Harmonics, air columns, and timbre

A stopper is a wall the air cannot cross, so the displacement s must vanish there: the closed end is a displacement node. Carry across the relation between the two descriptions of a sound wave, p = −B ∂s/∂x, and the same point looks different. The gradient of a sinusoid is steepest exactly where the sinusoid vanishes, so a displacement node is a pressure antinode — the closed end is where the pressure swing is largest, which is why a microphone pressed against it reads the loudest. An open end joins a reservoir of still air that cannot support an excess pressure, so p ≈ 0 there: a pressure node, and therefore a displacement antinode, the place where the air moves furthest. The two pictures describe one mode and sit a quarter wavelength apart in space. Reason with the pressure picture, because the physical statement at an opening is about pressure, not about how far the air moves; then convert once, at the end.

Take one mode of the linear wave equation, s(x, t) = A sin(kx + φ) cos(ωt), and impose the ends. Open at both: antinodes at x = 0 and x = L force φ = π/2 and then |cos kL| = 1, so kL = nπ, λₙ = 2L/n and fₙ = n v / (2L) for n = 1, 2, 3, … Stopped at x = 0: a node there forces φ = 0, and an antinode at x = L needs kL = (2m − 1)π/2, so λ = 4L/(2m − 1) and only the odd harmonics survive. Put numbers in with L = 0.500 m and v = 343 m s⁻¹. The open pipe gives 343, 686, 1029 Hz; the stopped pipe gives 171.5, 514.5, 857.5 Hz. Notice that both spectra are spaced by the same v/(2L) = 343 Hz — the stopped one merely starts at half of it. So the spacing of the peaks measures the length whichever end condition holds, and the position of the first peak tells you which condition it is.

Air just outside an opening is dragged along by the column, loading it with extra inertia, so the pressure node sits slightly outside the tube. The fix is an end correction: L_eff = L + NΔ with Δ ≈ 0.61a for a plain open end of bore radius a, or about 0.82a when the end is flanged into a large baffle. Take the 0.500 m tube again with a = 15 mm. Open at both ends it pays two corrections, L_eff = 0.5183 m, and f₁ = 343/(2 × 0.5183) = 330.9 Hz rather than 343 Hz — flat by 1200 log₂(343/330.9) = 62 cents, well over half a semitone and impossible to miss. Stopped, it pays one: L_eff = 0.50915 m and f₁ = 168.4 Hz, where an exact octave below 330.9 Hz would be 165.4 Hz. The stopped pipe therefore sits 31 cents sharp of the octave the idealisation promises. The correction scales as Δ/L ∝ a/L, so the same 15 mm bore radius in a 2.00 m pipe costs only 16 cents.

The mode shapes are orthogonal: ∫ sin(nπx/L) sin(mπx/L) dx across 0 to L is L/2 when n = m and exactly zero otherwise. So multiply the starting shape by sin(nπx/L), integrate, and every mode but the nth drops out, leaving Aₙ = (2/L) ∫ y(x, 0) sin(nπx/L) dx. One consequence does most of the work: carry that integral through for a string released from a point d and it comes out proportional to sin(nπd/L), so whenever d falls on a node of mode n the projection returns exactly zero and that mode cannot be fed at all. The same projection runs in an air column against each mode's pressure shape, so a small source a fraction p along a pipe open at both ends feeds harmonic n in proportion to |sin(nπp)|. At the middle, p = 0.5, every even harmonic vanishes even though the pipe allows them all. A clarinet's register key is this arithmetic made mechanical: a small hole near a pressure node of the third harmonic barely touches that mode while wrecking the fundamental, so the instrument jumps a twelfth.

Two instruments at the same pitch differ in the set of amplitudes {Aₙ}, not in the frequencies. To rank them fairly use energy, Eₙ = ¼ μ L ωₙ² Aₙ², which is proportional to n²Aₙ² because ωₙ ∝ n — a high partial of a given amplitude carries far more energy than a low one. Quote each partial relative to the fundamental as 10 log₁₀(Eₙ/E₁). Work it for a string plucked one fifth of the way along, where the projection integral gives Aₙ ∝ sin(nπ/5)/n². The amplitude ratios are A₂/A₁ = 0.404 and A₃/A₁ = 0.180, so in energy the second, third and fourth partials come out 1.8, 5.4 and 12.0 dB below the fundamental, and the fifth is missing outright. Move the pluck to the centre and every even partial goes, leaving a hollow tone from the same string at the same pitch. Pitch lives in the spacing of the partials; timbre lives in their relative sizes.

Three assumptions are doing quiet work, and each fails measurably. A real wire resists bending, which adds a restoring term growing with curvature and speeds up the short-wavelength modes: fₙ = n f₁ √(1 + Bn²). With B = 4.0 × 10⁻⁴, typical of a piano, the eighth partial runs 22 cents sharp of 8f₁, which is why tuners stretch the octaves rather than tune them pure. The end correction is itself frequency-dependent, shrinking once ka approaches 1, so a real pipe's upper modes also drift sharp of an exact series. And the open end does not reflect perfectly — that imperfection is the sound reaching you — so every mode has a finite lifetime and a resonance of finite width. Since the radiation resistance of a small opening rises as (ka)², the high partials leak away fastest, and a struck or plucked tone audibly dulls as it decays.