Standing Waves in Air Columns: Open & Closed Pipes

Blow across the mouth of an empty bottle and it hums a note. Pour in some water and blow again: the note rises. You have not changed the air, the temperature, or how hard you blow — you have only shortened the column of air that is free to vibrate, and the pitch obeyed. An organ works the same way on a grander scale, a rank of pipes each tuned by its length; a flute and a clarinet of nearly the same size sound an octave apart for a reason buried in their ends. This article does for the air in a pipe exactly what the previous one did for a stretched string: it asks which standing waves a bounded column of air is allowed to hold. The answer, again, is decided entirely at the boundaries — but a pipe has a new kind of end. A closed end, where the air is trapped against a wall, cannot move, so the vibration must have a displacement node there. An open end, where the air is free to slosh in and out, swings hardest, so it must be an antinode. Those two rules — node at a wall, antinode at an opening — are the whole subject. They split every pipe into two families. An open pipe, open at both ends, must be antinode-to-antinode, and it sounds the full harmonic series $f_n = n\,v/(2L)$ , every whole multiple. A closed pipe, stopped at one end, must run node-to-antinode, and it sounds only the odd harmonics $f_n = (2n-1)\,v/(4L)$ — its fundamental an octave below an open pipe of the same length, with the even harmonics simply missing. That single difference is why a clarinet and a flute have different voices, why a stopped organ pipe can be built half as tall, and why blowing harder on a bugle jumps between only certain notes.

Key takeaways

A pipe resonates when a standing sound wave fits its length under the end rules: a closed end is a displacement node, an open end is a displacement antinode.
Open pipe (both ends open): $f_n = n\,v/(2L)$ , $n = 1, 2, 3, \dots$ — all harmonics, exactly like a string.
Closed pipe (one end stopped): $f_n = (2n-1)\,v/(4L)$ — fundamental $v/4L$ , then only the odd harmonics $3f_1, 5f_1, \dots$
A closed pipe of length $L$ sounds an octave below an open pipe of the same length: $f_{1,\text{closed}} = \tfrac12 f_{1,\text{open}}$ .
The antinode sits slightly outside an open end, so the effective length is $L_{\text{eff}} = L + e$ with the end correction $e \approx 0.6r$ per open end.
Resonance frequencies rise with temperature, because the speed of sound $v$ rises while the pipe length stays fixed.
Key idea: a pipe's allowed notes are fixed by its two ends — antinode at an opening, node at a wall — which makes an open pipe sing every harmonic and a closed pipe sing only the odd ones, an octave lower for the same length.

The core insight

An air column resonates only at frequencies whose standing wave places a displacement antinode at each open end and a node at each closed end; the open pipe therefore sounds all harmonics $f_n = n\,v/(2L)$ , while the closed pipe sounds only odd harmonics $f_n = (2n-1)\,v/(4L)$ with a fundamental an octave below an open pipe of equal length.

Where this lives today

A pipe organ is a museum of equation (1) and equation (2) standing in a single loft: rank upon rank of open and stopped pipes, each cut to a length that places its fundamental on a musical note, with the stopped pipes sounding an octave below their height to reach the deep bass without a ten-metre pipe. Every wind instrument is a tunable air column — a flute an open pipe fingered to change $L$ , a clarinet a closed pipe with its odd-harmonic voice, a trombone an open pipe with a sliding length. Beyond music, the same resonance sets the boom of a subwoofer's ported enclosure, the howl of wind across a chimney or a car sunroof (a Helmholtz-style air resonance), and the frequencies engineers must design out of exhaust pipes and ventilation ducts to stop them from singing. The bottle you blew across to start this article is a closed air column, and its rising note as you added water is equation (2) with a shrinking $L$ .

The physical picture

Sound in a pipe is a longitudinal standing wave: the air oscillates back and forth along the pipe, and the standing-wave pattern is fixed by what the ends allow. It is easiest to track the displacement of the air — how far a slug of air moves from its rest position. At a closed end the air is pressed against a rigid wall and cannot move at all, so its displacement is zero for all time: a closed end is a displacement node. At an open end the air is free to move in and out with almost nothing resisting it, so it moves the most: an open end is a displacement antinode. These are the pipe's boundary conditions, and everything follows from fitting whole standing-wave shapes between them. (A subtlety worth stating early: the pressure wave is exactly out of step with the displacement wave — where displacement is largest, the pressure barely changes, and vice versa. A closed end, a displacement node, is a pressure antinode. We track displacement here because the end rules are cleanest that way.)

