Waves · Lesson 6 of 8
Air Columns and Resonance
“Use pressure and displacement boundary conditions to derive pipe modes, explain resonance and evaluate experimental data.”
• Explain why displacement nodes coincide with pressure antinodes and vice versa. • Derive allowed frequencies for pipes closed at one end and pipes open at both ends. • Identify resonant harmonics and distinguish harmonic number from overtone number. • Use successive resonance lengths and check whether measurements fit the stated idealisation.
Displacement and pressure tell different stories
An air-column wave is longitudinal. Let s(x,t) be the displacement of a small element along the tube. Draw displacement vertically on a graph only to display its signed value: the gas itself moves along the tube, not through the side wall. A pressure disturbance p′ is the pressure above or below the ambient value. It is related to how neighbouring displacements change the spacing between elements, rather than to the displacement of one element by itself.
To see the link, take a short tube slice of cross-sectional area S and equilibrium length Δx. Its ends move by s(x,t) and s(x + Δx,t). Its length change is the difference of these displacements. Divide by Δx and take a small slice: fractional volume change is ∂s/∂x, the spatial derivative at a fixed time. Positive slope means expansion; negative slope means compression. From the bulk-modulus definition, pressure change is −B times this strain.
At a displacement node, neighbouring air can converge towards and diverge away from the nearly stationary location, causing a large pressure oscillation. At a displacement antinode, neighbouring elements move nearly together, so local compression is smallest. This is why a displacement node is a pressure antinode, and a displacement antinode is a pressure node. These labels describe amplitude over time, not a permanently high pressure or a permanently displaced air element.
Closed end and open end impose different conditions
At a rigid closed end, air cannot move normally through the wall: longitudinal displacement is zero, giving a displacement node. Its pressure is free to oscillate and is typically a pressure antinode. At an ideal open end, the air pressure is held approximately at ambient by contact with the large surrounding atmosphere, giving a pressure node and displacement antinode. The ambient pressure itself is not zero; only its oscillating excess p′ is approximately zero. Real radiation and end corrections shift this ideal boundary slightly.
One closed end gives odd harmonics
Set x = 0 at the closed end and x = L at the open end. The form s ∝ sin(kx) already satisfies the closed-end node. To place an antinode at x = L, require |sin(kL)| = 1. Thus kL = (n + ½)π for n = 0,1,2,… . With k = 2π/λ, this means L = (2n + 1)λ/4. The shortest allowed column contains one quarter wavelength, then three quarters, five quarters, and so on.
Be careful with labels: the first overtone of a closed–open pipe is its second allowed mode, but the third harmonic, at 3f₁. The next is the fifth harmonic at 5f₁. The first six allowed patterns have harmonic numbers 1,3,5,7,9,11. Even multiples of the fundamental do not satisfy one-node/one-antinode end conditions for the same ideal pipe. Wavelength shortens as more half-wavelength segments are added between the ends.
Two open ends allow every integer harmonic
A pipe open at both ends has displacement antinodes at both ends. A convenient displacement pattern uses cos(kx), which already has an antinode at x = 0. Requiring another at L gives kL = nπ with n = 1,2,3,…, so L = nλ/2. This is the same wavelength sequence as a fixed-end string, even though the string endpoints are displacement nodes and the open pipe endpoints are displacement antinodes.
| System | End displacement conditions | Fundamental | Allowed harmonic numbers |
|---|---|---|---|
| String fixed at both ends | Node–node | v/(2L) | 1,2,3,4,… |
| Air column open at both ends | Antinode–antinode | v/(2L) | 1,2,3,4,… |
| Air column closed at one end | Node–antinode | v/(4L) | 1,3,5,7,… |
A source can drive a mode resonantly
A tuning fork produces an external periodic pressure variation. An air column can respond at the driving frequency even away from a natural mode, but the response becomes especially large when the driving frequency lies near one of the natural frequencies and energy is supplied in step with the oscillation. This is resonance. Damping and radiation limit the amplitude and give each resonance a finite width. The natural frequencies are selected by wave speed and boundary conditions; the source does not make every imposed frequency a normal mode.
Problem
A pipe is 30.0 cm long and open at both ends. Which mode can resonate with a 1.1 kHz source when sound speed is 330 m s⁻¹? What if one end is closed?
- 1.For the open pipe, f₁ = 330/[2(0.300)] = 550 Hz. The 1100 Hz source equals 2f₁, so it excites the second harmonic, the first overtone.
- 2.With one end closed, f₁ = 330/[4(0.300)] = 275 Hz. The allowed frequencies are 275, 825, 1375,… Hz, the odd multiples.
- 3.Although 1100 Hz is 4 × 275 Hz, the fourth harmonic is not an allowed ideal closed–open mode. Therefore the source does not give a normal-mode resonance in that configuration. This conclusion does not mean the pipe cannot have any forced response at all.
Problem
A 20 cm pipe is closed at one end and driven at 430 Hz. Use sound speed 340 m s⁻¹. Which resonance is approached, and would the same source resonate if both ends were open?
- 1.The ideal closed–open fundamental is 340/[4(0.20)] = 425 Hz. The 430 Hz source is close to that fundamental, so it can excite the first harmonic within a finite resonance width.
