider the air in a section of length L of a pipe of area A. Since all quantities are in phase, it moves as a rigid body with displacement ξ (a phasor) under the action of an unbalanced force pA, the difference in the forces on its ends. Since its mass is ρAL and the acceleration is -ω^{2}ξ, Newton's Law gives pA = -ω^{2}ρALξ. The velocity is v = jωξ, so p = jω(ρL/A)Q, or Z' = jω(ρL/A). Therefore, we identify the acoustical inductance as L' = ρL/A. Acoustical inductance is the consequence of the kinetic energy of the air. This energy is, in fact, on the average (1/2)(ρLA)Re[ω^{2}ξ^{2}] = (1/4)(ρL/A)|Q|^{2}, or Average[(1/2)L'Q^{2}], analogous to (1/2)LI^{2} for the electrical quantities.

Consider the air in a rigid container of volume V. Air can enter or leave at rate Q from an opening in the container, compressing or rarefying the air in the container, without producing significant velocity. The volume change of the air is ΔV (not a change in the size of the volume), so that the condensation s = -ΔV/V. In terms of Q, ΔV = Q/jω. Using the compressibility of air, ρc^{2}, we find that p = ρc^{2}Q/jωV. By comparison with the form of Z', we conclude that C' = V/ρc^{2}. Again, this is completely analogous with electrical capacitance, and the energy is the average of (1/2)C'p^{2}. Both L' and C' correspond to energy storage, not to its increase or dissipation.

In viscous flow, the particle velocity is proportional to the pressure gradient, say v = kp/L, or Q = vA = (kA/L)p. Therefore, the acoustic resistance R' = L/kA. The parameter k is the *permeability*. In cgs units, its dimensions are cm^{3}/s-g, or *darcy*. 1 darcy = 0.001 MKS unit, or m^{3}/s-g. Since p = R'Q, the power dissipation is P = Q^{2}R'. If Q is a peak value phasor, this must be divided by 2 for the average power, as usual.

If the element is a capillary tube, the flow is viscous if the Reynolds number N = ρvd/η is sufficiently small (less than 2000). η is the dynamic viscosity of air, 1.702 x 10^{-5} kg/m-s, which is approximately independent of pressure. Poiseuille's formula for flow in a capillary is Q = (πa^{4}/8ηL)p, where a is the radius of the capillary of length L. This gives the acoustic resistance immediately, R' = πa^{4}/8ηL. For a capillary of 1 mm diameter and 1 cm long, R' = 6.93 x 10^{6} kg/s-m^{4}. A bundle of 127 such capillaries has a total flow area of 1 cm^{2}, and R' = 5.46 x 10^{4} acoustic MKS ohms.

This element, of total area 1 cm^{2} and length 1 cm, also has acoustic inductance, which can be estimated at L' = ρL/A = 129.3 MKS units. At 1 kHz, the reactance will be 8.12 x 10^{5} acoustic MKS ohms. The resistance and inductance should be considered to be in series, though it looks as if the same pressure is applied to each. Actually, they experience the same Q. At high frequencies, the inductance dominates the resistance, while at low frequencies the resistance is active and is not short-circuited by an inductance.

We now have three kinds of circuit elements: acoustic resistance, inductance and capacitance, and can build many circuits by combining them. A simple circuit is shown in the diagram, and its analogous circuit at the right. This is a Helmholtz Resonator. The resonator looks like a Florence flask, and in fact a small Florence flask makes an excellent Helmholtz resonator. The pressure outside the resonator is zero, which is the reference pressure. The pressure p is at the neck of the resonator, with the capacitive volume V in one direction, and the inductive tube L of area A in the other. From this circuit, we find its parallel admittance Y = jωC' - 1/jωL' = 0/p must be zero to have a nonzero Q (the input Q to the circuit is zero). Therefore, ω^{2}L'C' = 1, from which we find the resonant frequency f = (c/2&pi)1/√(L'C'). Substituting the values of the parameters, f = (c/2π)√(A/VL). Since the velocity is nonzero for some distance beyond the end of the neck in both directions, a corrected value of L can be used. Experiment suggests that L' = L + 16a/3π, where a is the radius of the neck. For a 125 ml Florence flask, I measured L = 53 mm, a = 13.5 mm, and V = 130 cc. Using L' = 64 mm and c = 343 m/s, the resonant frequency was calculated to be 226 Hz. By comparison with a piano, the flask sounded the note A', or 220 Hz. Such close agreement is fortuitous, but shows that we are certainly on the right track.

The main energy loss mechanism is radiation from the mouth of the radiator. If this is assumed to radiate like a simple source in an infinite baffle, the radiation acoustic resistance is R' = ρck^{2}/2π = ρω^{2}/2πc. The rate of radiation is QR', of course, where Q is the volume velocity at the mouth. The quality factor Q (not to be confused with the volume velocity Q!) is Q = ωL'/R'. Substituting the values of the parameters, Q = 2π√(L^3V/A^{3}). For the 125-ml flask, this gives Q = 106, a typical value. The circuit analogy for a Helmholtz resonator is shown in the diagram. The resonator is driven by the pressure p_{a} applied at the mouth. The pressure amplitude p inside the resonator will be Q times the applied pressure, so the Helmholtz resonator is a sensitive detector of the frequency to which it is tuned, since Q is typically around 100, so the gain is about 40 dB. Helmholtz used this to very good effect in his study of physiological acoustics. For this use, the resonator is made with a small nipple that can be placed in the ear, or connected to the ear by tubing.

We assume that the pipe has a maximum transverse dimension much less than a wavelength, say smaller than a quarter-wavelength of the frequency transmitted, but its length may be indefinitely great. Because of the length, the phase is not uniform along the pipe, and we must take it into account. The cross-sectional shape of the pipe is not important; we simply require that it be compact and uniform along the pipe. Any bends in the cylinder are to be gradual; that is, they occupy many wavelengths. In fact, we will usually think of a straight circular cylinder of radius a for purposes of argument.

