ISEGORIA / MATH ENCYCLOPEDIA
Acoustics: strings, plates and moving sources
Sound is a small pressure wave, and almost everything interesting about it comes from boundaries and motion. Where a string is plucked decides which harmonics it can never play, and a bow sustains a single circulating kink that is neither pluck nor strike. A vibrating plate draws its own standing waves in sand. And a source that moves squeezes its wavefronts ahead of it until, past the speed of sound, they fold into a cone.
Before you begin: The wave equation, Fourier series and standing waves
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. Where you pluck decides what you hear
A string fixed at both ends can vibrate only in modes sin(nπx/L), at frequencies n times the fundamental. Pluck it at a point and the initial triangle is a sum of those modes with amplitudes proportional to sin(nπβ)/n², where β is the pluck point as a fraction of the length: any mode with a node at the pluck point cannot be excited at all. A narrow hammer blow gives every mode the same kick, so the harmonics fall off only as 1/n. A bow does something different again: the string sticks to the bow, then slips back, and the result is Helmholtz motion, a single sharp corner running round two parabolas once per period. Choose the excitation, drag the point along the string, and compare the spectra and the force the string puts on the bridge.
Worked example. A violin A string at 440 Hz plucked at \(\beta=1/5\) has no 5th, 10th or 15th harmonic (2200, 4400 and 6600 Hz), and 43.8% of its vibrational energy is in the fundamental; plucked at the middle, only odd harmonics sound and the fundamental holds \(8/\pi^2=81.1\%\). Bowed at \(\beta=0.1\), the string under the bow moves with the bow for 90% of each period and slips back at 9 times the bow speed, so a bow drawn at 0.2 m/s gives a slip at 1.8 m/s. In ideal Helmholtz motion the fundamental holds \(6/\pi^2=60.8\%\) of the energy whatever \(\beta\) is.
Watch out. The string here is ideal: perfectly flexible, lossless and fixed to rigid ends, so every harmonic is an exact multiple of the fundamental and the shape repeats for ever. Real strings are slightly stiff, which raises the upper partials (strongly so in pianos), and the bridge is not rigid, which is how sound gets out. The Helmholtz corner is drawn perfectly sharp; on a real violin it is rounded by the bow’s width and the string’s stiffness, and Raman showed that the stick phase carries small ripples. The time axis is slowed enormously.
Why is the 5th harmonic missing when you pluck at one fifth of the length?
Mode 5 has nodes at \(x=L/5,2L/5,\dots\), so its shape is zero exactly where the string was pulled. The amplitude of a mode is the overlap \(\int y_0(x)\sin(5\pi x/L)\,dx\) of the initial shape with the mode, and for a triangle this overlap is proportional to the mode’s value at the corner, \(\sin(5\pi\beta)=\sin\pi=0\). The same holds for the 10th, 15th and every multiple, which all share that node.
2. Chladni figures
In 1787 Ernst Chladni scattered sand on a metal plate clamped at its centre and bowed its edge. The plate rang at one of its natural frequencies, the sand danced away from the parts that moved and settled along the nodal lines that stayed still, drawing the standing wave. A square plate with free edges has no exact formula for its modes, but the classic approximation cos(mπx)cos(nπy) ± cos(nπx)cos(mπy) captures the patterns: each product alone has straight nodal lines, and the plus and minus combinations bend them into rings, diagonals and stars. Choose a mode in the gallery, flip the sign, and press Play to sprinkle the sand.
Worked example. The three lowest modes of a free square plate show why the approximation is only a guide. Mode (1, 1), with a cross of nodal lines along the midlines, is the plate twisting. The pair \((2,0)\pm(0,2)\) gives two famous figures: the minus sign makes \(\cos2\pi x=\cos2\pi y\), whose nodal lines are the two diagonals, and the plus sign gives a closed ring. Here both have \(m^2+n^2=4\), twice the twisting mode. Leissa’s accurate values for a plate with Poisson ratio 0.3 put them at about 1.46 and 1.80 times the twisting mode: the free edges split the pair.
Watch out. The patterns are the classic approximation, not exact solutions: the products of cosines have zero slope at the edges, as a free edge needs, but they do not make the bending moment and twisting force vanish there as well, so frequencies are only roughly right and some patterns are distorted near the edges. The sand is a random walk whose step grows with the local amplitude, which is enough to herd it onto the nodal lines; real sand bounces ballistically, and very fine powder can instead collect at the antinodes, carried by air currents, as Faraday showed.
Why do the + and − versions of a mode draw different figures at the same frequency?
Because \(\cos m\pi x\cos n\pi y\) and \(\cos n\pi x\cos m\pi y\) are two different standing waves with the same \(m^2+n^2\), so in the approximation they share a frequency, and any combination of them is also a mode. The symmetry of the square picks out the two combinations that are symmetric and antisymmetric under swapping \(x\) and \(y\), and these draw different nodal lines; the antisymmetric one always has the diagonal \(x=y\) as a nodal line. On a real plate the free edges couple the two products and push the two combinations to slightly different frequencies.
3. The Doppler effect and the Mach cone
A source that moves keeps emitting crests from new places, so the crests ahead of it bunch up and those behind spread out: an approaching siren sounds higher and a receding one lower. What the listener hears depends on the angle θ between the source’s motion and the line to the listener at the moment the sound was emitted, f′ = f/(1 − M cos θ), where M is the source speed divided by the speed of sound. At M = 1 the crests ahead pile up into a single wall. Beyond it the source outruns its own sound, the crests fold into a cone of half-angle μ with sin μ = 1/M, and the listener hears nothing until the cone sweeps past with a boom, then two sounds at once, one of them backwards. Change the Mach number and drag the listener.
Worked example. An ambulance at 25 m/s sounding 700 Hz has \(M=25/343=0.0729\): heard head-on it is \(700/(1-0.0729)=755\) Hz, and going away \(700/1.0729=652\) Hz, a drop of about two and a half semitones. Concorde at Mach 2 had a Mach angle \(\mu=30^\circ\); flying at 15 km, its boom reached a listener on the ground when it was already \(15\sqrt{M^2-1}=26\) km beyond them.
Watch out. The source moves in a straight line at constant speed through still air with a uniform speed of sound, and only the timing of the crests is modelled, not their loudness. The formula uses the angle at emission: by the time the sound arrives the source has moved on, so the angle you would see is different. Real booms come from the pressure of a solid body, are shaped by temperature layers in the atmosphere, and arrive as an N-shaped double crack; and a listener who moves also hears a Doppler shift, by a different formula.
Why does a supersonic plane’s sound arrive backwards after the boom?
Because the plane outruns its sound, sound emitted later can arrive earlier. The arrival time \(t_{\rm arr}(t_e)\) then has a minimum, at the emission for which \(\cos\theta=1/M\), and that minimum is the boom, where sound from a stretch of the path arrives at once. After it each arrival time has two emissions: one later than the boom emission, heard in the right order, and one earlier, for which \(dt_{\rm arr}/dt_e<0\), so the sounds of the earlier approach arrive in reverse order.