Standing waves in pipes — open and closed ends — Physics

Before this

This leads to

Resonance

Two pipes, nearly the same length, an octave apart

A flute and a clarinet are both tubes of air about 60 cm long. You might expect them to have the same lowest note. They do not. The flute's lowest note is about 262 Hz. The clarinet's is about 147 Hz, nearly an octave lower (Acoustics Today, 2018).

There is a second difference. Blow harder on a flute with the same fingers, and the note jumps up an octave: the frequency doubles. On a clarinet, the player presses a small key and the note jumps further, to three times the frequency. Musicians call that jump a twelfth.

Both puzzles have one cause. The flute is open at both ends. The clarinet behaves like a pipe closed at one end, because the reed at the mouthpiece acts like a closed end. The last lecture ended with a promise: "a pipe with open or closed ends gives a different list of harmonics, and a different sound." This lecture works out that list.

What the ends of a pipe do to the air

Sound in a pipe is a longitudinal wave. The air moves backwards and forwards along the pipe, not from side to side. But the standing-wave idea is exactly the same as on a string. Waves reflect from the ends, travel both ways, and add up to fixed nodes and antinodes. So the question is the same as before: what does each end force on the air?

A closed end. The air touching a solid wall cannot move into the wall. So the displacement of the air there is always zero: a closed end is a node. You have met this already. In the standing waves lecture, Kemal heard quiet places in front of a hard wall. The wall was a closed end.

An open end. At an open end the air is free to move in and out, into the open air outside. So it moves the most: an open end is an antinode.

Some books describe the same pipe using pressure instead of displacement. Then everything swaps: an open end is a pressure node, because the pressure there stays at the pressure of the room, and a closed end is a pressure antinode. Both descriptions are correct. In this course we describe the displacement of the air, so a closed end is a node and an open end is an antinode.

Stop and think

A student says: "At the closed end, the air is squashed against the wall, so it must be where the air moves most." What is wrong with this?

Show the answer

The student has mixed up pressure and movement. At a closed end the air is squashed and stretched the most, so the pressure changes most there: a pressure antinode. But it is squashed precisely because it cannot move. Its displacement is always zero: a displacement node. A closed end is a node for movement and an antinode for pressure. Always say which one you mean.

A pipe open at both ends

An open pipe has an antinode at each end. From the standing waves lecture, neighbouring antinodes are half a wavelength apart, $\frac{λ}{2}$ . So, exactly as for a string, the pipe must hold a whole number of half-wavelengths. The only difference is that the pattern is turned inside out: a string has nodes at its ends, and an open pipe has antinodes.

L = n \times \frac{λ _{n}}{2} so λ_{n} = \frac{2 L}{n} and f_{n} = \frac{n v}{2 L}, n = 1, 2, 3, \dots

Here $v$ is the speed of sound in the air in the pipe, about 343 m/s at 20 °C. It is the same for every harmonic. So an open pipe has every harmonic: $f_{1}, 2 f_{1}, 3 f_{1}, \dots$ , the same list as a string.

The displacement of the air in the first three possible patterns of each pipe. The solid line shows the air's displacement at one moment, and the dashed line half a period later. Dark dots are nodes. Top: a pipe open at both ends, with an antinode at each end. Bottom: a pipe closed at the left end, with a node at the closed end and an antinode at the open end.

A pipe closed at one end

Now close one end. There is a node at the closed end and an antinode at the open end. This is the hard part of the lecture, so go slowly.

From the standing waves lecture, a node and its nearest antinode are a quarter of a wavelength apart, $\frac{λ}{4}$ . So the shortest pattern that fits, the fundamental, is one quarter-wavelength: $L = \frac{λ _{1}}{4}$ , so $λ_{1} = 4 L$ . That is twice the wavelength of an open pipe of the same length. So the frequency is half: $f_{1} = \frac{v}{4 L}$ , one octave lower.