We shall assume that conditions are uniform in any cross-sectional plane, and that the particle displacement is parallel to the axis of the pipe. For sufficiently low frequencies, only such modes will be propagating modes. For all other modes, the pipe will be a high-pass filter. The uniform mode will propagate down to zero frequency, so it is the acoustical analogue of the electromagnetic TEM mode in a transmission line. There are, indeed, higher modes with transverse nodal surfaces in pipes, but we shall not consider these. As in the electrical case, they require field calculations. We can specify the TEM mode by a single coordinate, which can be chosen in many ways.

Let ξ(z,t) be the displacement of the fluid initially at the coordinate z. Then the fluid in a slice of thickness dz finds itself in a slice of thickness (1 + ∂ξ/∂z)dz when disturbed by the wave, so that the condensation s = -∂ξ/∂z, and the pressure p = -ρc^{2}(∂ξ/∂z). The force per unit area acting on a slice of thickness dz is -(∂p/∂z)dz and its mass is ρdz, so its acceleration is ∂^{2}ξ/∂t^{2} = (1/c^{2})(∂^{2}ξ/∂z^{2}). This is the one-dimensional wave equation with phase velocity c.

We note that this is exactly the wave equation for propagation of a plane wave in unlimited space, so we conclude that in pipe propagation a limited portion of wavefront is selected, which propagates in the pipe without spreading or attenuation, so long as we neglect the small effects at the surface of the pipe. This conclusion is very closely true, and explains the utility of speaking tubes and other such applications. Speaking tubes are (or at least were) of great use on ships. They consisted of a uniform tube with a plug in each end in the form of a whistle. When you wished to attract the attention of the party at the other end, you pulled out the plug and blew into the tube, causing a whistle at the other end. This was the original "blower."

We now assume harmonic solutions ξe^{jωt}, where ω is the angular frequency 2πf. If this solution is substituted in the wave equation, we find that tbe equation for the amplitude ξ becomes d^{2}ξ/dx^{2} + k^{2}ξ = 0 (the one-dimensional Helmholtz's equation), with solutions e^{-jkz}, where k is the propagation constant, with dimensions m^{-1}. Also, k = ±ω/c. Therefore, k is also 2πλ, where λ is the wavelength. Therefore, our harmonic solution is now ξe^{j(ωt - kz)}, where ξ is a constant phasor. If k is positive, the wave is travelling in the +z direction; if k is negative, the wave is moving in the opposite direction. A complex k can also handle the matter of attenuation if necessary.

Now differentiation with respect to the time becomes multiplication by jω, and differentiation with respect to space is multiplication by -jk. The pressure, therefore, is p = jρc^{2}kξ (the exponential factor is understood). The particle velocity is v = jωξ, so the ratio p/v = ρc. This ratio is the *wave impedance* z, which is real, showing that p and v are in phase (or 180° out of phase) in a plane wave. For a pipe of area A, we are interested in the ratio p/Q = p/(vA) = ρc/A, which is the acoustical impedance we discussed above. This is the *characteristic impedance* of the pipe. The average energy flow in the pipe is (1/2)Re(pQ*) = (|p|^{2}/2ρc)A. The factor in parentheses is just the intensity of a plane wave in W/m^{2}.

An ideal electric transmission line has a series inductance L, and a shunt capacitance C, per unit length. The phase velocity is c = 1/√(LC) and the characteristic impedance is Z_{o} = √(L/C). A unit length of the acoustic pipe will have an inductance L' = ρ/A and a capacitance C' = A/ρc^{2}, which we find simply by setting the length L = 1. Then, by the analogy, the phase velocity is 1/√(1/c^{2}) = c, and the characteristic impedance is ρc/A, precisely what we found in the preceding paragraph.

Let us consider the end of the pipe at z = 0, and the source at z = -L. In general, there will be waves travelling in both directions, so that p = Ae^{-jkz} + Be^{jkz} and Q = (1/Z_{o}')(Ae^{-jkz} - Be^{jkz}). The ratio of these two quantities is the acoustical impedance at the point z. At the termination z = 0, then, we have p = A + B and Q = (A - B)/Z_{o}'. If the pipe is terminated by an impedance Z, then Z = Z_{o}'[(A + B)/(A - B)]. If Z/Z_{o} = r, then this equation can be solved for the ratio B/A: B/A = (r - 1)/(r + 1). If the pipe is terminated in its characteristic impedance (which is an acoustic resistance), then r = 1 and B/A = 0. That is, there is no reflected wave. This is a well-known property of electrical transmission lines.

Let us now suppose that the end of the pipe is open. This means that p/Q = 0, so r = 0, and B/A = -1. The pressure is then p = A(1 - 1) = 0, as required, but the volume velocity is a maximum, at 2A/Z_{o}. Waves are totally reflected in this case. On the other hand, if the end of the pipe is shut, Q = 0, so Z = ∞, r = ∞, and B = A. The wave is again totally reflected, but at z = 0 p = 2A and Q = 0. We might have thought that the wave would simply flow out of an open end of the pipe, but this does not happen to any degree. The wave is reflected by the discontinuity at the end of the pipe instead. In order to couple the pipe effectively with the free air, we need a *horn* to reduce the discontinuity by giving the pipe something to work into. Otherwise, reflection reduces the pressure at the end of the pipe to a small value, which radiates very poorly.

