SCIENCE · INVESTIGATE AND EXPLAIN

Match pressure and displacement at tube boundaries

A tube boundary constrains pressure and displacement differently. At an ideal open end the pressure variation is approximately zero while displacement can have an antinode. At a rigid closed end displacement is zero while pressure has an antinode. Confusing these two fields reverses the interpretation of an otherwise familiar standing-wave sketch.

Learning outcome: Compare both topologies at lengths .5,1,1.5 m. Sketch pressure and displacement nodes independently and predict the next resonance before revealing it. Explain why a 1:3 ratio suggests but does not uniquely prove a closed/open tube if some modes could be missing from a measured record.

Starting recap

Separate local oscillation from propagation, amplitude from its square, and a calibration fit from a fresh prediction. Use declared phase, medium frame and boundary conditions.

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Try

At c=340 m/s and L=1 m, an open/open tube has first resonances 170,340,510,680 Hz. Closing one end changes them to 85,255,425,595 Hz. The second resonance is three times the first, not twice. A fixed/fixed string of length 1 m and speed 40 m/s instead begins at 20 Hz.

Open the experiment

Predict

For an ideal closed/open tube, the first overtone is:

Your prediction and reason

Experiment

Compare both topologies at lengths .5,1,1.5 m. Sketch pressure and displacement nodes independently and predict the next resonance before revealing it. Explain why a 1:3 ratio suggests but does not uniquely prove a closed/open tube if some modes could be missing from a measured record.

Match displacement and pressure boundary conditions

Worked teaching model · synthetic values · Ideal uniform air column speed 340 m/s, negligible end correction and losses. Closed end at x=0 when selected; open end at x=L. Open ends are pressure nodes and displacement antinodes. Four allowed resonances; normalized profiles show only the fundamental spatial mode.

Effective tube length: 1 m Topology: 0 open/open, 1 closed/open: 1 code

Reset keeps your prediction, captures and practice history.

Fundamental spatial amplitudes: pressure is not displacement

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: displacement coefficient (solid).2: pressure coefficient (dashed). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Fundamental spatial amplitudes: pressure is not displacement

displacement coefficient

position in tube (m)	normalized signed spatial coefficient
0	0
0.0125	0.0196337
0.025	0.0392598
0.0375	0.0588708
0.05	0.0784591
0.0625	0.0980171
0.075	0.117537
0.0875	0.137012
0.1	0.156434
0.1125	0.175796
0.125	0.19509
0.1375	0.214309
0.15	0.233445
0.1625	0.252492
0.175	0.27144
0.1875	0.290285
0.2	0.309017
0.2125	0.32763
0.225	0.346117
0.2375	0.36447
0.25	0.382683
0.2625	0.400749
0.275	0.41866
0.2875	0.436409
0.3	0.45399
0.3125	0.471397
0.325	0.488621
0.3375	0.505657
0.35	0.522499
0.3625	0.539138
0.375	0.55557
0.3875	0.571788
0.4	0.587785
0.4125	0.603556
0.425	0.619094
0.4375	0.634393
0.45	0.649448
0.4625	0.664252
0.475	0.678801
0.4875	0.693087
0.5	0.707107
0.5125	0.720854
0.525	0.734323
0.5375	0.747508
0.55	0.760406
0.5625	0.77301
0.575	0.785317
0.5875	0.797321
0.6	0.809017
0.6125	0.820401
0.625	0.83147
0.6375	0.842217
0.65	0.85264
0.6625	0.862734
0.675	0.872496
0.6875	0.881921
0.7	0.891007
0.7125	0.899748
0.725	0.908143
0.7375	0.916188
0.75	0.92388
0.7625	0.931215
0.775	0.938191
0.7875	0.944806
0.8	0.951057
0.8125	0.95694
0.825	0.962455
0.8375	0.967599
0.85	0.97237
0.8625	0.976766
0.875	0.980785
0.8875	0.984427
0.9	0.987688
0.9125	0.990569
0.925	0.993068
0.9375	0.995185
0.95	0.996917
0.9625	0.998266
0.975	0.999229
0.9875	0.999807
1	1

