Physics / Modes of Vibration of an Air Column
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EXPERIMENT 40 · PHYSICS LABORATORY

Modes of Vibration of an Air Column theory

Understand modes of vibration of an air column: the equations, assumptions, supporting concepts and measurement method.

THE WORKING RELATIONSHIP
L = \left(\frac{v}{2}\right)\left(\frac{1}{f_1}\right) - 2c \iff y = m x + b \iff v = 2 m, \quad c = -\frac{b}{2}

Keep the physical quantities, graph variables and units consistent.

01Symmetric Open-Open Boundary Conditions and Normal Modes

The acoustic behavior of a cylindrical duct open at both ends represents a classic Sturm-Liouville boundary-value problem in continuum mechanics. Expressing the longitudinal acoustic velocity potential as ψ(x, t) = [A cos(k x) + B sin(k x)] e^(i ω t), the acoustic pressure perturbation is given by p(x, t) = −ρ₀ (∂ψ/∂t) = −i ω ρ₀ ψ(x, t), while the longitudinal air particle displacement is u(x, t) = −∂ψ/∂x. At an open aperture communicating with an unconfined atmospheric reservoir, the acoustic pressure perturbation must vanish to first order at the effective acoustic boundaries: p(0, t) = 0 and p(L_eff, t) = 0. Applying p(0, t) = 0 enforces the cosine amplitude coefficient to zero (A = 0), leaving ψ(x) = B sin(k x). Applying the second boundary condition at x = L_eff dictates sin(k L_eff) = 0, which requires k L_eff = n π for integer n = 1, 2, 3, ... Substituting the acoustic wavenumber k = 2π / λ = 2π f / v, we obtain the fundamental quantization condition for open pipes: L_eff = n (λ_n / 2) = n [v / (2 f_n)]. Rearranging for the modal resonance frequencies yields the harmonic series f_n = n [v / (2 L_eff)]. The fundamental mode (n = 1) corresponds to an acoustic half-wavelength spanning the pipe: L_eff = λ₁ / 2, with an acoustic displacement node (pressure antinode) situated exactly at the midpoint x = L_eff / 2. Because n spans all positive integers, an open pipe produces a rich harmonic spectrum containing both even and odd multiples of the fundamental frequency.

L_{\text{eff}} = n \frac{\lambda_n}{2} = n \frac{v}{2 f_n} \implies f_n = n \frac{v}{2 L_{\text{eff}}} \quad (n = 1, 2, 3, 4, \dots)
Why does an open-open pipe generate all integer harmonics (1f, 2f, 3f, 4f, ...) while a closed-open pipe generates only odd harmonics?

Because both ends of an open pipe have identical boundary conditions (pressure nodes), the pipe can accommodate any integer number of half-wavelengths: L_eff = n(λ/2). In contrast, a closed pipe has asymmetric boundaries (a pressure node at one end and a pressure antinode at the other), requiring an odd number of quarter-wavelengths: L_eff = (2m − 1)(λ/4).

02Dual-Aperture Acoustic Radiation and the 2c Rayleigh Correction

In an open organ pipe or cylindrical resonance tube, sound waves radiating from both open ends encounter ambient atmospheric inertia. At each aperture, the oscillating air column must accelerate an external 'virtual air piston' or hemispherical mass of exterior air, creating an acoustic radiation mass (inertance). In 1877, Lord Rayleigh proved that for an unflanged circular tube of radius r, the radiation inertance shifts the acoustic pressure node outward by an effective distance c ≈ 0.6133 r beyond each physical lip. Because an open pipe possesses two identical radiating apertures—one at each end—the total acoustic correction is exactly twice the single-aperture value: ΔL_total = 2c ≈ 2 × (0.6133 r) = 1.2266 r. Thus, the effective acoustic length of an open pipe of physical length L is L_eff = L + 2c = L + 1.2266 r. For a typical university apparatus with bore diameter 4.00 cm (radius r = 2.00 cm), the total dual end correction is 2c ≈ 1.2266 × 2.00 cm = 2.453 cm (approximately 1.23 cm per open end). If this dual end correction were neglected in an open tube resonating at 512 Hz (physical length L ≈ 31.1 cm), the calculated sound speed would underestimate the true velocity by 2.45 / 31.1 ≈ 7.9%. Linear regression of L against 1/f cleanly isolates 2c as the vertical intercept, eliminating this massive systematic error.

L_{\text{eff}} = L + 2c = L + 1.2266 \cdot r, \qquad 2c = -b \iff c = -\frac{b}{2}
Why is the total end correction for an open pipe twice that of a closed pipe with the same bore radius?

An open pipe has two open radiating terminations that communicate with the ambient atmosphere, each contributing an acoustic end correction c ≈ 0.61 r. A closed pipe has only one open aperture, so its end correction is simply c.

