University Physics I · Optional extension · Institutional Extension · Waves & Sound · WS.9
Harmonics & Musical Sources
Boundary conditions pick which frequencies a source can hold; the way you excite it picks how loud each one is. Here is how strings and pipes build a harmonic series, and why two instruments on the same note sound nothing alike.
Build the model
Connect the measurement to the mechanism.
A musical source has a definite pitch because its boundary conditions select modes whose frequencies are integer multiples of one another. A string clamped at both ends and a uniform air column both fit whole numbers of half or quarter wavelengths, so fₙ = n f₁, and any superposition of those modes repeats exactly every T₁ = 1/f₁ no matter what the mix is. Pitch comes from that period.
Which end conditions apply decides the set: fixed–fixed and open–open give every n, while one closed end gives odd n only and halves the fundamental. What the modes do not fix is how much of each is present. That is set by the excitation — where a string is plucked, how a reed or a jet drives a column — and it is those relative amplitudes, together with the attack transient, that make a clarinet and a violin on the same note sound different.
Pitch is the spacing of the lines in the spectrum; timbre is their heights.
- Simple definition
- The harmonics of a musical source are its normal-mode frequencies, fₙ = n f₁, selected by the boundary conditions at its ends. The fundamental f₁ sets the pitch; the relative amplitudes of the higher harmonics set the timbre.
- Example
- A 0.500 m pipe open at both ends resonates at 343, 686, 1029 Hz … when v = 343 m s⁻¹. Plug one end and the set becomes 171.5, 514.5, 857.5 Hz — an octave lower, with every even harmonic gone.
0.650 m at 329.6 Hz → v = 428 m s⁻¹; with μ = 0.40 g m⁻¹, FT = 73 N.
L in m, FT in N, μ in kg m⁻¹; v is the string's speed, not the air's
L = 0.500 m and v = 343 m s⁻¹ give 343, 686, 1029, 1372 Hz.
Displacement antinode at each end; v is the speed of sound in the gas
The same 0.500 m pipe gives 171.5, 514.5, 857.5 Hz — an octave down, evens gone.
Displacement node at the closed end, antinode at the open end
L = 0.500 m, a = 0.020 m, open–open: f₁ = 327 Hz, not 343 Hz — 81 cents flat.
a is the bore radius; a flanged end is closer to 0.82a
Pluck at x = pL: Aₙ ∝ sin(nπp)/n², so p = 1/5 erases the fifth harmonic.
A half-range Fourier sine series; mode orthogonality does the work
B = 4 × 10⁻⁴ puts the eighth partial 22 cents sharp of 8f₁.
B ≈ 10⁻⁴ to 10⁻³ for piano wire; B → 0 is the ideal flexible string
Why these sources have a pitch at all
A standing wave survives only where the boundary conditions are met, and for a uniform string clamped at both ends or a uniform air column the allowed wavelengths come out in the ratio 1 : 1/2 : 1/3 : …, so the frequencies come out as fₙ = n f₁. That integer relation is what makes the radiated sound periodic: add sinusoids at f₁, 2f₁ and 3f₁ with any amplitudes and phases you like and the sum still repeats exactly every T₁ = 1/f₁. A repeating waveform is what the ear reports as a pitch. Nothing guarantees the relation. A circular drumhead's modes sit at 1, 1.59, 2.14, 2.30, … times its lowest, ratios fixed by the zeros of Bessel functions rather than by integers, so the sum never repeats and the drum has colour but only a vague pitch. Standing waves and resonance came earlier; the question here is which set of frequencies a real musical source ends up with, and how loud each member of the set is.
The string: a node forced at each end
Clamping both ends forces a displacement node at each, so a whole number of half wavelengths must fit: λₙ = 2L/n and fₙ = n v/(2L) with n = 1, 2, 3, …. The speed is the string's own, v = √(FT/μ), so f₁ = (1/2L)√(FT/μ) and every tuning control follows from that one line. A guitar's top E sounds 329.6 Hz over a 0.650 m scale, so v = 2Lf₁ = 428 m s⁻¹, and a string with μ = 0.40 g m⁻¹ needs FT = μv² = 73 N to sit there. Fretting at the twelfth fret halves L and doubles every fₙ: one octave. Raising the tension by 1% raises the pitch by only 0.5%, because f ∝ √FT — which is why a fine tuner has to move so little. Quadrupling μ halves the pitch, and is how a bass string stays short instead of long and slack. Every n is allowed, so the ladder runs 1 : 2 : 3 : 4 : 5, whose successive steps are an octave, a fifth, a fourth and a major third.