Sound is attenuated in pipes by several mechanisms. Perhaps the smallest effect is absorption in the volume of the medium, which we shall assume is air. Classical absorption, due to viscosity and heat conduction, is quite small and usually negligible, since it is proportional to the square of the frequency. In moist air, absorption due to vibrational relaxation in oxygen catalyzed by the water vapor can be considerable, up to about 0.01 neper/m. Some energy can be lost in the elasticity of the pipe, and by sound propagating into its mass, but this is also quite small if the pipe is reasonably rigid. The largest contribution is probably from viscosity and heat conduction at the surface of the pipe, in the boundary layer. If we calculate the losses due to viscosity alone, and use an effective viscosity that includes the effect of heat conduction, we will get the losses due to both. The dynamic viscosity of air is about 1.702 x 10^{-5} kg/m-s, and the effective viscosity is 1.93 times greater, or 3.28 x 10^{-5} kg/m-s.

The wall attenuation in neper/m is α = (1/ac)√(ηω/2ρ), where a is the radius of the pipe, or α = 2.76 x 10^{-5} √f/a. The equivalent acoustic resistance per metre is R' = (1/πa^{3}c)√(2ρηω). Substituting the values of the parameters, this is R' = 2.22 x 10^{-5} √f/a^{3}. The amplitude attenuation in a transmission line is α = cR'C'/2, where R' and C' are per unit length.

The neper unit refers to exponential attenuation of, say, the pressure according to p = p_{o}e^{-αx}, where αx is the attenuation in neper. The intensity varies as I = I_{o}e^{-2αx}. Since dB = 10 log(I/I_{o}), dB = 8.686 x neper. For example, 0.01 neper/m is 0.08686 dB/m or about 87 dB/km. Note that neper refers to amplitude or pressure attenuation, not power attenuation, unless otherwise stated, while dB always refers to power.

The wall attenuation in a speaking tube of 5 cm (2") diameter will be about 0.0349 neper/m, or 30.3 dB/100 m, by the above formula, at 1000 Hz. This gives some idea of the effective range of a speaking tube. The neck of the Helmholtz resonator discussed above, with a = 0.00675 m, and at 226 Hz, has R' = 1085 MKS per m, or R' = 57.5 MKS. This is to be compared to the radiation resistance of 1254 MKS. The wall attenuation is indeed negligible, as we claimed. Measured acoustic attenuation is notoriously greater than theoretical values, but at least our numbers are lower limits.

Absorption also has a small effect on the phase velocity, just as the series resistance has an effect in an electrical transmission line, and the effect can be calculated from the circuit analogy, using the formulas for the electrical transmission line. The result is v = c[1 - (1/2a)√(2η/ρω)]. At STP, v = c(1 - 1.421 x 10^{-3}/a√f). For our speaking tube of 5 cm diameter, v = 0.998c at 1000 Hz, and v = 0.992c at 50 Hz.

If a pipe is terminated at both ends so that the reflection coefficient has magnitude unity, it is clear that the length of pipe will form a resonant system at certain frequencies. The amplitude of vibration is maintained by *blowing* the pipe. In the case of organ pipes, the pipe can be considered open at the end at which it is blown, while the other end can be *open*, or *stopped*. An open pipe will have a pressure node at both ends, so it will be a half-wavelength long. A stopped pipe will have a pressure node at the open end, and a pressure loop at the closed end, so it will be a quarter of a wavelength long. In both cases, these are the lowest, or *fundamental* notes sounded by the pipes. An open pipe will also resonate at all harmonics of the fundamental frequency f, at 2f, 3f and so on. The stopped pipe will resonate only at the odd harmonics, 3f, 5f and so on. This gives stopped and open pipes different qualities. The exact length of pipe for a certain note depends on how it is blown, and other factors, so the simple estimates are only approximate. In any case, an "end correction" of about 0.6a should be added to the length at an open end, where a is the radius of a cylindrical pipe.

Another way of looking at resonance is to consider pressure pulses moving along the pipe. These pulses represent wave packets that are a superposition of waves of different frequencies. Since the propagation is nondispersive, the pulses will remain sharp, without spreading, and move with the group velocity, which here is the phase velocity c. The boundary condition at the end of an open pipe is that the pressure is zero, while at the end of a closed pipe the velocity must be zero. In the diagram, consider the pulse marked t = t_{1}, which is travelling to the right with velocity c. The arrow marked v shows the direction of the particle velocity, which is in phase with the pressure. Now imagine a similar pulse, but of the opposite pressure, approaching from the right at velocity c, which will arrive at the end of the pipe at the same instant as the positive pulse. The situation at one time while the pulses overlap is shown, and the dot represents the zero pressure at the boundary. This pressure will remain zero as the pulses pass by. When the original pulse is completely beyond the boundary, the reflected pulse will be completely inside, and will reach the position marked t = t_{2} at a later time. A positive pulse is reflected with opposite amplitude, as a negative pulse, or pressure is replaced by rarefaction. If the pulses outside the boundary were real, and there were really no boundary, then the pressure would remain zero at the original boundary, and conditions inside the pipe would be exactly the same.

The bottom part of the diagram shows the case when the boundary condition is v = 0. Now the pressure at the boundary rises to twice the pressure in the pulse, but the velocity is always zero. A positive pulse is reflected as a positive pulse, a negative as a negative. These properties of pulse reflection are good to know. They show, for example, that a positive pulse in air reflected at a water surface returns as a positive pulse, while a pulse from in the water reflected at the surface returns as a negative pulse.

Now suppose a positive pulse sets out from one open end of a pipe towards the other open end. A time t = L/c is required to reach the other end. It is reflected as a negative pulse, which then returns down the pipe in another interval of L/c. This negative pulse is reflected as a positive pulse, and we are back exactly where we started, having gone through a complete cycle of events. The time taken is 2L/c, so the frequency is c/2L, corresponding to a wavelength of 2L. If we launch n pulses in a time 2L/c, the frequency will be nc/2L.