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

pressure coefficient

position in tube (m)	normalized signed spatial coefficient
0	-1
0.0125	-0.999807
0.025	-0.999229
0.0375	-0.998266
0.05	-0.996917
0.0625	-0.995185
0.075	-0.993068
0.0875	-0.990569
0.1	-0.987688
0.1125	-0.984427
0.125	-0.980785
0.1375	-0.976766
0.15	-0.97237
0.1625	-0.967599
0.175	-0.962455
0.1875	-0.95694
0.2	-0.951057
0.2125	-0.944806
0.225	-0.938191
0.2375	-0.931215
0.25	-0.92388
0.2625	-0.916188
0.275	-0.908143
0.2875	-0.899748
0.3	-0.891007
0.3125	-0.881921
0.325	-0.872496
0.3375	-0.862734
0.35	-0.85264
0.3625	-0.842217
0.375	-0.83147
0.3875	-0.820401
0.4	-0.809017
0.4125	-0.797321
0.425	-0.785317
0.4375	-0.77301
0.45	-0.760406
0.4625	-0.747508
0.475	-0.734323
0.4875	-0.720854
0.5	-0.707107
0.5125	-0.693087
0.525	-0.678801
0.5375	-0.664252
0.55	-0.649448
0.5625	-0.634393
0.575	-0.619094
0.5875	-0.603556
0.6	-0.587785
0.6125	-0.571788
0.625	-0.55557
0.6375	-0.539138
0.65	-0.522499
0.6625	-0.505657
0.675	-0.488621
0.6875	-0.471397
0.7	-0.45399
0.7125	-0.436409
0.725	-0.41866
0.7375	-0.400749
0.75	-0.382683
0.7625	-0.36447
0.775	-0.346117
0.7875	-0.32763
0.8	-0.309017
0.8125	-0.290285
0.825	-0.27144
0.8375	-0.252492
0.85	-0.233445
0.8625	-0.214309
0.875	-0.19509
0.8875	-0.175796
0.9	-0.156434
0.9125	-0.137012
0.925	-0.117537
0.9375	-0.0980171
0.95	-0.0784591
0.9625	-0.0588708
0.975	-0.0392598
0.9875	-0.0196337
1	-6.12323e-17

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
Quantity	Value	Unit
Allowed resonance 1	85	Hz
Allowed resonance 2	255	Hz
Allowed resonance 3	425	Hz
Allowed resonance 4	595	Hz
Fundamental wavelength	4	m
Next-to-first frequency ratio	3	ratio

The first overtone is the second allowed resonance, not always the second harmonic.

Open/open displacement antinodes at both ends allow lambda_n=2L/n. Closed/open displacement node at x=0 and antinode at L allow lambda_n=4L/(2n−1). Thus the latter contains only odd harmonics in this ideal model. Pressure nodes and displacement antinodes exchange locations. For the chosen standing-mode displacement xi=X(x)cos(omega t), pressure is proportional to −Xprime(x)cos(omega t). The plotted normalized signed coefficients use this common temporal factor; pressure and displacement coefficients are spatially shifted, while particle velocity is in temporal quadrature. Effective lengths neglect end correction; real mode frequencies may depart from these ratios.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

The first overtone is the second allowed resonance, not always the second harmonic.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Define the observable

Build the bounded model

For a tube open at both ends, the allowed wavelengths are 2L/n and frequencies are nc/(2L), with n=1,2,3 and so on. For one closed and one open end, wavelengths are 4L/(2n−1) and frequencies are (2n−1)c/(4L). The ordinal resonance number n is therefore not always the harmonic number. The first overtone of a closed/open tube is its second allowed resonance but its third harmonic.

Interpret the evidence

The normalized signed spatial coefficients show the fundamental only. For displacement ξ=X(x)cos(ωt), pressure is proportional to −Xprime(x)cos(ωt): both share the chosen temporal factor. Thus open/open coefficients can be cos(kx) and +sin(kx), while closed/open coefficients can be sin(kx) and −cos(kx). Their spatial nodes differ; particle velocity, rather than displacement, is in temporal quadrature with pressure. The model uses a supplied sound speed and an effective length with end correction neglected. Real openings, losses and geometry shift resonances. A stretched string fixed at both ends instead has displacement nodes at both ends and fn=n√(T/μ)/(2L); identical frequency ratios can arise from different boundary physics.

Worked original case

Predict and investigate

Changed-case practice

Try the changed case before feedback

For a closed/open tube with effective length .5 m and c=340 m/s, find its second allowed resonance frequency in Hz.

Numerical result in the requested units

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original prose and cases checked 30 October 2026 against the primary source. These are silent, synthetic wave models: no sound playback, exposure prescription, hearing test or physical apparatus is requested. A correct public number is not independent mastery; retain first explanations and a changed-case review in the course evidence workspace.

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Which fresh observation separates a failed physical model from a fitted calibration parameter?

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