Figure

Open pipe and closed pipe drawn side by side, showing the displacement antinode at each open end and the node at the closed end, for the fundamental and the next allowed mode of each

The open pipe: all harmonics

Take a pipe open at both ends. Each end must be a displacement antinode. The shortest standing wave with an antinode at each end has half a wavelength spanning the pipe — a single node in the middle — so $L = \lambda_1/2$ , giving $\lambda_1 = 2L$ . Fit more half-loops and the pattern is the same as the string:

L = n\,\frac{\lambda_n}{2} \quad\Rightarrow\quad \lambda_n = \frac{2L}{n} \quad\Rightarrow\quad f_n = \frac{v}{\lambda_n} = n\,\frac{v}{2L}, \qquad n = 1, 2, 3, \dots \qquad (1)

where: $v$ — the speed of sound in the air inside the pipe; $L$ — the pipe length; $f_n$ — the $n$ -th harmonic. The open pipe sounds every whole-number multiple of its fundamental — the complete, gap-free harmonic series, exactly as a string does. Take a pipe of length $L = 0.50\ \mathrm{m}$ and, using the room-temperature value $v = 340\ \mathrm{m\,s^{-1}}$ (state your value — this article uses $340\ \mathrm{m\,s^{-1}}$ throughout except where noted):

f_1 = \frac{v}{2L} = \frac{340}{2 \times 0.50} = 340\ \mathrm{Hz}, \quad f_2 = 680\ \mathrm{Hz}, \quad f_3 = 1020\ \mathrm{Hz}, \dots

A flute is essentially an open pipe (its embouchure hole acts as an open end), which is why it overblows to the octave — the second harmonic, $2f_1$ — when you blow harder.

The closed pipe: odd harmonics only

Now stop one end. That end must be a displacement node; the open end is still an antinode. The shortest wave that runs node-to-antinode fits a quarter of a wavelength into the pipe, so $L = \lambda_1/4$ and $\lambda_1 = 4L$ — twice as long a wavelength as the open pipe of the same length, hence half the frequency. Fitting more node-to-antinode shapes adds a half-loop each time, so the length is always an odd number of quarter-wavelengths:

L = (2n-1)\,\frac{\lambda_n}{4} \quad\Rightarrow\quad \lambda_n = \frac{4L}{2n-1} \quad\Rightarrow\quad f_n = (2n-1)\,\frac{v}{4L}, \qquad n = 1, 2, 3, \dots \qquad (2)

where: $(2n-1)$ runs through the odd numbers $1, 3, 5, \dots$ The fundamental is $f_1 = v/4L$ , and the higher modes are $3f_1, 5f_1, 7f_1, \dots$ — the even harmonics are physically impossible, because a standing wave with an even number of quarter-loops would put an antinode at the closed wall, which the wall forbids. For the same $L = 0.50\ \mathrm{m}$ pipe:

f_1 = \frac{v}{4L} = \frac{340}{4 \times 0.50} = 170\ \mathrm{Hz}, \quad f_3 = 3f_1 = 510\ \mathrm{Hz}, \quad f_5 = 5f_1 = 850\ \mathrm{Hz}, \dots

Compare the two fundamentals directly:

\frac{f_{1,\text{closed}}}{f_{1,\text{open}}} = \frac{v/4L}{v/2L} = \frac{1}{2} \qquad (3)

The closed pipe sings an octave lower for the same length — or, read the other way, a stopped organ pipe reaches a given bass note at half the height of an open one, which is exactly why cathedral organ builders stop their longest pipes to save space. A clarinet behaves as a closed pipe (the reed end is effectively closed), so it overblows to the twelfth — the third harmonic, $3f_1$ — not the octave, and its bottom register is unusually low for its length. Those are the audible fingerprints of equation (2).