Let's repeat the experiment, but with the other end of the pipe closed. The pulse is reflected as a positive pulse and returns in a time 2L/c. However, it is reflected at the starting end as a negative pulse, so things are not back to what they were. In another 2L/c, the negative pulse returns, and is reflected as a positive pulse. Therefore, things have repeated in a time 4L/c, so the frequency is c/4L and the wavelength is 4L. By stopping the pipe, we have doubled the wavelength and halved the frequency. Explanation of the harmonics is left as an exercise for the reader. In this case, the frequencies are 3c/4L, 5c/4L and so on. Note that we cannot launch another positive pulse at the instant the first one returns, since they would cancel each other.

For example, an open pipe two feet long resonates at a wavelength of 4 x 0.3048 m = 1.219 m, or f = 272 Hz (neglecting end corrections). The 2-foot pipe was indeed the standard for musical scales, and differences arose because of the differing lengths of the foot in different countries. Its note was taken as the piano's middle C, in the usual notation c'. The octave beginning with c' was called the "two-foot" octave. A conference of German scientists at Stuttgart in 1834 decided on c' = 264 Hz, which is still the general musical standard. The German foot was a little longer than the English foot. The physicist took c' = 256 because that was a power of 2, so all the octaves of C would come out even. The frequency of a' (a "fifth" above c') is more often quoted; the Stuttgart agreement makes a' = 440, the French Academy adopted a" = 435, and the physicist used a' = 427. Over the years, a' has varied from 370 to 567 Hz, with churches and the military favoring higher values, voices the lower. Mozart used a' = 421.6. Much classical music is now performed at pitches higher than intended by its composers, causing extra strain on voices. Musical harmony depends only on relative frequencies, so absolute pitch is of no significance. Incidentally, most of these pitches were preserved as tuning forks.

Note that the pitch of a pipe depends on the temperature through the dependence of the speed of sound on temperature. An increase by 33°C from 0°C sharpens the pitch of a pipe by a full semitone. All music instruments that depend on the resonance of an air column are similarly affected. If the pitch is determined mainly by the resonant frequency of a reed, the frequency will not change, as with the harmonica. Usually, there are means to adjust the pitch slightly to make up for temperature variations. Thermal expansion has a negligible effect on wind instruments, but it can decrease the tension in a string, thereby flatting its note. A large change in temperature can put an orchestra seriously out of tune, if no steps are taken to retune.

Pipes may be sounded, or blown, by a jet of air that strikes a sharp edge, as in organ pipes and the flute. An organ flue pipe is shown in the diagram. Vortexes are shed by the jet in a regular way, creating the "edge tone" when the jet meets the sharp edge. The edge tone is then amplified by resonance. This method favors the fundamental note, and gives a soft tone. The method of blowing changes the resonant frequency of the pipe by a small amount. Another method of blowing uses a *reed*, so called because the original was a length of reed slit at one end and pressed lightly between the lips. The interaction of the vibrating air with the reed is in proper phase to reinforce the vibration by admitting puffs of air. Reeds may also be of metal or wood, just closing an aperture. A metal reed usually has only one frequency of vibration (as in a mouth harmonica), and gives a bright tone. Reeds create sharp puffs containing many harmonics, something that is usually desirable. The lips of a player can control a reed to select different modes of vibration. Finally, the lips themselves may form a double reed, often with the aid of a cup-shaped mouthpiece. The resonances of the pipe control the frequencies, but the player can select which frequencies are sounded. A pipe blown by air is called a *flue* pipe, one by a reed is a *reed* pipe. Incidentally, a reed pipe usually has a pressure antinode at the reed, instead of a pressure node, as in a flue pipe.

A steam whistle is shown at the left. The steam enters from the bottom, passing into a volume that supplies an annular jet. The jet strikes the sharp lip of the bell of the whistle, and the edge tone excites stopped-pipe vibrations. The frequency is approximately f = c/4L, where the velocity of sound in steam (M = 18) should be used. If we take T = 473 K, c = 555 m/s, a whistle sounding 660 Hz would have a length L = 21 cm, or about 8.26". The whistle can be blown with air, in which case the frequency will be only 395 Hz.

If two organ pipes of the same length stand side-by-side, they form two coupled oscillators that can vibrate in symmetrical and antisymmetrical modes. In the symmetrical mode, the pipes are in phase and sound a slighly lower note. In the antisymmetrical mode, the resonant frequency is a little higher, and the pipes are in opposite phase, air leaving the top of one at the same time that air is sucked into the top of the other, so that the air is passed from tube to tube. When such pipes are blown at the same time, they rapidly enter the antisymmetric mode and are practically silent, though the air in each tube is in full vibration.

The circular, or annular, resonator shown in the diagram can be considered as two open organ pipes joined end to end. Since a node is no longer enforced at the end of the pipes, the fundamental wavelength is equal to the circumference of the radiator. A fundamental mode excited with a velocity node at c, perhaps at a diaphragm D for exciting the resonator, has another node diametrically opposite at point a. There are velocity loops at b and d. The air rushes back and forth between a and c. In any mode, the nodes and loops are spaced at λ/4.

Examples of such a resonator may be formed by architectural passages forming a closed loop, but since the circumference of such a loop may be large, the fundamental frequency will be infrasonic. For example, a loop 500 ft in length will resonate at about 2.2 Hz. Such circuits may explain certain observations of infrasonic signals, since they will occur in sewers and mines as well as in buildings.

The human voice may be likened to a reed pipe, with the vocal cords in the *larynx* corresponding to the reed, and producing the *larynx tone*, which is modified by the *mouth tone* made by varying conditions at the other end of the pipe represented by the throat, mouth and nose. In whispering, only the mouth tone is heard. The mouth tone produces the high frequencies that carry the information, while the larynx tone provides the power. Whistling produces tones in a completely different way. Air exiting between the lips forms an unstable jet, which excites a resonator composed of the mouth and the passage between the lips, the frequency controlled mainly by the volume of the mouth cavity as adjusted by the tongue. The lips do not vibrate like a reed, as Rayleigh determined by experiment.