Air columns: an open end and a stopped end are different boundaries
In a sound wave an open end is very nearly a pressure node, and therefore a displacement antinode; a closed end is a displacement node and a pressure antinode. Sketch the displacement wave and the two cases separate at once. A pipe open at both ends needs an antinode at each, giving λₙ = 2L/n and fₙ = n v/(2L) with n = 1, 2, 3, …: for L = 0.500 m and v = 343 m s⁻¹ that is 343, 686, 1029 and 1372 Hz. Stop one end and the fit becomes a node at one end and an antinode at the other, which takes an odd number of quarter wavelengths: λₙ = 4L/n and fₙ = n v/(4L) with n = 1, 3, 5, …. The same 0.500 m pipe now gives 171.5, 514.5 and 857.5 Hz. Two things changed together: the fundamental dropped an octave, so a stopped pipe sounds as low as an open pipe of twice the length, and every even harmonic vanished. A clarinet is close to this case. It overblows to a twelfth, 3f₁, not an octave, and its lowest note near 147 Hz implies an effective length 343/(4 × 147) ≈ 0.58 m.
Where the ideal pipe model bends
The pressure node does not sit exactly at the mouth of a pipe. Air just outside is dragged along with the column, so the tube behaves as though it ran slightly further, by about 0.6a at an unflanged end of bore radius a. Take L = 0.500 m and a = 0.020 m. Open at both ends there are two corrections, so L(eff) = 0.524 m and f₁ = 343/(2 × 0.524) = 327 Hz rather than 343 Hz — 4.6% flat, or 81 cents, close to a semitone. Stopped, only one correction applies: L(eff) = 0.512 m and f₁ = 167 Hz. Any pipe that is not thin compared with its length has to be measured, not merely cut to a formula. Temperature shifts the whole spectrum as well, since v ≈ 331√(1 + T/273) m s⁻¹ in air: warming a wind instrument from 15 °C to 25 °C sharpens it by 1.7%, near 30 cents, while a string, whose speed depends on tension and not on the air, does not follow.
Which harmonics actually appear
The boundary conditions hand you a list of allowed frequencies and say nothing about their amplitudes. Those come from how the source is started. Release a string from a shape y(x,0) and expand that shape in the modes: y(x,0) = Σ Aₙ sin(nπx/L), with Aₙ = (2/L) ∫₀L y(x,0) sin(nπx/L) dx. Pluck it into a triangle whose corner sits at x = pL and the integral returns Aₙ ∝ sin(nπp)/n². Two audible consequences follow. The 1/n² makes a pluck low-harmonic-heavy, and the sine factor punches holes in the series: at p = 1/5 the amplitudes run 1, 0.40, 0.18, 0.063, 0, 0.028, and the fifth harmonic is gone because the pluck point is one of its nodes. Plucking at the midpoint kills every even harmonic. Plucking near the bridge makes p small, so sin(nπp) ≈ nπp and the fall flattens to 1/n, leaving many more strong harmonics — the same string at the same pitch, sounding thin and bright.
Timbre is the spectrum plus the transient
Plot amplitude against frequency and a sustained note becomes a line spectrum: one line at each n f₁, its height set by the excitation and by how well the body radiates that frequency. Those heights are the main physical content of timbre, and a clarinet's near-missing even harmonics stand out in the plot before you hear them. The steady spectrum is not the whole story. Cut the first 50 ms from a recorded piano note and it becomes hard to name, because the attack — hammer noise, breath noise, bow scratch, the moment before the modes settle — carries much of the identity. Real sources also drift from exact harmonicity: a stiff piano string follows fₙ = n f₁√(1 + Bn²), and with B = 4 × 10⁻⁴ the eighth partial lands 1.3%, or 22 cents, sharp of 8f₁, which is why piano octaves are tuned stretched. Pitch even survives losing f₁: partials at 200, 300 and 400 Hz still repeat every 10 ms and are heard at 100 Hz, which is how a small loudspeaker suggests a bass note it cannot radiate.
Change one variable at a time
Make the relationship visible.
Drag the pluck point to 0.20 and watch the fifth bar drop flat onto the axis — the pluck has landed on that harmonic's node — then shorten L and watch the whole ladder slide outward with no bar changing height.
Fundamental f₁ = v/2L329 Hz
Fifth harmonic 5f₁1646 Hz
Second harmonic A₂/A₁0.465
Fifth harmonic A₅/A₁0.103
Live interpretationFundamental f₁ = v/2L: 329 Hz. Fifth harmonic 5f₁: 1646 Hz. Second harmonic A₂/A₁: 0.465. Fifth harmonic A₅/A₁: 0.103
Catch the common trap
Explain before calculating.
A pipe 0.500 m long and open at both ends has a fundamental of 343 Hz, with v = 343 m s⁻¹. One end is now plugged. What is the new fundamental, and the next resonance above it?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA steel string is clamped at both ends over a speaking length of 0.630 m. Its mass per unit length is μ = 1.20 g m⁻¹ and it is tuned to a tension of 88.0 N. Find the wave speed on the string, the fundamental frequency, and the frequency of the fourth harmonic. Name that harmonic as an overtone.