Figure

The harmonic ladders of an open pipe and a closed pipe of the same length placed side by side, showing the open pipe carrying every harmonic and the closed pipe carrying only the odd ones

The end correction

There is one honest complication. An open end is not a perfectly sharp antinode: the vibrating air pokes a little way out of the pipe before it turns around, so the antinode sits slightly beyond the physical opening. The pipe therefore behaves as if it were a touch longer than it measures, and we patch this with an end correction $e$ added to the length:

L_{\text{eff}} = L + e, \qquad e \approx 0.6\,r = 0.3\,d \qquad (4)

where: $r$ is the pipe's radius and $d = 2r$ its diameter. The correction applies once per open end, so a closed pipe (one open end) uses $L_{\text{eff}} = L + e$ , while an open pipe (two open ends) uses $L_{\text{eff}} = L + 2e$ . Take a pipe of radius $r = 0.02\ \mathrm{m}$ (diameter $0.04\ \mathrm{m}$ ): the correction is $e = 0.6 \times 0.02 = 0.012\ \mathrm{m}$ , so a closed pipe of measured length $0.50\ \mathrm{m}$ has $L_{\text{eff}} = 0.50 + 0.012 = 0.512\ \mathrm{m}$ . Using the slightly more precise $v = 343\ \mathrm{m\,s^{-1}}$ , the corrected fundamental is

f_1 = \frac{v}{4L_{\text{eff}}} = \frac{343}{4 \times 0.512} = 167.5\ \mathrm{Hz},

against the uncorrected $343/(4 \times 0.50) = 171.5\ \mathrm{Hz}$ . The correction lowers the predicted pitch by about $4\ \mathrm{Hz}$ — small, but exactly the discrepancy a careful resonance-tube experiment measures, and the whole point of the two-length method in the resonance-tube archetype later in this chapter. For the same pipe used as an open pipe, both ends contribute: $L_{\text{eff}} = L + 1.2r = 0.50 + 0.024 = 0.524\ \mathrm{m}$ .

Derivation AThe mode formulas from the end rules

The goal is to obtain both master formulas — $f_n = n\,v/(2L)$ for the open pipe and $f_n = (2n-1)\,v/(4L)$ for the closed pipe — from nothing but the two boundary rules and the geometry of fitting standing waves. Nothing here needs calculus: we are counting how many quarter- and half-wavelengths fit between a given pair of ends, then converting to frequency with $f = v/\lambda$ . The strategy is to handle the two end-pairings separately, because that difference is the entire physics.

Step 1. Fix the end rules. Open end: air free to move, so a displacement antinode. Closed end: air pinned by the wall, so a displacement node. A resonant standing wave must satisfy the rule at each end simultaneously.

Step 2. Open pipe, antinode-to-antinode. The shortest such wave has an antinode at each end and one node between them: that is half a wavelength, so $L = \lambda_1/2$ . Adding a full half-loop each time keeps antinodes at both ends, so $L = n\lambda_n/2$ , giving $\lambda_n = 2L/n$ and $f_n = v/\lambda_n = n\,v/(2L)$ for $n = 1, 2, 3, \dots$ — every harmonic.

Step 3. Closed pipe, node-to-antinode. The shortest wave has a node at the wall and an antinode at the opening: that is a quarter wavelength, $L = \lambda_1/4$ . Each further resonance adds a half-loop (a node and an antinode), so the pipe always holds an odd number of quarter-wavelengths: $L = (2n-1)\lambda_n/4$ , giving $\lambda_n = 4L/(2n-1)$ and $f_n = (2n-1)\,v/(4L)$ for $n = 1, 2, 3, \dots$ — the odd harmonics only.

Step 4. Read the numbers for $L = 0.50\ \mathrm{m}$ , $v = 340\ \mathrm{m\,s^{-1}}$ . Open: $f_1 = 340$ , $f_2 = 680$ , $f_3 = 1020\ \mathrm{Hz}$ . Closed: $f_1 = 170$ , then $3f_1 = 510$ , $5f_1 = 850\ \mathrm{Hz}$ . The closed fundamental is exactly half the open fundamental. Verified: open f₁ = 340/(2·0.5) = 340 Hz, f₂ = 680, f₃ = 1020; closed f₁ = 340/(4·0.5) = 170 Hz, 3f₁ = 510, 5f₁ = 850 — recomputed in verify_hs2-air-column-resonance.py.