As an example of a resonating pipe as a music instrument, consider the *Waldhorn*, or *French horn*, a long brass tube usually wound into a coil for convenience. The shape is loosely called conical, but it is really a hyperbolic shape. The 4-metre length of the horn, an open pipe, has its fundamental resonance at about 40 Hz, a very deep E flat. The fundamental, and the octave at 80 Hz, are not used. Useful notes begin at the next octave, e flat at 160 Hz, which is the 4th harmonic. The next octave is the 8th harmonic, so there are only four notes available in the octave above e flat. The 8th harmonic is e' flat, 320 Hz, and in its octave there are eight useful notes, up to the 16th harmonic at e" flat, 640 Hz. This abundance of higher harmonics is used in many musical instruments. Most of the resonant pipes used in musical instruments are open pipes, which generate all harmonics.

The composer for the French horn had to be careful to use only those notes the horn could sound. By stuffing his hand into the bell of the horn, the player could sharpen a note a half-tone or so, and this helped a little. Finally, piston valves were applied that added different lengths to the horn so that enough notes became available to play anything. Another means of doing this was with a slide, which was actually used on trumpets and other horns, but has survived only in the trombone. It is very fortunate that a horn player need play only one note at a time, selecting it carefully from a large heap of wrong ones.

A different method of varying the pipe length is used not only in flutes, but also in instruments descended from whistles, which have holes in the side of the pipe. A hole, as we have seen, is an acoustical inductance shunting the transmission line. Its inductance is so small that it acts as a short circuit, causing a nearly total reflection and changing the resonant length of the pipe. The precise effect of these fingering holes is difficult to analyze, but the gross effect is clear. Note that sound is radiated from the holes, since there is substantial Q going in and out there.

A special and interesting way of exciting a resonator is by means of heat. This requires that heat be added or extracted at the proper phase of the oscillation, and at the proper locations. The *singing flame*, devised soon after the discovery of hydrogen gas, is illustrated in the diagram, with typical dimensions in mm. The resonator is the open-open tube, which in its fundamental has a velocity node at its centre. A small hydrogen flame is introduced into the bottom of the tube, and when it reaches some position like that shown, the resonator breaks into continuous sound. A tube of these dimensions should sound about an E flat above middle C. Whether the arrangement oscillates or not depends very critically on the length of the supply tube to the flame. As shown, it should be less than a quarter wavelength in hydrogen (about 1 m in this case) at the frequency of oscillation. The flame does not sing with a longer supply tube until its length increases by a wavelength. Other inflammable gases, such as natural gas, can also be used, but it is more difficult to produce oscillation.

When the flame is observed stroboscopically (as was done first by Wheatstone) it is found to vary at the frequency of oscillation, even descending into the supply tube at the pressure minima. Its phase with respect to the oscillation is very important, so that on the average it delivers more heat when the pressure is high at the node, and less when the pressure is low, encouraging the oscillations. A role is also played by the current of heated air in the tube, that continually brings in cool air. The interplay of these conditions makes the most effective position of the flame some distance below the node. It is not an easy phenomenon to analyze, and gave rise to an extended discussion.

A related effect that makes an even better demonstration is Rijke's tube, which I call his "Boomer." It was discovered in 1859, and is quite reliable in operation, not requiring the fiddling associated with singing flames. The gas flame is used to heat a doubled iron gauze placed in the lower part of the tube to bright red heat. When the flame is withdrawn, the tube breaks out into a very loud sound, in this case at around 100 Hz. This remarkable sound continues as long as the gauze is hot, about 10 seconds. In this case the excitation depends sensitively on the upward current of air. The hot gauze is placed where there are both pressure and velocity variations, and the net effect is to add heat in the proper phase. If the hot gauze is placed near the top of the tube, resonance cannot be excited. However, Rijke showed that if this gauze were cold instead, oscillation was again excited. The steady current in this case is downward.

These singing tubes are excellent illustrations of the creation of oscillations by forcing a mechanical oscillator in the proper phase by forces depending on its own oscillation, which is a closed feedback loop of gain greater than unity. Heat is only one agency that may be active in these systems, and it acts rather strongly on gases. I have a saucepan that oscillates on a hot burner, reminiscent of Trevelyan's Rocker, an old demonstration.

Some interesting problems involve branching pipes, which are analogous to connecting transmission lines. A pipe is characterized by its cross-sectional area, which multiplies the particle velocity to get the volume velocity. At a junction of several pipes, the pressure p must be the same in all pipes, while the sum of the volume velocities must be zero. Generally, in each pipe there will be waves propagating in opposite directions that can be specified by their peak overpressures. If the waves are harmonic, then the acoustical impedance Z' can be used to relate pressure and volume velocity. For a wave travelling to right (towards larger z) p = Z'Q, while for a wave travelling to the left, p = -Z'Q. This minus sign must always be used. It expresses the fact that the particle velocity is in the direction of propagation for a positive pressure. For a pipe, Z' = ρc/A. In these problems, ρc is usually a common factor and can be understood.

We can look at waves that are not necessarily harmonic by using the vector potential φ to describe the wave instead of the pressure. The meaning of the vector potential is explained in Soundwaves. The particle velocity is the gradient of the vector potential, so for our one-dimensional problems, v = ∂φ/∂x. The pressure is given by p = -ρ∂φ/∂t. For a wave travelling to the right, φ = φ(z - ct). Therefore, ∂φ/∂z = φ' and ∂φ/∂t = -cφ', so that both v and p can be expressed in terms of the derivative φ' of φ with respect to its argument, as v = φ' and p = ±ρcφ'. This is a very convenient description, and it is easy to keep the signs correct.