To fit the next pattern, add another node and antinode inside the pipe. That adds half a wavelength, so the pipe now holds three quarter-wavelengths: $L = \frac{3 λ}{4}$ . The frequency is $\frac{3 v}{4 L} = 3 f_{1}$ . The pattern after that holds five quarter-wavelengths, at $5 f_{1}$ .

Why can a closed pipe never hold two, four or six quarter-wavelengths? Because an even number of quarter-wavelengths is a whole number of half-wavelengths. Half a wavelength always joins two places of the same kind: node to node, or antinode to antinode. But this pipe needs a node at one end and an antinode at the other. Only an odd number of quarters can join two places of different kinds.

That is it. That is the whole idea. A closed pipe has only odd harmonics, because its two ends are different.

Standing waves in pipes

Open at both ends (antinode at each end): $λ_{n} = \frac{2 L}{n}$ and $f_{n} = \frac{n v}{2 L} = n f_{1}$ , for $n = 1, 2, 3, \dots$ All harmonics.

Closed at one end (node at the closed end, antinode at the open end): $λ_{n} = \frac{4 L}{n}$ and $f_{n} = \frac{n v}{4 L} = n f_{1}$ , for $n = 1, 3, 5, \dots$ Odd harmonics only.

A closed pipe's fundamental is half that of an open pipe of the same length: one octave lower.

Notice the numbering. In a closed pipe the second possible frequency is the 3rd harmonic, not "the second harmonic". The harmonic number $n$ is always the multiple of $f_{1}$ . There is no 2nd harmonic in a closed pipe at all.

Now return to the hook. Treat both instruments as ideal 60 cm pipes. The flute, open at both ends, would have $f_{1} = \frac{343}{2 \times 0.60} \approx 286$ Hz. The clarinet, closed at one end, would have $f_{1} = \frac{343}{4 \times 0.60} \approx 143$ Hz, an octave lower. The real notes, 262 Hz and 147 Hz, are not exactly an octave apart, because neither instrument is a perfect cylinder: the clarinet's flared bell changes its lowest note, for example. But the model explains the big difference. It also explains the jumps. A flute's next pattern is at $2 f_{1}$ , an octave up. A clarinet has no $2 f_{1}$ , so its next pattern is at $3 f_{1}$ : the twelfth.

Predict — closing one end

An open pipe 0.50 m long sounds its fundamental at about 343 Hz. You put a cap on one end. What is the new lowest note? Choose one, then read the explanation.

The four answers students most often give, and the right one.

About 686 Hz, an octave higher, because the pipe is "shorter" for the air. Not right.
About 343 Hz, unchanged, because the length is the same. Not right.
About 172 Hz, an octave lower. Right: the fundamental now fits a quarter-wavelength, so λ = 4L = 2.0 m and f = 343 ÷ 2.0 ≈ 172 Hz.
About 1029 Hz, the 3rd harmonic. Not right: that is the second possible note, not the lowest.

What should you have learned? Capping a pipe does not change its length. It changes the pattern that fits inside it. A quarter-wavelength instead of a half-wavelength doubles the wavelength and halves the frequency.

A real pipe is slightly longer than it looks

One detail makes real pipes behave a little differently. The air just outside an open end also moves with the wave. So the antinode is not exactly at the end of the pipe. It is a short distance beyond it. That distance is called the end correction. For a plain pipe it is about 0.6 times the radius of the pipe (Britannica). A closed end has no end correction: the node is exactly at the wall.

So a closed pipe behaves as if its length were $L + e$ , where $e$ is the end correction. For a narrow school tube, $e$ is less than a centimetre. It matters in careful experiments, and the classic experiment below removes it neatly.

Real-world model — the resonance tube

A glass tube of radius 1.5 cm stands in water, so it is a pipe closed at the bottom by the water surface. A tuning fork sounds above it. Raise the tube to make the air column longer, and listen for the loud places. Then try another fork. (A simplified model: the loudness rises sharply when a pattern fits.)