03Linear Regression Formulation for Open Pipe Resonance

To determine the speed of sound and verify Rayleigh's dual end correction without circular calibration assumptions, experimentalists measure the fundamental physical resonance length L across multiple known acoustic frequencies f₁. Expressing the fundamental half-wave condition L + 2c = λ₁ / 2 = v / (2 f₁) and rearranging yields the canonical linear regression model: L = (v / 2) (1 / f₁) − 2c. Defining x = 1 / f₁ (in seconds) as the independent variable and y = L (in metres) as the dependent variable, the data conform to the standard straight-line equation y = m x + b. Here, the regression slope m represents half the propagation speed of sound: m = v / 2, which allows the velocity to be calculated as v = 2m. Crucially, the slope m is completely independent of the tube end corrections. The vertical intercept b represents the negative total dual end correction: b = −2c, so that the single-aperture correction is c = −b / 2. By comparing this experimental intercept with Rayleigh's theoretical prediction 2c = 1.2266 r, students verify both the validity of acoustic aperture theory and the thermodynamic speed of sound with sub-percent accuracy.

L = m \left(\frac{1}{f_1}\right) + b \implies m = \frac{v}{2} \iff v = 2 m, \qquad b = -2c \iff c = -\frac{b}{2}
How does the slope of the resonance curve for an open pipe (L vs 1/f) compare with that of a closed pipe?

The slope for an open pipe is m_open = v/2, which is exactly twice the slope for a closed pipe m_closed = v/4. This reflects the fundamental physical difference that an open pipe accommodates a half-wavelength (λ/2) while a closed pipe accommodates a quarter-wavelength (λ/4).

04Overtone Series, Acoustic Timbre, and Organ Pipe Design

The distinction between open and closed air columns lies at the very heart of musical acoustics, organ design, and vocal tract phonetics. The subjective perception of sound quality—known as musical timbre or tone color—is determined by the harmonic spectrum (the relative amplitudes of the overtone frequencies). In an open organ pipe (such as the Principal or Flute stops of a pipe organ), the air column produces all integer harmonics: f, 2f, 3f, 4f, 5f, 6f, ... The presence of strong even harmonics (especially the octave 2f and double-octave 4f) imparts a bright, rich, warm, and open tone. Conversely, in a closed organ pipe (known as a Stopped Diapason or Gedackt), the rigid upper stopper suppresses all even harmonics, allowing only odd harmonics to sound: f, 3f, 5f, 7f, ... The complete absence of the octave (2f) and presence of a prominent twelfth (3f) gives stopped pipes a hollow, woody, clarinet-like, and sombre timbre. Furthermore, because a stopped pipe requires only an effective quarter-wavelength (L_eff ≈ λ/4) to produce a given pitch, a stopped organ pipe requires only half the physical height of an open pipe producing the same fundamental frequency, saving valuable vertical space in church pipe organ chambers.

f_{\text{open}} \in \{1f_1, 2f_1, 3f_1, 4f_1, \dots\}, \qquad f_{\text{closed}} \in \{1f_1, 3f_1, 5f_1, 7f_1, \dots\}
Why do open organ pipes sound bright and full, whereas closed (stopped) organ pipes sound hollow and nasal?

Open pipes produce a full harmonic series containing all integer multiples (1f, 2f, 3f, 4f, ...), providing rich acoustic timbre. Closed pipes completely lack even harmonics (2f, 4f, 6f, ...), leaving only odd harmonics (1f, 3f, 5f, ...), which creates a characteristically hollow, clarinet-like sound.

05Acoustic Pressure vs Particle Displacement Wave Profiles

A rigorous understanding of acoustic standing waves requires clear differentiation between the acoustic pressure field p(x, t) and the particle displacement field u(x, t). These two physical fields are spatially and temporally in quadrature (separated by a 90° or π/2 phase shift). In an open pipe, the boundary conditions enforce pressure nodes (p = 0) at both open ends. By Euler's fluid momentum equation ∂p/∂x = −ρ₀ (∂²u/∂t²), the spatial derivative of pressure drives fluid particle acceleration; where pressure is spatially flat (p = 0), particle acceleration and displacement achieve maximum amplitude (displacement antinodes). Conversely, at the physical center of an open pipe resonating in its fundamental mode (x = L_eff / 2), air molecules arriving simultaneously from left and right collide, compressing the fluid to create an acoustic pressure antinode (|p| = max), but their physical net displacement is constrained to zero by symmetry (displacement node u = 0). When a student inserts a dust powder (such as lycopodium or cork dust as in Kundt's tube), the dust is agitated violently and swept away from displacement antinodes, gathering into quiet ridges at displacement nodes, providing visible physical proof of internal wave geometry.

p(x, t) = p_0 \cos(k x) \sin(\omega t), \qquad u(x, t) = u_0 \sin(k x) \cos(\omega t)
Where does maximum acoustic pressure fluctuation occur inside an open organ pipe resonating in its fundamental mode?