Go deeper · Derivation B · 1 minWhy the closed pipe drops its even harmonics and drops an octave

Derivation BWhy the closed pipe drops its even harmonics and drops an octave

Here we make precise the two most-tested facts about the closed pipe: it is missing every even harmonic, and its fundamental is an octave below the open pipe of the same length. Both come from the same source — the node forced at the stopped end — and the goal is to show that these are not two separate rules but one, seen twice. The strategy is to test whether an even harmonic could ever satisfy the closed-end rule, then compare fundamentals directly.

Step 1. Ask whether a would-be even harmonic can live in a closed pipe. An even harmonic $2f_1 = 2 \times v/4L = v/2L$ needs $\lambda = 2L$ , i.e. $L = \lambda/2$ — a half-wavelength in the pipe. But a half-wavelength that starts as a node at the wall ends as a node at the opening, and the opening must be an antinode. Contradiction: the pattern is forbidden. Every even multiple fails the same way, so only odd multiples survive.

Step 2. Confirm the survivors are exactly the odd ones. The allowed lengths are odd numbers of quarter-wavelengths, $L = (2n-1)\lambda/4$ , which gives frequencies $(2n-1)v/4L = 1, 3, 5, \dots$ times $v/4L$ . So the spectrum is $f_1, 3f_1, 5f_1, \dots$ with $f_1 = v/4L$ .

Step 3. Compare fundamentals of equal-length pipes. Open: $f_{1,\text{open}} = v/2L$ . Closed: $f_{1,\text{closed}} = v/4L$ . Their ratio is $(v/4L)/(v/2L) = 1/2$ — the closed pipe is one octave lower. With $L = 0.50\ \mathrm{m}$ , $v = 340\ \mathrm{m\,s^{-1}}$ : open $340\ \mathrm{Hz}$ , closed $170\ \mathrm{Hz}$ .

Step 4. State the musical consequence, which is a good memory hook. An open flute overblows to the octave (its next mode is $2f_1$ ); a closed clarinet overblows to the twelfth (its next mode is $3f_1$ , because $2f_1$ does not exist). The "missing evens" are audible. Verified: the ratio f₁,closed / f₁,open = (v/4L)/(v/2L) = 0.5 exactly, and 340 Hz vs 170 Hz for L = 0.50 m — recomputed in verify_hs2-air-column-resonance.py.

Limiting cases

case	result	meaning
open pipe, $n = 1$	$\lambda_1 = 2L$ , $f_1 = v/2L$	fundamental: half a wave in the pipe, antinode at each end
closed pipe, $n = 1$	$\lambda_1 = 4L$ , $f_1 = v/4L$	fundamental: a quarter wave, node at the wall
same $L$ , open vs closed	$f_{1,\text{closed}} = \tfrac12 f_{1,\text{open}}$	the closed pipe is an octave lower for the same length
closed pipe spectrum	$f_1, 3f_1, 5f_1, \dots$	even harmonics forbidden by the closed-end node
add end correction	$L \to L + e$ (closed), $L + 2e$ (open)	the pitch drops slightly; the antinode sits outside the opening
raise the temperature	$f_n$ rises with $v$	the pipe length is fixed, but the speed of sound climbs

Common trapA closed pipe is not just a shorter open pipe

The tempting mistake is to treat every pipe with $f_n = n\,v/(2L)$ and only change $L$ . A closed pipe obeys $f_n = (2n-1)\,v/(4L)$ — a $4L$ in the denominator, not $2L$ , and only the odd multiples. Two consequences trip students up. First, the closed fundamental is $v/4L$ , half the open fundamental, so the same length gives an octave-lower note. Second, the closed pipe is missing its even harmonics entirely, so its overtones jump $f_1 \to 3f_1 \to 5f_1$ , never touching $2f_1$ or $4f_1$ . And do not put a node at an open end or an antinode at a closed one — that inverts the whole analysis. A quick self-check: a closed pipe's fundamental wavelength is four times its length; an open pipe's is twice its length.

Worked examples

Board level — open and closed pipes of the same length (CBSE and ICSE pattern)

A pipe is $0.50\ \mathrm{m}$ long. Taking the speed of sound as $340\ \mathrm{m\,s^{-1}}$ , find the first three resonant frequencies (a) if it is open at both ends, and (b) if it is closed at one end. Ignore end corrections.

