Acoustics & Music Theory
why it sounds like that
Concept

Boundary condition — where it appears

What a wave is required to do at the edge of the region it occupies, which is what selects the modes that region has. Change the condition at one end of a tube and the whole series of resonances changes with it.

Named by 16 essays across 4 fields — each of them below, with the objects they name alongside it.

What a 60 cm tube supports, by how its ends are closed. The first 6 modes of an open cylinder and a stopped cylinder, all of the same acoustic length. An open tube supports every whole multiple of the fundamental and reaches the second mode an octave up. A cylinder stopped at one end supports only the odd multiples and reaches its second mode 1902 cents up, which is a twelfth. Its fundamental is also an octave below the others, because it fits a quarter of a wavelength where they fit a half.

A tube that skips every other partial

Stop one end of a cylinder and half its modes vanish. That single fact about where the pressure has to be decides that a clarinet sounds hollow, that it plays an octave below its length suggests, and that it must cover nineteen semitones with fingers before it can overblow — while every other woodwind covers twelve.

instruments · Air column What a 60 cm tube supports, by how its ends are closed. The first 6 modes of a stopped cylinder and a cone, all of the same acoustic length. A cone supports every whole multiple of the fundamental and reaches the second mode an octave up. A cylinder stopped at one end supports only the odd multiples and reaches its second mode 1902 cents up, which is a twelfth. Its fundamental is also an octave below the others, because it fits a quarter of a wavelength where they fit a half.

A cone is not a cylinder

A saxophone has a reed at a closed end, exactly as a clarinet does, and it overblows at the octave rather than the twelfth. If the previous essay's argument were about reeds that would refute it. It is about geometry, and a cone closed at its apex has the complete harmonic series for a reason that takes one line of algebra and is genuinely surprising.

instruments · Air column The end correction, for a bore of radius 7.5 mm. How flat a tube sounds against what its physical length alone would predict, because the wave carries on past the opening before it turns round. The correction is 4.6 mm at every note — Levine and Schwinger's 0.6133 times the radius for an unflanged end — and the error it causes is 13 cents on a 60 cm sounding length and 52 cents on 15 cm. It is the same millimetres in both cases.

The tube ends after it ends

A wave does not turn round at the opening. It carries on into the room for about six-tenths of the bore radius and reflects there, so every tube is acoustically longer than it is. The correction is a fixed number of millimetres against a wavelength that halves every octave — a rounding error at the bottom of an instrument's range and most of a semitone at the top.

instruments · Air column One register vent, 12 fingerings. Where the second mode's pressure node sits for each fingering of a stopped tube, against a single register hole drilled 12 cm from the mouthpiece. The node is a third of the way along the sounding length, so it moves every time a hole is opened, and the vent's error runs from -38 to -1 cents across the range. A perfect register system would need one hole per fingering. The number of holes actually fitted is one, and the leftover is a design decision rather than a fault.

One hole doing a dozen jobs

A register key works by forcing a pressure node where the second mode already has one, which kills the fundamental and leaves the mode above. The node sits a fixed fraction along the sounding length — and the sounding length changes with every fingering, while the hole stays where it was drilled. The leftover error is computable, and it is why the throat notes are the ones players complain about.

instruments · Tone holes What is struck, and where its partials land. Partial ratios for an ideal string, an ideal membrane, a kettledrum, drawn on a logarithmic axis so that a whole-number ratio is a fixed distance from the last. An ideal string: 1, 2, 3, 4, 5, 6. An ideal membrane: 1, 1.59, 2.14, 2.29, 2.65, 2.92. A kettledrum: 1, 1.50, 1.99, 2.44, 2.89. The membrane's are Bessel zeros — a derivation, and they land nowhere near the whole numbers. The kettledrum's have been pulled toward 2 : 3 : 4 : 5 by the enclosed air, on a fundamental an octave below the lowest partial present.

What a drum is doing instead

An ideal membrane's modes are zeros of Bessel functions — 1, 1.59, 2.14, 2.30 — which support no common fundamental, so a drum rings without a note. A timpani is that problem solved — the kettle's air and the radiation load drag four modes onto 1, 1.5, 2, 2.5, and the pitch a timpanist tunes is the fundamental those four imply and none of them is.

instruments · Missing fundamental The flow through the larynx, over two periods of a 110 Hz note. Volume flow against time, in Rosenberg's two-half-cosine model of the glottal pulse — a slow opening, a faster closing, and a closed phase during which no air passes at all. M1 — chest is open for 50 per cent of each period and opens 2.4 times as slowly as it closes. Nothing here is a displacement: the folds are a valve on a steady stream of air, and the flat stretches are the moments they are shut. At 110 Hz each period lasts 9.1 milliseconds, of which 4.5 is silence.

