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core29 min studyLesson 195 in Physics

Find the sound modes of an ideal pipe

Use pressure and particle-velocity boundary conditions to determine pipe resonances and their limits.

Before you begin
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The question for today

What you’ll understand

  • Translate rigid and pressure-release ends into acoustic boundary conditions.
  • Calculate open–open and closed–open modal frequencies.
  • Distinguish an ideal eigenfrequency from a driven resonance with losses.

A pipe selects certain spatial patterns because its boundaries must be satisfied repeatedly. The most useful first step is not memorizing a list of harmonics but deciding which acoustic quantity vanishes at each end. Pressure and particle velocity have complementary nodes. We will derive ideal one-dimensional pipe modes, compare two boundary choices, and explain why measured resonances need not occur at exactly the ideal frequencies.

A closed end fixes velocity, not pressure

An ideal rigid end cannot move normally, so the fluid’s normal particle velocity vanishes there. The linear momentum equation then makes the spatial pressure derivative vanish at the wall for an oscillating mode. Pressure can be an antinode rather than a node. An ideal open end is approximated as a pressure-release boundary: its pressure perturbation is zero relative to the exterior. The ambient equilibrium pressure is still present. Confusing perturbation pressure with total pressure makes the open-end condition seem physically impossible.

For a uniform pipe whose transverse dimensions are small relative to wavelength, approximate the field as one-dimensional. Write pressure as a spatial function times a sinusoidal time factor. The wave equation gives a second-order spatial equation whose solutions are sine and cosine. Boundary conditions select the allowed wavenumbers. This is an eigenvalue problem: a freely oscillating ideal mode exists only for particular wavenumbers, while its overall amplitude remains undetermined until excitation or initial conditions are specified.

P′′(x)+k2P(x)=0,u=0⇒P′=0,p1=0⇒P=0P^{\prime\prime}(x)+k^2P(x)=0,\qquad u=0\Rightarrow P^{\prime}=0,\qquad p_1=0\Rightarrow P=0
P(x) is the spatial pressure amplitude. The rigid-end implication uses harmonic motion with nonzero frequency and the linear momentum equation.↔ Scroll wide equations horizontally.

Count wavelengths between unlike boundaries

For two pressure-release ends, the spatial pressure must vanish at both ends. A nonzero sine solution fits an integer number of half wavelengths into the length, producing frequencies nc/(2L) for positive integers n. Two rigid ends have the same positive-frequency spacing, although pressure nodes and antinodes exchange positions. A zero-frequency uniform pressure shift in the rigid case is not an oscillatory sound mode and should not be counted as an audible fundamental.

For a rigid end at x=0 and a pressure-release end at x=L, choose a cosine pressure shape to satisfy the derivative condition at zero. The first pressure zero occurs a quarter wavelength away. Higher allowed lengths contain an odd number of quarter wavelengths, so the ideal frequencies are odd multiples of c/(4L). This odd-harmonic result relies on these contrasting boundary conditions; it is not a universal property of every pipe, cavity, or wind instrument.

fnopen−open=nc2L (n=1,2,…),fmclosed−open=(2m+1)c4L (m=0,1,…)f_n^{\mathrm{open-open}}=\frac{nc}{2L}\ (n=1,2,\ldots),\qquad f_m^{\mathrm{closed-open}}=\frac{(2m+1)c}{4L}\ (m=0,1,\ldots)
The model assumes a uniform one-dimensional lossless pipe with ideal rigid or pressure-release ends and negligible end correction.↔ Scroll wide equations horizontally.

A resonator also needs coupling and damping

Real open ends radiate into the surrounding fluid, so the pressure node lies approximately outside the geometrical opening. An end correction changes the effective acoustic length. Viscous and thermal boundary layers dissipate energy, while radiation carries energy away. A driven response therefore has finite amplitude and width rather than an infinite peak. A mode’s decay rate and its coupling to a source help determine which resonance is visible; eigenfrequency alone does not set the strength heard by a listener.

The oscillator lab offers an honest analogy for one isolated damped resonance. Its controls are damping ratio and natural angular frequency, with fixed initial displacement one and initial velocity zero. Predict how raising damping changes the decay before pressing Play, and compare underdamped with critical response. It is a free single-coordinate oscillator, not a pipe-mode solver or a driven resonance spectrum. The separate wave lab’s same-direction components would also be insufficient to make a spatial standing wave.

Q≃ω02Γ,A(t)∝e−ΓtQ\simeq\frac{\omega_0}{2\Gamma},\qquad A(t)\propto e^{-\Gamma t}
For a weakly damped isolated resonance, Γ is the amplitude decay rate; this approximate Q relation does not identify a complete acoustic loss mechanism.↔ Scroll wide equations horizontally.

Work it through

Compare the first three pipe modes

Use c=320 m/s and L=0.8 m, neglecting end corrections. Find the first three positive frequencies for an open–open pipe and for a closed–open pipe of the same length.

  1. For open–open boundaries, the fundamental is c/(2L)=320/1.6=200 Hz. Positive integer multiples give 200, 400, and 600 Hz.

  2. For closed–open boundaries, the fundamental is c/(4L)=320/3.2=100 Hz. The next allowed multipliers are three and five, giving 300 and 500 Hz.

  3. Check the lowest wavelength: the open–open fundamental has λ=2L=1.6 m, while the closed–open fundamental has λ=4L=3.2 m. Both satisfy fλ=c.

The result

The sequences begin 200, 400, 600 Hz and 100, 300, 500 Hz. The missing even multiples belong to the ideal closed–open boundary problem.

The living notebook

Explore the model

Interactive lab

Interactive lab · oscillator

Inspect an isolated mode’s decay analogy

The lab varies damping ratio ζ and natural angular frequency ω₀ of a unit-mass oscillator starting at displacement one and velocity zero. Predict whether a larger ζ produces ringing before playing.

Inspect an isolated mode’s decay analogyUnit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Current readings: Displacement x(t): 1; Velocity v(t): 0; Regime: underdamped. All parameters are adjustable with the labeled controls below.02.557.510−1.15−0.5700.571.15Time · sDisplacementunderdamped
Displacement x(t)Current time
Displacement x(t)
1
Velocity v(t)
0
Regime
underdamped
t = 0 s

Unit mass, x(0) = 1 and v(0) = 0. The analytic solution includes underdamped, critically damped and overdamped cases; time is measured in seconds. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThis free oscillator illustrates damping regimes for a single coordinate; it does not calculate pipe shapes, radiation loss, or a driven frequency response.

Pause & reason

Check your understanding

Which ideal condition belongs at a rigid closed end of a pipe?

Your turn

Try it yourself

An ideal closed–open pipe has fundamental 150 Hz at a fixed sound speed. What are its next two allowed frequencies? If its effective length grows by 10%, what happens to the fundamental?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use odd multiples for the spectrum and inverse-length scaling for a fixed mode number.

Compare with the solution

The next two allowed frequencies are 450 and 750 Hz. Increasing effective length by a factor 1.1 divides every ideal modal frequency by 1.1, so the fundamental becomes about 136.4 Hz. A 10% length increase gives about a 9.1% frequency decrease, not exactly 10%.

Keep the idea

Take it with you

  • Rigid ends constrain normal velocity; pressure-release ends constrain pressure perturbation.
  • Allowed frequencies follow from the spatial boundary problem.
  • Real resonances acquire end corrections, finite widths, and excitation-dependent strengths.

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