Step 1. Open pipe uses $f_n = n\,v/(2L)$ , all harmonics. Fundamental: $f_1 = 340/(2 \times 0.50) = \boxed{340\ \mathrm{Hz}}$ . Then $f_2 = 2f_1 = \boxed{680\ \mathrm{Hz}}$ and $f_3 = 3f_1 = \boxed{1020\ \mathrm{Hz}}$ .

Step 2. Closed pipe uses $f_n = (2n-1)\,v/(4L)$ , odd harmonics only. Fundamental: $f_1 = 340/(4 \times 0.50) = \boxed{170\ \mathrm{Hz}}$ .

Step 3. The next two closed resonances are the odd multiples: $3f_1 = \boxed{510\ \mathrm{Hz}}$ and $5f_1 = \boxed{850\ \mathrm{Hz}}$ — skipping $340$ and $680$ , which are the forbidden even harmonics.

Step 4. Sense-check with equation (3): the closed fundamental ( $170$ ) is exactly half the open fundamental ( $340$ ), the octave drop it must show. If your closed fundamental had come out equal to or above the open one, you used $2L$ where you needed $4L$ .

Board level — the temperature twist (JEE-Main and NEET pattern)

An open pipe sounds a fundamental of $340\ \mathrm{Hz}$ on a day when $v = 340\ \mathrm{m\,s^{-1}}$ . On a hotter day the speed of sound is $346\ \mathrm{m\,s^{-1}}$ . What is the new fundamental? Does shortening the pipe or heating the room raise the pitch more reliably?

Step 1. The pipe length is fixed; only $v$ changed. Since $f_1 = v/(2L)$ and $L$ is constant, $f_1 \propto v$ , so the new fundamental scales with the speed: $f_1' = 340 \times (346/340) = \boxed{346\ \mathrm{Hz}}$ .

Step 2. Read the physics: an organ goes sharp as the hall warms up, because the air carries sound faster while the metal pipes barely change length. This is why organs are tuned at performance temperature.

Step 3. Compare the two levers. Heating from $340$ to $346\ \mathrm{m\,s^{-1}}$ is a $1.8\%$ rise in pitch — small and drifting. Halving the length would double the pitch. Length is the strong, controllable dial; temperature is a small, unwanted drift you correct for.

Competitive level — resonance-tube reading with end correction (JEE and NEET lab pattern)

A tube closed at the lower end by water has an air column of measured length $0.50\ \mathrm{m}$ and internal radius $2.0\ \mathrm{cm}$ . Taking $v = 343\ \mathrm{m\,s^{-1}}$ , find the fundamental frequency with and without the end correction, and comment on the size of the error.

Step 1. End correction from equation (4): $e = 0.6r = 0.6 \times 0.020 = 0.012\ \mathrm{m}$ . It applies once (one open end), so $L_{\text{eff}} = 0.50 + 0.012 = 0.512\ \mathrm{m}$ .

Step 2. Corrected fundamental (closed pipe, $4L_{\text{eff}}$ ): $f_1 = 343/(4 \times 0.512) = \boxed{167.5\ \mathrm{Hz}}$ .

Step 3. Uncorrected, using the raw length: $f_1 = 343/(4 \times 0.50) = \boxed{171.5\ \mathrm{Hz}}$ .

Step 4. The correction lowers the prediction by about $4\ \mathrm{Hz}$ , roughly $2\%$ — small but systematic, and always in the same direction (the true note is lower than the naive formula says). Ignoring it in a resonance-tube experiment biases the measured speed of sound, which is exactly why the standard method uses two resonance lengths and subtracts them to cancel $e$ (the resonance-tube archetype does this). Timing note: about 90 seconds. The one trap is applying the correction twice — this is a closed pipe with a single open end, so $L_{\text{eff}} = L + e$ , not $L + 2e$ . Add $2e$ only for a pipe open at both ends.

Exam traps

$4L$ for a closed pipe, $2L$ for an open one. The commonest single error is using $f = n\,v/(2L)$ for a stopped pipe. A closed pipe's fundamental wavelength is $4L$ , its fundamental is $v/4L$ , and its harmonics are the odd multiples only. Write "closed = quarter wave = $4L$ " at the top of your working.