The pipe of AB of area A_{1}in the diagram branches at the point B into branch BCE of area A_{2} and branch BDE of area A_{3}, which reunite at point E. At the junction B, pressures must be the same in all three pipes, and the sum of the volume velocities must be zero. Lower case φ represents waves travelling to the right, while Φ represents waves travelling to the left. We assume that there are no waves approaching B from the right in the branches. In terms of the vector potentials, the pressure condition is φ_{1}' - Φ_{1}' = φ_{2}' = φ_{1}', while the volume velocity condition is (φ_{1}' + Φ_{1}')A_{1} = φ_{2}'A_{2} + φ_{3}'A_{3}. These equations yield Φ_{1}' = [(A_{2} + A_{3} - A_{1})/(A_{2} + A_{3} + A_{1}]φ_{1}', and φ_{2}' = φ_{3}' = [2A_{1}/(A_{1} + A_{2} + A_{3})]φ_{1}. The only properties of the pipes that enter here are their areas. The three equations allow us to determine three unknown vector potentials at this three-way junction. If n pipes meet at a junction, we will have n equations and can find n vector potentials.

If the area is constant, A_{1} = A_{2} + A_{3}, then Φ_{1}' = 0, and so there is no reflection, and φ_{2}' = φ_{3} = φ_{3}', so all the vector potentials are equal. If the lengths of the branches BCE and BDE are equal, and the area of the pipe EF is equal to the area of AB, then the waves recombine at E to the original wave. The branching has had no effect. This is easily seen be dividing AB and EF longitudinally into equal areas; then the waves pass through with no reflection.

The boundary conditions at the end of an open or a closed pipe of p = 0 or v = 0 are better realized when the diameter of the pipe is small compared to the wavelength. A rigid cap gives v = 0 quite well, but the zero pressure at the end of an open pipe is less rigorous, leading to the well-known end corrections for an open organ pipe. It is very easy to see from the vector potential that an open end leaves φ unchanged, but its sign is reversed at a closed end, from the relation of the vector potential to the velocity and pressure. We have assumed nothing about the time dependence in the vector potential analysis, only that it depends on the combination z ± ct, so our analysis applies to arbitrary pulses as well as to harmonic waves.

Now let's consider harmonic waves of angular frequency ω, with wavelength λ = 2πc/ω. If branch BCE is of a different length than branch BDE, the waves will arrive at E in different phases, complicating conditions at the junction. This problem can be attacked exactly like the analogous electromagnetic problem. If branch BCE is an even number of half wavelengths longer than BDE (that is, an integral number of wavelengths), the phases will be the same as when the distances were equal, so the waves will recombine without reflection and continue along EF. However, if BCE is an odd number of half-wavelengths longer than BDE, the phase will be exactly reversed, equivalent to multiplying the vector potential by -1, and so they will add to give φ = 0, or p = 0, at E. The superposition (called "interference" for historical reasons) has resulted in the annihilation of the wave.

Interference cannot create or dissipate energy, only move it around, so the energy in the waves must be reflected at E, move backwards through ECB and EDB, recombine at B, and return through BA. It should be quite clear that we have total reflection, and Φ = φ as on reflection from an open end. Indeed, dividing the pipe longitudinally into two as we did above shows the the waves in each branch are individually reflected at E, where p = 0. When one says that the two waves moving along BCE and BDE meet at E and annul each other, it should be remembered that there is more to the story, and that the waves are totally reflected.

If we are dealing with pulses instead of harmonic waves, the pulse will arrive first at E on the shorter path, and be partially reflected there, and partially transmitted to the other branch as well as on EF, presuming that the whole pulse arrives before the pulse on the longer branch. When the pulse travelling along the longer branch arrives, it will be treated the same way. There will be successive reflections to complicate matters, especially in the ideal case when the attenuation is zero, so pulses will bounce back and forth until the energy is all dissipated (not absorbed, however, but sent towards A and F).

The conical pipe or horn is another system for which we can find an exact solution. Just as the cylindrical pipe cut a cross section out of a plane wavefront, and the boundary condition of zero normal velocity at the surface of the pipe was satisfied, a cone cuts a cross section out of a spherical wavefront, and the boundary condition on the velocity is again satisfied, because the velocity is radial. Therefore, spherical wave solutions like p = (A/r)e^{j(ωt - kr)} for a wave propagating toward +r are valid, and |k| = ω/c = 2π/λ. We note that the conical horn has no low cutoff frequency, as does the exponential horn.

The trick to analyzing a conical pipe is to measure distance z, a coordinate corresponding to r, from the vertex of the cone of which it is a section. If we have a pipe of radius a_{1} at one end, and a_{2} at the other, and of length L, then z_{1} = a_{1}L/(a_{1} + a_{2}). If m = (a_{2} - a_{1})/L, then the cross-sectional area is A(z) = πm^{2}z^{2}.

The ratio of pressure to velocity, the wave impedance, is that of a spherical wave, or p/v = ρc(1 + j/kz)[k^{2}z^{2}/(1 + k^{2}z^{2}]. If we divide by A(z), we find the acoustic impedance at any position z. The acoustic impedance is higher at the small end, or *throat*, and lower at the large end, or *mouth*, so the conical horn is an impedance transducer.

Waves can travel in both directions in the horn, so that in general p = (Ae^{-jkz} + Be^{jkz})/z (times e^{jωt}, of course). The ratio B/A is the reflection coefficient at the mouth. If the diameter of the mouth is not small compared to a wavelength, B/A will be small, and the horn will be an efficient radiator. For most practical conical horns, this condition is not well satisfied, because it would require a very long horn. The *exponential horn* is as short as possible for a large exit end, so it is normally used instead of a conical horn. Nevertheless, conical horns can do exactly the same thing.