Length of the air column, L: 0.100 m

A resonance tube of radius 1.5 cm (end correction about 0.009 m), closed at the bottom by water. The table gives the air-column lengths at which each fork's sound becomes loud, taking the speed of sound as 343 m/s.

Fork	1st loud length	2nd loud length	Difference = λ/2
512 Hz	≈ 0.158 m	≈ 0.493 m	≈ 0.335 m
384 Hz	≈ 0.214 m	≈ 0.661 m	≈ 0.447 m
256 Hz	≈ 0.326 m	≈ 0.996 m	≈ 0.670 m

The first loud length is a quarter-wavelength minus the end correction. The second is three quarters minus the same correction. So their difference is exactly half a wavelength, and the end correction cancels.

What should you have seen? Each fork gives loud places at definite lengths, and nowhere in between. The first loud length is a little less than a quarter-wavelength, because the antinode sits just above the top of the tube. The gap between neighbouring loud lengths is always exactly half a wavelength. Lower forks have longer wavelengths, so their loud places are further apart.

Working with pipes

Worked Example 1 — Open and closed organ pipes (invented)

An organ has two pipes, each 0.85 m long. One is open at both ends. The other is closed at the top. Find the first three frequencies each can sound. Take $v = 343 m/s$ .

Open: $f_{1} = \frac{v}{2 L} = \frac{343}{1.70} \approx 202 Hz$ . Every harmonic is allowed, so $f_{2} = 2 f_{1} \approx 404 Hz$ and $f_{3} = 3 f_{1} \approx 605 Hz$ .

Closed: $f_{1} = \frac{v}{4 L} = \frac{343}{3.40} \approx 101 Hz$ . Only odd harmonics, so the next two are $3 f_{1} \approx 303 Hz$ and $5 f_{1} \approx 504 Hz$ .

Check: the closed pipe's fundamental is half the open pipe's, $202 \div 2 = 101$ . ✓ Organ builders use this. A closed ("stopped") pipe sounds the same low note as an open pipe twice as long, so it saves space.

Worked Example 2 — Open or closed? (invented)

A pipe 0.60 m long sounds a note of 286 Hz. Is it open or closed, and which harmonic is this?

Try both, because one test alone can mislead. If open: $f_{1} = \frac{343}{1.20} \approx 286 Hz$ . So 286 Hz is $1 \times f_{1}$ : the fundamental. Possible.

If closed: $f_{1} = \frac{343}{2.40} \approx 143 Hz$ . Then $286 \div 143 \approx 2$ . But $n = 2$ is not allowed in a closed pipe. Impossible.

So the pipe is open at both ends, sounding its fundamental. Check: $λ = 2 L = 1.20 m$ , and $f λ = 286 \times 1.20 \approx 343 m/s$ . ✓

Worked Example 3 — How long is a clarinet?

A clarinet's lowest note is about 147 Hz, with all its holes closed. Treat it as a closed pipe. Estimate the length of its air column. Then find the length that sounds 294 Hz, an octave higher.

For the fundamental, $L = \frac{λ}{4} = \frac{v}{4 f _{1}} = \frac{343}{4 \times 147} \approx 0.583 m$ . A clarinet is about 60 cm long (HyperPhysics), so the model fits well.

For 294 Hz: $L = \frac{343}{4 \times 294} \approx 0.292 m$ , half as long. The clarinet does not get shorter, of course. The player opens tone holes. An open hole lets the air move freely there, so it acts like the open end of a shorter pipe. A clarinettist plays notes near this one with roughly half the instrument (UNSW Music Acoustics).

Check the octave: half the length, double the frequency, $147 \times 2 = 294$ . ✓

Worked Example 4 — Speed of sound from a resonance tube

With a 512 Hz fork, a resonance tube is loud at air-column lengths of 0.158 m and 0.493 m. Find the speed of sound and the end correction.

Neighbouring loud lengths are half a wavelength apart: $\frac{λ}{2} = 0.493 - 0.158 = 0.335 m$ . So $λ = 0.670 m$ , and $v = f λ = 512 \times 0.670 \approx 343 m/s$ .