Even harmonics belong only to the open pipe. A closed pipe cannot sound $2f_1$ , $4f_1$ , $6f_1$ . If a question asks for "the third resonance of a closed pipe", it wants $5f_1$ (the third surviving mode: $f_1, 3f_1, 5f_1$ ), not $3f_1$ — read whether it means the third harmonic or the third resonance.

Node at a wall, antinode at an opening — never the reverse. Putting an antinode at the closed end (or a node at the open end) inverts every formula. Track displacement: the air cannot move at a wall (node) and moves freely at an opening (antinode).

End correction is per open end. Closed pipe: $L_{\text{eff}} = L + e$ . Open pipe: $L_{\text{eff}} = L + 2e$ , because both ends leak. And $e \approx 0.6r$ uses the radius; a diameter given in the question needs $e \approx 0.3d$ .

Resonance rises with temperature, not with pressure. Warm air carries sound faster, so $f_n \propto v$ climbs as the room heats — the pipe length is fixed. Pressure alone does not change $v$ (it cancels in $v = \sqrt{\gamma P/\rho}$ ), so "higher pressure, higher pitch" is wrong. This is a favourite JEE distractor.

Board scope. Open and closed pipes and their harmonics are on CBSE, ICSE, JEE, NEET, AP Physics 2 and IB (both levels); IGCSE 0625 has no standing waves at all. The end-correction numericals, though, are an India-facing lab drill — do not spend AP or IB revision time on $0.6r$ corrections, which those syllabuses do not assess.

Origins · 1 minArnold Sommerfeld and how this was discovered

Arnold Sommerfeld

1868–1951 · Munich

Arnold Sommerfeld was the great teacher of early quantum physics — four of his Munich students won Nobel Prizes — but his tie to this article runs through his lecture volume Partial Differential Equations in Physics, still a standard reference. There he laid out the canonical way to solve the wave equation subject to boundary conditions, and the vibrating air column is precisely that kind of boundary-value problem: a wave equation inside the pipe, with the antinode-and-node rules imposed at the ends. The end correction — the fact that an open end is not a mathematically perfect boundary — is exactly the sort of subtlety his boundary analysis was built to handle. Sommerfeld did not build organ pipes or study music; he mapped the mathematics that governs every bounded wave, of which the resonating pipe is one of the friendliest examples. He also gave his name to the radiation condition and the fine-structure constant.

History

1636

— Marin Mersenne relates an organ pipe's pitch to its length and records that a stopped pipe sounds an octave below an open pipe of equal length.

1701

— Joseph Sauveur coins the words node, antinode and acoustics, and measures the overtones of pipes and strings, founding the science he named.

1759

— Joseph-Louis Lagrange, in his memoir on the propagation of sound, treats the vibrating air column as a boundary-value problem and recovers its modes.

1787

— Ernst Chladni's experiments on vibrating plates make nodal patterns visible, confirming that a bounded body rings only in discrete modes.

1877

— Lord Rayleigh's Theory of Sound systematises the acoustics of pipes, including the end correction of roughly 0.6 times the radius.

Going further

Texts. NCERT Physics Class 11, Ch. 15 ("Waves"), sections on standing waves in air columns and organ pipes · H. C. Verma, Concepts of Physics Vol. 1, "Sound Waves", organ pipes and resonance columns · Halliday, Resnick and Walker, "Waves — II", sound in pipes · Cambridge A-Level 9702 topic 8 (stationary waves in air columns).

Provenance. Standard results. Recompute: open pipe $f_1 = 340/(2\times0.5) = 340\ \mathrm{Hz}$ , $f_2 = 680$ , $f_3 = 1020$ ; closed pipe $f_1 = 340/(4\times0.5) = 170\ \mathrm{Hz}$ , $3f_1 = 510$ , $5f_1 = 850$ ; ratio $f_{1,\text{closed}}/f_{1,\text{open}} = 0.5$ ; end correction $e = 0.6\times0.02 = 0.012\ \mathrm{m}$ , closed $L_{\text{eff}} = 0.512\ \mathrm{m}$ , $f_1 = 343/(4\times0.512) = 167.5\ \mathrm{Hz}$ vs uncorrected $343/(4\times0.5) = 171.5\ \mathrm{Hz}$ ; open $L_{\text{eff}} = 0.50 + 1.2\times0.02 = 0.524\ \mathrm{m}$ ; temperature scaling $340 \times 346/340 = 346\ \mathrm{Hz}$ .