- 2.Exact equality is not present in the rounded data; do not rewrite 425 as exactly 430. Whether the response is strong enough depends on damping and the actual end correction.
- 3.The open–open fundamental would be 340/[2(0.20)] = 850 Hz, with higher modes at integer multiples. A 430 Hz source is not near any of those ideal modes.
Successive resonance lengths measure wavelength
With the source frequency fixed, move a piston or change the water level to vary the closed–open air-column length. The fundamental has effective length λ/4; the next allowed mode has 3λ/4. Each further resonance adds another λ/2, so consecutive resonant lengths differ by λ/2. If an approximately constant end correction e makes the effective length L + e, subtraction removes e. This is often more reliable than assuming a single measured length equals exactly λ/4.
Problem
A 340 Hz fork produces reported closed–open resonances at lengths 25.5 cm and 79.3 cm. Estimate speed from their separation, and test whether both lengths fit the no-end-correction idealisation.
- 1.The length separation is 0.793 − 0.255 = 0.538 m. If the resonances are consecutive, λ = 2(0.538) = 1.076 m and v = 340(1.076) = 365.84 m s⁻¹.
- 2.A first resonance of 0.255 m alone, with no end correction, would give v = 4fL = 346.8 m s⁻¹. A second resonance at 0.793 m alone would give v = 4fL/3 = 359.49 m s⁻¹. These disagree: the two raw lengths are not exactly in the ideal 1:3 ratio.
- 3.The pair can be fitted by a common effective-length correction e: λ/4 = 0.269 m and e = 0.269 − 0.255 = 0.014 m. Then 0.793 + 0.014 = 0.807 m = 3λ/4.
- 4.Thus 365.84 m s⁻¹ is the spacing estimate under the consecutive-resonance assumption, while an exact neglect of edge effects makes the given pair inconsistent. Report the inconsistency instead of presenting 347 m s⁻¹ as if it used both measurements.
Rod and membrane boundaries use the same idea
A steel rod clamped at its middle has a longitudinal displacement node at its centre. At either free end the longitudinal stress vanishes; the fundamental has displacement antinodes there. Each half rod spans a node–antinode quarter wavelength, so a rod of total length L has λ = 2L. A 1.00 m rod with fundamental 2.53 kHz therefore gives v = 2L f = 2(1.00)(2530) = 5060 m s⁻¹ = 5.06 km s⁻¹. This is longitudinal vibration of the rod, not the transverse motion of a wire stretched under external tension.
A tabla membrane is fixed around its circular rim, so displacement must vanish along that circumference. Different two-dimensional patterns satisfy that boundary and have their own normal frequencies. Calculating them requires more mathematics than the one-dimensional string and pipe examples, but the organising principle is unchanged: the restoring dynamics and the boundary conditions together select the modes. A membrane’s frequencies should not be assumed to be the same integer sequence as an ideal string.
Quick Quiz
Quiz
At an ideal rigid closed end of an air column, there is
Air cannot move through the wall, but it can be compressed near it.
The first overtone of an ideal closed–open pipe has frequency
Only odd harmonics are allowed; the next mode is the third harmonic.
An open–open pipe has fundamental 500 Hz. Which frequency is an allowed mode?
1500 Hz is its third harmonic, 3 × 500 Hz.
Successive closed–open resonances at fixed frequency differ in effective length by
Each added mode adds two quarter-wavelength segments.
At an ideal open pipe end, “pressure node” means
The pressure stays approximately atmospheric while the air can move.
Practice Problems
- Derive p′ = −B ∂s/∂x from the expansion of a short air slice. Use a standing sinusoidal displacement to show why pressure and displacement nodes exchange positions.
- Derive the wavelength and frequency sequences for a closed–open pipe. Draw displacement patterns for its first three allowed modes and label their harmonic and overtone numbers.
- A 0.30 m pipe carries sound at 330 m s⁻¹. List its first four frequencies when open at both ends and its first four allowed frequencies when closed at one end. Test a 1100 Hz source in each case.
- For a 0.20 m pipe closed at one end with sound speed 340 m s⁻¹, test a 430 Hz source. Distinguish approximate resonance from exact equality and repeat the test for both ends open.
- A 340 Hz fork produces successive reported resonances at 0.255 m and 0.793 m. Calculate the spacing-based speed; then test each length separately with zero end correction and explain the inconsistency.
- A 1.00 m rod clamped at its middle has longitudinal fundamental 2.53 kHz. Sketch the displacement node and antinodes, infer the wavelength and calculate the rod-wave speed.
- Explain how the fixed circular rim of a tabla selects membrane modes. State why the one-dimensional string harmonic formula cannot simply be copied for a membrane.
- Explain how a pipe can respond at a driving frequency away from a natural mode while showing a much stronger response near resonance.
Key Takeaways
• Closed ends are displacement nodes and pressure antinodes; ideal open ends reverse those roles. • Closed–open pipes allow odd harmonics with f₁ = v/(4L); open–open pipes allow all integer harmonics with f₁ = v/(2L). • Resonance is a large driven response near a natural frequency, limited by losses. • Consecutive resonance spacing gives λ/2. Measurements must be checked against the assumed boundary model.