The other instrument with a reed

The folds do not vibrate the way a string does. They open and shut across a steady stream of air, once per period, and what leaves the larynx is a train of flow pulses with a closed phase in it. Everything said about the voice's tone is a statement about the shape of that pulse — and the shape has two numbers in it.

instruments · The voice Two mechanisms, the notes both of them make, and the seam. The frequency range of each laryngeal mechanism for an adult male voice, on a logarithmic axis, with the band both can produce shaded. M1 — chest runs 82–349 Hz and M2 — falsetto runs 220–698 Hz, so 799 cents of the range — 8.0 semitones — can be sung either way. The two dots inside that band are the measured signature that this is a bifurcation rather than a threshold: the change upward happens at 330 Hz and the change downward at 294 Hz, 200 cents lower. A threshold is crossed at the same place in both directions and this is not.

Two mechanisms, and the seam between them

Every singer has a place in the range where the voice changes character, and eight semitones of it can be produced either way. The measurement that settles what kind of a place it is takes ten seconds: the change upward happens two hundred cents higher than the change downward, and a threshold cannot do that.

instruments · The voice What gets out of an opening, for 4 openings. The fraction of the wave's energy radiated at an open end against frequency, in the baffled-piston model — the radiation resistance of a circular piston, normalised to the tube's own impedance. Each curve runs from nothing at the bottom, where the opening is far smaller than a wavelength and the wave simply turns round, to everything above ka ≈ 2. Half the energy leaves at 6364 Hz for a flute's embouchure end (radius 10 mm), 2015 Hz for a clarinet's bell (radius 30 mm), 975 Hz for a trumpet's bell (radius 62 mm), 403 Hz for a horn's bell (radius 150 mm). The crossover goes as one over the radius, so the widest and narrowest here are 15.8 times apart in frequency. The same number decides how strongly the tube resonates and how much sound it makes, which is why a bell cannot brighten an instrument without also weakening its own resonances.

The bell decides what gets out

A tube resonates because the wave turns round at the open end, and it is audible because some of the wave does not. Those are the same number with opposite signs. One quantity — the size of the opening against a wavelength — decides how loud an instrument is, how bright it is and how directional it is, and a bell moves the boundary rather than removing it.

timbre · Air column One key, nineteen notes, one right answer. A register hole disturbs a mode in proportion to the square of that mode's pressure at the hole, so the place that spoils the fundamental and leaves the third harmonic alone is the third harmonic's own pressure node — a third of the way down whatever length is sounding. The length changes with every fingering and the key does not move, so the two curves cross at one note. At A3 the key sits almost exactly on the node and the twelfth speaks cleanly; at A♭4 it is at 62 per cent of the tube and disturbs the third harmonic by 96 per cent of what an antinode would, which is the throat of the instrument and is exactly where players say the notes are worst.

The hole that spoils a note

A tone hole shortens the tube. A register hole does the opposite job: it is small enough to shorten nothing and is placed where it will wreck the fundamental's resonance and leave the third harmonic's alone, so the note jumps a twelfth instead of retuning. The place that does both is a pressure node of the harmonic being kept — a third of the way along whatever length is sounding — and the length changes with every fingering while the key does not. One key is at the right place for exactly one note, and the note it is worst for is in the throat of the instrument, which is where players say the instrument is worst.

instruments · Tone holes Where the page ends a phrase, and where the ear does. Twinkle, twinkle with two sets of phrase boundaries on it. The lower curve is a local boundary detector — a peak in how much the interval and the note length change from one to the next, with nothing in it about bar lines or harmony — and the marks above it are where the notation puts the phrase ends. It finds 100 per cent of them and 2 boundaries the page does not have. Where the two agree it is because a long note is sitting at the join; where they disagree the page is marking a grammatical unit and the detector is finding a perceptual one.

Where a phrase ends

Run a boundary detector over the three tunes used throughout and it agrees with the notated phrasing on one of them perfectly and on another almost not at all. The reason is which cue each tune uses: Twinkle's phrases all end on a long note, so a duration-weighted detector finds five of five with no false alarms; Ode to Joy's run on in crotchets and its phrasing is in the intervals, where a duration detector finds one of three and a pitch detector finds all three and eight others. No fixed weighting serves both, and the published one is worse on each tune than the single cue that tune uses.

form · Phrase How near the node a hammer has to land. The level of partial 7, relative to the loudest partial of the same note, against how far the hammer is from that partial's node — for a contact 2.6 per cent of the speaking length wide, which is 16 mm on a 620 mm string. At the node exactly the partial is absent whatever the width, because a mode shape is odd about its own node and integrating an odd function symmetrically gives zero. Twenty decibels of suppression needs the hammer within 2.82 mm, and thirty needs 0.89 mm. The width itself barely matters in the musical range: its sinc reaches its first zero only at partial 77.