Prerequisites. Normal Modes and Harmonics on a String — the same boundary-condition logic for a fixed-fixed string, of which the pipe is the acoustic twin. Leads to. ARCHETYPE: Resonance-Tube and Pipe Problems — the exam drill, including the two-length end-correction method · Speed of Sound: Newton–Laplace and Factors — where the $v$ used here comes from.

Go deeper (UG). The air column as an eigenvalue problem of the wave equation, and the physical origin of the end correction, are developed in Normal Modes and Standing Waves on a String (the same mathematics with different boundary conditions) in the university mechanics track. Enrichment only.

Why is a closed end a displacement node but a pressure antinode?

Because displacement and pressure in a sound wave are a quarter-cycle out of step. At the closed wall the air is jammed up and cannot move, so its displacement is zero — a node. But that is exactly where the air is most compressed and rarefied as the wave pushes against the wall, so the pressure swings hardest there — a pressure antinode. The two descriptions are consistent, just phrased in different variables. We build the end rules from displacement because "the air can't move at a wall, and moves freely at an opening" is the most intuitive statement. If your textbook draws pressure waves instead, every node and antinode simply swaps — the physics is identical.

Why does a clarinet sound an octave lower than a flute of about the same length?

Because a clarinet is a closed pipe and a flute is an open one. The flute is open at both ends (the mouth hole and the far end), so its fundamental fits half a wavelength and is $v/2L$ . The clarinet's reed end is effectively closed, so its fundamental fits only a quarter wavelength and is $v/4L$ — half the frequency, an octave lower, for a similar length. The same closed-end rule is why a clarinet overblows to the twelfth (its next note is $3f_1$ ) rather than the octave, giving it that distinctive break between registers. Two instruments, one boundary condition apart.

What exactly is the end correction, and why 0.6 times the radius?

The air just inside an open end does not stop moving precisely at the opening — it carries a little way out into the surrounding air before it turns around, so the pressure antinode sits slightly outside the pipe. The pipe therefore resonates as if it were a bit longer than it measures, by an amount called the end correction, $e \approx 0.6r$ , set by how far that extra plug of air reaches, which scales with the pipe's radius. Lord Rayleigh worked out the coefficient; for a plain open end it is close to $0.6$ times the radius. It applies once per open end, so a stopped pipe adds $e$ and an open pipe adds $2e$ . In careful experiments you cancel it by taking two resonance lengths and subtracting.

If I blow harder, why does the note sometimes jump to a higher one?

You are exciting a higher mode — overblowing. Blowing gently favours the fundamental; blow harder (or change your lip or breath) and the air column locks onto its next allowed resonance instead. For an open pipe like a flute that next resonance is $2f_1$ , one octave up. For a closed pipe like a clarinet the next resonance is $3f_1$ (there is no $2f_1$ ), so it jumps a twelfth. A bugle, which has no valves, plays a tune only by overblowing between its harmonics — every note it can sound is a member of one pipe's harmonic series, which is why bugle calls use just a handful of pitches.

Does the gas inside the pipe matter, or only the length?

It matters through the speed of sound. The resonance is $f_n \propto v$ , and $v$ depends on the gas: sound travels faster in a light, hot gas than in a heavy, cold one. Breathe helium and your vocal tract — an air column — resonates at higher frequencies because $v$ in helium is nearly three times that in air, which is the squeaky-voice trick (the pitch of your vocal folds barely changes, but the resonances that shape the sound shift way up). Warm the air and $v$ rises, so the pitch drifts sharp. The length sets the wavelengths the pipe allows; the gas sets how fast those wavelengths turn into frequencies.

Our resonance-tube reading did not match v = 340 m/s. What went wrong?

Almost certainly the end correction, and it is not a mistake — it is real physics. If you compute the speed from a single resonance length using $v = 4fL$ , you ignore the little bit of length $e$ that the antinode pokes outside the tube, so your $L$ is too short and your $v$ comes out low by a few per cent. The fix is the standard two-length method: find two successive resonances and subtract their lengths, which cancels $e$ exactly and gives $v = 2f(L_2 - L_1)$ . The next-chapter resonance-tube archetype is built entirely around that subtraction. A smaller error can come from using the wrong temperature's value of $v$ .