If the mouth of the horn is small, B/A will be close to -1 for an open horn, and *standing waves* will be established in the horn. Then, p = (A/z)sin k(z - z_{1}) will be a typical solution with a pressure node at the throat. Since the pressure must also be zero at the mouth, k(z_{2} - z_{1}) = nπ. This gives the condition (2/λ)L = n, or λ = 2L/n, and f = n(c/2L), the same as for an open cylindrical pipe. The resonances are not affected by the flare of the horn. A curiosity is that the pressure loops are not halfway between the nodes. Taking the derivative of p with respect to z and setting it equal to zero, we find tan ku = ku + kz_{1}, where u is z - z_{1}, the distance from the mouth. Only when kz_{1} is very large, so that tan ku is very large, and ku = π/2 or u = L/2 for the fundamental.

We have mainly been concerned with conical pipes that are frustums of a cone, but some pipes may be actual cones, and include the vertex (such as the oboe). Standing waves will then be of the form p = (A sin kz + B cos kz)/z. Since z = 0 is included, we must have B = 0. There must be another mode of p at the mouth, so kL = nπ, from which λ = 2L/n, as for an open-open cylindrical pipe. Note that the vertex of the pipe does not act as a closed end. On the other hand, if the mouth z = L is closed, p must be a maximum there. Taking the derivative of p with respect to z and setting it equal to zero, we find tan kL = kL, a familiar transcendental equation. The smallest root is kL = 4.49340946 = 1.430π. Succeeding roots are 2.46π, 3.47π, 4.48π, 5.58π, and so on. The fundamental wavelength is 1.398L rather than 2L, and the overtones are not multiples of the fundamental. The differences between the frequencies, however, soon becomes a constant, c/2L. A superposition of these overtones may be perceived as this low fundamental frequency.

Rayleigh shows that for a conical pipe of length L and an average z of Z, the wavelengths of the normal modes for a closed-closed pipe are λ = (2L/n)[1 - 4(L/nπZ)^{2}] instead of λ = 2L/n for a cylindrical pipe, provided Z is much greater than L. The effect of conicality is only of second order. If Z is not much greater than L, we must solve tan kL = kL/[1 + (k^{2}/4)(Z^{2} - L^{2}] for k. If Z = L, as is approximately true for the cheerleader's horn, then this becomes tan kL = kL, as above. Of course, the cheerleader's horn is not closed.

We have seen that pipes open (or closed) at both ends are resonant at all harmonics of the fundamental, but pipes closed at one end and open at the other are resonant only at odd harmonics of the fundamental. It is often assumed that the clarinet, with a cylindrical bore, resonates best at the odd harmonics, while the oboe, with a conical bore, resonates at all harmonics of the fundamental. This difference are principally due to the end conditions, not to the shape of the bore. Indeed, flutes have a cylindrical bore, but emit all the harmonics because the mouth hole is a pressure node. In a clarinet, the reed acts as a closed end. In the oboe, the reed end acts as an open end because the bore is conical. The detailed situation is actually a bit more complicated. In the clarinet, the even harmonics are generally stronger than the odd harmonics, but both occur to some degree. The first three harmonics of the oboe have been observed to be very weak, but the higher harmonics are present, the 4th and 5th especially strong. The fundamental in a musical note is often weak or absent, but filled in mentally by the aural sense. The flute is one of the rare instruments to make a note composed of a strong fundamental and successively weaker overtones in the classic picture.

It is interesting to compare our exact solution with an approximate method that can handle pipes of arbitrary dependence of cross-sectional area on distance along the pipe. A pipe is shown at the right where the area is A(z). The air in a distance dz where the area is A moves to a distance (1 + dξ/dz)dz to where the area is A' = A + (dA/dz)ξ, approximately. ξ is the particle displacement of the air. The change in volume of this air is ΔV = (A + dA/dz)(1 + dξ/dz)dz - Adz = (dA/dz)ξ + [A(dξ/dz) + (dA/dz)(dξ/dz)]dz. Although dz is infinitesimal, ξ need not be. However, we presume that the nonlinear term is small compared to the rest, so that the change of volume is approximately ΔV = [d(Aξ)/dz]dz.

We can now relate the change in volume to the change in pressure, through p = ρc^{2}s = ρc^{2}(-ΔV/V). ΔV is given above, and V = Adz. Therefore, p = -(ρc^{2}/A)[d(Aξ)/dz]. The force on a slice of air of thickness dz is -A(dp/dz)dz, and its mass is ρAdz, so Newton's Law gives us (ρAdz)(d^{2}ξ/dt^{2}) = ρc^{2}A (d/dz)[(1/A)d(Aξ)/dz]dz. Removing identical factors gives d^{2}ξ/dt^{2} = c^{2}(d/dz)[(1/A)d(Aξ)/dz]. Since ξ is a function of both x and t, all these derivatives should be partials, of course. If A is a constant, as in a cylindrical pipe, then this reduces to the ordinary wave equation, with its well-known solutions.

For a conical pipe, A is proportional to z^{2}, so the wave equation we have just derived becomes ∂^{2}ξ/∂t^{2} = c^{2}(∂/∂z)[(1/z^{2})(∂/∂z)(z^{2}&xi)]. If z is replaced by r, then the right-hand side is just div grad ξ for ξ(r,t), so we recover the wave equation in three dimensions. That is, the approximate method gives the exact wave equation. If y(z,t)/z is substituted for ξ, we obtain ∂^{2}y/∂t^{2} = c^{2}∂^{2}y/∂x^{2}, so that y(z,t) satisfies the ordinary wave equation. Rayleigh calls this Euler's equation.