The hammer is not a point either

The comb of missing partials is computed throughout for a contact of no width, and a piano hammer is sixteen millimetres of felt on a string of six hundred. Integrating the mode shape across the felt turns out not to fill the null in: a mode is odd about its own node, so a symmetric contact centred on it gives an exact zero however wide it is. What destroys the null is being in the wrong place, and the tolerance is 2.8 millimetres for twenty decibels of suppression and 0.89 for thirty. The design decision found at the outset is a manufacturing tolerance.

instruments · Excitation point 4 bores of one length, and the series each supports. cylinder, cone, exponential horn, Bessel horn — every one of them 148 cm long, 5.5 mm at the throat and 62 mm at the mouth, so the only thing that differs is the shape between the two. Beside each is the series of resonances Webster's equation gives it, and the number is how far that series is from evenly spaced, in cents. cylinder: 0.0 cents, with each mode sitting at minus 0.50 of a spacing off a whole multiple; cone: 9.6 cents, with each mode sitting at minus 0.12 of a spacing off a whole multiple; exponential horn: 22.8 cents, with each mode sitting at minus 0.16 of a spacing off a whole multiple; Bessel horn: 3.7 cents, with each mode sitting at minus 0.11 of a spacing off a whole multiple.

Only two shapes make a series

A tube's length sets where its modes are and its shape sets whether they are a series at all — and of every shape a maker could choose, exactly two give evenly spaced modes. Neither of them is the shape of a trumpet. Webster's equation says how far the others miss by, and the answer is a quarter of a semitone for the obvious ones and four cents for the family brass bells actually belong to.

instruments · Bore profile What a trumpet mouthpiece does to the series. Each bore's modes before a mouthpiece is fitted and after, on the axis that says which harmonic each mode is: zero means the m-th mode sits at m times the spacing and is the m-th harmonic. Every shape starts near -0.14 and ends near 0.05. The cup holds 3.0 millilitres against a throat 3.7 mm across and pops at 642 hertz. What does not improve is the regularity: the spacing stays as uneven as the flare left it.

What the mouthpiece is actually for

A cup and a throat look like a comfort fitting and are a component. Put a real one on the front of a real bore in the horn equation and the whole mode series slides sideways by about a seventh of its own spacing — which is exactly the distance between a series whose second resonance is 1.85 times the spacing and one whose second resonance is the second harmonic. The flare decides whether the modes are evenly spaced; the mouthpiece decides which harmonic each one is.

timbre · Bore profile The hand closing, and the note it is holding. The fourth resonance of a 370 cm horn against how much of the bore the hand occludes, at 97 per cent of the way along where the radius is 36 mm. Nothing much happens for the first ninety per cent. The note then falls to -401 cents — most of a fourth — over the next few, and between 99.5 and 99.90 per cent it jumps to 96 cents SHARP, because the lowest resonance has left the bottom of the range and every mode has taken the place of the one below it. The series' ratios are unchanged across the jump.

The hand goes in, and the note jumps

A horn player's hand closes the bell and the pitch falls — 19 cents, then 55, then 132, then four hundred, accelerating the whole way. Then, in the last half per cent of closure, it stops falling and lands a semitone above where it started. An earlier essay guessed the mechanism was the boundary condition changing kind and the series going odd-only. It is not. The series never changes at all.

tuning · Air column The first five peaks, followed as the hand closes. Each line is one member of the series, tracked by its rank rather than by its frequency, and each dot's size is that peak's height. The lowest peak falls from 38 hertz to 34 as the hand closes and then jumps to 45, which is the renumbering computed earlier: past the wall the series is one member shorter at the bottom and every peak has taken the place of the one below it. The dots shrink through the middle of the travel and grow again at the far end, so the transition costs the player support as well as pitch — and the cost is temporary, which is why a fully stopped horn is a usable instrument and a nearly stopped one is not.

A resonance has a strength as well as a frequency

What eleven earlier essays drew is a row of frequencies, because the solver behind it has no losses and a lossless resonance has no width. Put the losses in and every one of them acquires a height and a Q — and the hand closing a horn's bell turns out to take away nine and a half per cent of the instrument's total support before giving all of it back, in a window a few per cent wide where the horn is genuinely hard to play.

tuning · Air column Second order and fourth order, from the same sequence of wavenumbers. Frequency against wavenumber for a string and for a bar, with each object's own wavenumbers marked on the axis. A string's equation is second order in space, so frequency goes as wavenumber and the straight line carries its modes at 1, 2, 3, 4. A bar's is fourth order, so frequency goes as the SQUARE of wavenumber and the parabola carries the same kind of sequence to 1, 2.757, 5.404, 8.933. Both sequences of wavenumbers are arithmetic — a string's are the even multiples of π/2L and a bar's the odd ones, except that the first is 3.011 rather than 3 — and squaring an arithmetic sequence is the whole of why one object has a fundamental and the other has none. Every number is derived from the boundary condition; none is measured.

A bar's partials are the odd numbers, squared

A string's wave equation is second order in space and a bar's is fourth, so a string's frequencies go as its wavenumbers and a bar's go as their squares. Both objects fit a nearly arithmetic sequence of wavenumbers between their ends; squaring one is the whole of why a marimba key has no note in it. The ratios are 1 : 2.756 : 5.404 : 8.933, which are the odd squares over nine, uniformly 12.9 cents flat, and the 12.9 cents is one root that refused to be 3π/2.

instruments · Struck bar

Named alongside it

The objects these essays reach for when they reach for this one.

BoreStanding waveHarmonic seriesRegisterNormal modeOverblowingPartialResonanceBrassEnd correctionHorn equationInharmonicity

All concepts