Maximum acoustic pressure fluctuation (the pressure antinode) occurs at the exact geometric center of the pipe (x = L_eff / 2), where fluid particle displacement is zero (displacement node).

06Temperature Dependence and Organ Pitch Sensitivity

Because the speed of sound in air is governed by thermodynamic temperature through v(T) = √(γ R T / M) ≈ 331.30 + 0.606 t m/s, any change in ambient room temperature directly alters the resonant frequencies of air columns. For an open organ pipe of fixed physical length L, the fundamental frequency is f₁ = v / [2(L + 2c)]. Differentiating with respect to temperature t yields df₁ / dt = [1 / 2(L + 2c)] (dv / dt) = [f₁ / v] (dv / dt). Substituting dv / dt ≈ 0.606 m s⁻¹ K⁻¹ and v ≈ 343 m/s gives a relative frequency temperature coefficient of (1 / f₁) (df₁ / dt) ≈ 0.606 / 343 ≈ +0.00177 K⁻¹ (+0.18% per °C). For a 10 °C temperature rise in a concert hall, the pitch of all open and closed organ pipes rises by approximately 1.8%, which corresponds to nearly 30 cents (almost a third of a semitone!). By contrast, stringed instruments (such as pianos and violins) expand slightly with heat while their tension decreases, causing their pitch to drop as temperature rises. This opposite temperature dependence explains why organists and orchestral musicians must carefully stabilize auditorium climate before tuning and performance.

\frac{df_1}{dt} = \frac{f_1}{v} \frac{dv}{dt} \approx +0.00177 \cdot f_1 \quad (\text{Hz} \cdot \text{K}^{-1}), \qquad \Delta \text{cents} \approx 1200 \log_2\left(1 + \frac{\Delta v}{v}\right)
What happens to the pitch of a pipe organ when the temperature of a church sanctuary rises by 5 °C?

The speed of sound in air increases by approximately 0.6 m/s per °C (about +3.0 m/s for 5 °C). Because resonant frequency is directly proportional to sound speed (f = v / 2L_eff), the pitch of all pipes rises by approximately 0.9% (about 15 cents sharp).

07Acoustic Aperture Impedance: From Helmholtz to Levine and Schwinger

The precise calculation of the acoustic end correction has a rich history in mathematical physics. In 1860, Hermann von Helmholtz first applied potential theory to acoustic apertures, proving that an unflanged pipe must exhibit an end correction bounded by (π / 4) r ≈ 0.785 r. In 1877, Lord Rayleigh used energy variational principles to calculate the upper and lower bounds for circular pipes, demonstrating that a pipe terminating in an infinite flat baffle (a flanged pipe) has an end correction of c_flanged = (8 / 3π) r ≈ 0.8488 r. The exact analytical solution for an unflanged, thin-walled circular pipe in an infinite free medium was finally achieved in 1948 by Harold Levine and Julian Schwinger using Wiener-Hopf integral equation factorization. Levine and Schwinger demonstrated that at low frequencies (ka = 2π r / λ << 1), the reflection coefficient magnitude at the open mouth is |R| = exp(−2 k c) ≈ 1 − (ka)² / 2, with the exact asymptotic end correction being c_unflanged = 0.6133 r. At higher frequencies, radiation into exterior space increases, the reflection coefficient drops, and the end correction decreases slightly, transitioning into an acoustic horn radiation regime.

c_{\text{unflanged}} = 0.6133 \cdot r, \qquad c_{\text{flanged}} = 0.8216 \cdot r, \qquad |R| \approx e^{-2 k c}
What did Levine and Schwinger prove regarding the acoustic reflection coefficient at the open mouth of a pipe at very low frequencies?

They proved that as frequency approaches zero (ka << 1), the magnitude of the acoustic reflection coefficient approaches unity (|R| → 1), with total phase shift reflecting an apparent acoustic pipe lengthening of exactly c = 0.6133 r.

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Connect the equations to a measurement.

The open air column experiment demonstrates the harmonic mode structure of acoustic waveguides with symmetric open-open boundary conditions. Because acoustic pressure drops to ambient atmospheric levels outside both apertures, displacement antinodes form at both ends, requiring a total dual end correction of 2c ≈ 1.2266 r. Ordinary least-squares regression of fundamental tube length L against reciprocal frequency 1/f isolates the propagation velocity v = 2m = 343.2 m s⁻¹ and the dual-end aperture offset 2c = 2.45 cm, confirming the half-wave resonance eigenvalue L + 2c = λ/2.