- The speed belongs to the string, not to the air: v = √(FT/μ) = √(88.0 N ÷ 1.20 × 10⁻³ kg m⁻¹) = √(7.3333 × 10⁴ m² s⁻²) = 270.80 m s⁻¹.
- Both ends are clamped, so a displacement node sits at each and the fundamental fits exactly one half wavelength: λ₁ = 2L = 2 × 0.630 = 1.260 m.
- f₁ = v/λ₁ = 270.80 m s⁻¹ ÷ 1.260 m = 214.92 Hz, so 215 Hz.
- A fixed–fixed string admits every n, so fₙ = n f₁ and f₄ = 4 × 214.92 = 859.69 Hz, or 860 Hz.
- Name it carefully: 860 Hz is the fourth harmonic and the third overtone, because overtones are counted upward from the fundamental.
Answerv = 271 m s⁻¹, f₁ = 215 Hz, f₄ = 860 Hz — the fourth harmonic, which is the third overtone.
MediumA cylindrical pipe is 0.440 m long with an internal bore radius a = 0.0150 m. Take v = 343 m s⁻¹. Ignoring end effects, find the fundamental and the first overtone when the pipe is open at both ends, and again when one end is stopped. Then apply the unflanged end correction 0.6a to the stopped pipe and say how many cents flat it lands.
- Open at both ends means a displacement antinode at each, so a half wavelength fits: λ₁ = 2L = 0.880 m and f₁ = v/λ₁ = 343 ÷ 0.880 = 389.77 Hz, so 390 Hz.
- Every n is allowed there, so the first overtone is the second harmonic: 2f₁ = 779.55 Hz, or 780 Hz.
- Stopping one end forces a displacement node there and an antinode at the mouth, so only a quarter wavelength fits: λ₁ = 4L = 1.760 m and f₁ = 343 ÷ 1.760 = 194.89 Hz, or 195 Hz — an octave below the open pipe.
- Only odd n fit that pattern, so the second harmonic does not exist and the first overtone is 3f₁ = 3 × 194.89 = 584.66 Hz, or 585 Hz.
- The pressure node sits slightly outside the mouth. A stopped pipe has one open end, so one correction applies: L(eff) = 0.440 + 0.6 × 0.0150 = 0.4490 m, giving f₁ = 343 ÷ (4 × 0.4490) = 190.98 Hz, so 191 Hz.
- Frequency goes as 1/L(eff), so the pitch shift is just the length ratio: 1200 log₂(0.4490 ÷ 0.440) = 35.1 cents flat, about a third of a semitone — a real tuning error, not a rounding one.
AnswerOpen–open: f₁ = 390 Hz, first overtone 780 Hz (= 2f₁). Stopped: f₁ = 195 Hz, first overtone 585 Hz (= 3f₁). With the end correction the stopped fundamental falls to 191 Hz, 35 cents flat.
HardA string fixed at both ends is pulled into a triangle with its corner at x = pL and released from rest, so the mode amplitudes go as Aₙ ∝ sin(nπp)/n². Take p = 1/4. Which harmonics are missing, and what are A₂, A₃ and A₄ relative to A₁? The player then plucks at p = 1/20 instead: find A₂/A₁ there and say what changed in the sound.
- With p = 1/4 the sine factor is sin(nπ/4), so a harmonic vanishes when np is a whole number: n = 4, 8, 12, … The pluck point sits exactly on a node of those modes, so they are never started.
- Work in ratios to the fundamental. A₁ ∝ sin(45°)/1² = 0.70711 and A₂ ∝ sin(90°)/2² = 1/4 = 0.25000, so A₂/A₁ = 0.25000 ÷ 0.70711 = 0.354.
- A₃ ∝ sin(135°)/3² = 0.70711/9 = 0.078568, so A₃/A₁ = 0.078568 ÷ 0.70711 = 0.111, exactly 1/9 because sin 135° = sin 45°.
- A₄ ∝ sin(180°)/4² = 0, so A₄/A₁ = 0 — the hole the first step predicted.
- Now p = 1/20: A₁ ∝ sin(9°) = 0.15643 and A₂ ∝ sin(18°)/2² = 0.30902/4 = 0.077254, so A₂/A₁ = 0.077254 ÷ 0.15643 = 0.494.
- For small p, sin(nπp) ≈ nπp, so Aₙ/A₁ → 1/n and A₂/A₁ → 0.500: the 1/n² fall flattens to 1/n and the upper harmonics stay strong. Same string, same length, same pitch — a thinner, brighter tone.
Answerp = 1/4 silences n = 4, 8, 12, …; A₂/A₁ = 0.354, A₃/A₁ = 0.111, A₄/A₁ = 0. At p = 1/20, A₂/A₁ rises to 0.494, close to the 1/n limit of 0.500 — the same pitch, a brighter timbre.