If the time dependence is e^{jωt}, the amplitude equation becomes (d/dz)[(1/A)d(Aξ)/dz] + k^{2}, where k = ω/c. For the conical horn, A is proportional to z^{2}, so we get (d/dz)[(1/z^{2})d(z^{2}ξ)/dz] + k^{2}ξ = 0. Now let ξ(z) = y(z)/z. The amplitude equation becomes d^{2}y/dz^{2} + k^{2}y = 0, of which the solutions are y = e^{±jkz}. We recover the solutions ξ = (1/z)(Ae^{-jkz} + Be^{jkz}).

Consider a wave moving toward +z, with ξ = (A/z)e^{j(ωt - kz)}. The pressure in this wave is p = -ρc^{2}[dξ/dz + (2/z)ξ]. This expression differs slightly from that for spherical waves. Now, dξ/dz = -jkξ - (1/z)ξ, so p = -ρc^{2}[-jk + (1/z)]ξ. The velocity v = jωξ, so p/v = ρc[1 + j/kz]. This differs from the exact result by the factor k^{2}z^{2}/(1 + k^{2}z^{2}). If kz >> 1, this factor is practically unity. The explanation for this difference seems to be that the wavefronts are plane in the approximate solution, but are spherical in the exact solution. Since z is the radius of curvature of the wavefront, kr is 2π times the ratio of the radius of curvature to the wavelength. If this ratio is large, the approximate and exact methods give the same results. This factor could be used to correct for the assumption of plane wavefronts, if the actual curvature could be estimated.

The approximate method is used to analyze the exponential horn in Speakers. The analysis of cylindrical and conical pipes above is valid for any size of pipe relative to the wavelength, not just for pipes whose diameter is small compared to a wavelength, as even Rayleigh states. However, a small pipe propagates only the analogue of the TEM mode, so higher modes are guaranteed to be absent (actually, rapidly attenuated).

Acoustical systems may be combined with mechanical systems, coupled by the mutual forces between them. An excellent example is the loudspeaker, where an electrical system is further combined with the mechanical. It is well-known that mechanical systems can also be described by a circuit analogy, with force ↔ V and velocity ↔ I. Force times velocity is power, of course. The ratio of force to velocity for time variation e^{jωt} is the *mechanical impedance* Z, with MKS units of kg/s, the MKS mechanical ohm.

If a mechanical system has mass m kg, stiffness s N/m, and resistance r N-s/m, and moves under an applied force f N in the direction of increasing x, then the equation of motion is m(d^{2}x/dt^{2}) = f - r(dx/dt) - sx. In the frequency domain, v = jωx, so this equation becomes jωm = f - rv - (s/jω)v, or f = (r + jωm + s/jω)v. Therefore, the mechanical impedance is Z = r + j(ωm - s/ω), and r is analogous to resistance, m to inductance, and 1/s to capacitance. Stiffness s is defined by f = -sx, while *compliance* c = 1/s is defined by x = -cf.

A simple but instructive example of coupled mechanical and acoustical systems is shown in the diagram. The mechanical system consists of a diaphragm of area A and mass m, suspended at its periphery by a support that contributes stiffness s and resistance r. An external applied force f drives the diaphragm. Its displacement is x and its velocity is v. The acoustical system consists of a volume V part of whose periphery is formed by the diaphragm of area A, which produces a volume velocity Q = Av. The pressure, acting on the diaphragm, exerts a force f' = pA on the mechanical system. Of course, the mechanical system exerts an equal and opposite force on the acoustical system.

The mechanical impedance Z relates the total force acting on the diaphragm to its velocity through f - f' = Zv. Therefore, f = Zv + pA. The acoustical impedance Z' relates the pressure to the volume velocity through p = Z'Q. Therefore, f = Zv + AZ'Q. However, Q = Av, so f = Zv + A^{2}Z'v, or f = (Z + A^{2}Z')v. The combined system behaves as a mechanical oscillator with mechanical impedance Z + A^{2}Z'. In this case, Z' = 1/jωC', and C' = V/ρc^{2}, or A^{2}Z' = A^{2}ρc^{2}/jωV. The mechanical impedance is then r + jωm + (1/jω)[s + A^{2}ρc^{2}/V]. The resistance and mass are the same as in the absence of V, but now the stiffness has been increased by A^{2}ρc^{2}/V. In the circuit analogy, the compliance c' = V/A^{2}ρc^{2} has been connected in series with the original compliance c = 1/s.

The equivalent mechanical circuit of the combined systems is shown in the diagram. It is easy to see that the resonant frequency has been raised, and easy to calculate by how much. Acoustical impedances have been converted to mechanical impedances by multiplication by A^{2}, and the variables are f and v. A is, in general, the area of interaction of the two systems. One may, of course, go the other way and convert mechanical to acoustical impedances by dividing by A^{2}. Then the variables are p and Q. In the case of moving-coil speakers, the conversion factor is (BL)^{2} instead of A^{2}, as explained in Speakers, and the impedance transferred into the electrical system is called the *motional impedance*. In this case, we have three coupled systems with different variables: electrical, mechanical and acoustical.

Lord Rayleigh, *The Theory of Sound*, 2nd ed., Vol. II (New York: Dover, 1945). Conical pipes are treated in Art. 281, p. 114.

L. E. Kinsler and A. R. Frey, *Fundamentals of Acoustics*, 2nd ed. (New York: John Wiley & Sons, 1962. See especially Chapters 7-10.

H. von Helmholtz, *On The Sensations of Tone* (New York: Dover, 1954). The translation by Ellis of Helmholtz's *Tonempfindungen* of 1877. Clarinets and oboes are mentioned on p. 99.

Sir W. Bragg, *The World of Sound* (New York: Dover, 1968; republication of the 1920 edition). Royal Institution Christmas lectures from 1919 covering the world of acoustics.

A. Wood, *Acoustics*, 2nd ed. (New York: Dover, 1966).

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Composed by J. B. Calvert

Created 16 September 2003

Last revised 30 January 2009