For the first loud length, $L_{1} + e = \frac{λ}{4} = 0.1675 m$ . So $e \approx 0.1675 - 0.158 \approx 0.0095 m$ , about 1 cm.

Check: for a tube of radius 1.5 cm, $0.6 \times 0.015 = 0.009 m$ . ✓ Taking the difference of two lengths is the trick: the unknown end correction cancels.

Three classic mistakes

Trap 1 — A node at the open end

An open end is a displacement antinode: the air there moves the most. A closed end is a displacement node. If your book uses pressure, the words swap. Decide which one you are describing, then stay with it.

Trap 2 — Calling a closed pipe's second note the 2nd harmonic

The second note of a closed pipe is $3 f_{1}$ , the 3rd harmonic. A harmonic's number is always its multiple of $f_{1}$ , so a closed pipe has no 2nd, 4th or 6th harmonic. Some books call $3 f_{1}$ "the first overtone": as on a string, overtones are counted differently.

Trap 3 — Using λ = 2L for a closed pipe

For a closed pipe, the fundamental is a quarter-wavelength: $λ_{1} = 4 L$ , not $2 L$ . Using $2 L$ makes every answer an octave too high. Draw the pattern first, count the quarter-wavelengths, then write the formula.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck. Take the speed of sound as 343 m/s.

A pipe 0.50 m long is open at both ends. (a) Find its first three harmonic frequencies. (b) Draw the displacement pattern of the 2nd harmonic, and mark the nodes and antinodes.

Hint 1
An open pipe has an antinode at each end, so $f_{1} = \frac{v}{2 L}$ .

Hint 2
The 2nd harmonic fits two half-wavelengths, so there are two nodes inside.

Full answer
(a) $f_{1} = \frac{343}{1.00} = 343 Hz$ , $f_{2} = 686 Hz$ , $f_{3} = 1029 Hz$ . (b) Antinodes at both ends and in the middle, with nodes a quarter and three quarters of the way along: A N A N A. Check: $λ_{2} = L = 0.50 m$ , and $686 \times 0.50 = 343 m/s$ . ✓
A pipe closed at one end has two neighbouring resonant frequencies of 165 Hz and 275 Hz. (a) Find its fundamental frequency. (b) Which harmonics are these? (c) Find the length of the pipe.

Hint 1
A closed pipe has only odd harmonics. Neighbouring ones differ by $2 f_{1}$ .

Hint 2
$275 - 165 = 110 = 2 f_{1}$ . Then use $L = \frac{v}{4 f _{1}}$ .

Full answer
(a) $2 f_{1} = 110$ , so $f_{1} = 55 Hz$ . (b) $165 = 3 \times 55$ and $275 = 5 \times 55$ : the 3rd and 5th harmonics. (c) $L = \frac{343}{4 \times 55} \approx 1.56 m$ . Check: both are odd multiples of 55. ✓
A pipe 1.225 m long sounds two neighbouring resonances, at 210 Hz and 350 Hz. Is it open at both ends or closed at one end? Explain.

Hint 1
For an open pipe, neighbouring resonances differ by $f_{1}$ . For a closed pipe, they differ by $2 f_{1}$ .

Hint 2
Work out $f_{1}$ both ways from the gap of 140 Hz, and test whether 210 is an allowed multiple.

Full answer
The gap is 140 Hz. If open, $f_{1} = 140$ and $210 \div 140 = 1.5$ , not a whole number: impossible. If closed, $f_{1} = 70$ , and $210 = 3 \times 70$ , $350 = 5 \times 70$ : the 3rd and 5th harmonics. So it is closed at one end. Check with the length: $f_{1} = \frac{343}{4 \times 1.225} = 70 Hz$ . ✓
Your ear canal is a tube about 2.5 cm long, open to the air at one end and closed by the eardrum at the other. (a) Find its fundamental frequency. (b) Find the next frequency at which it resonates. (c) Human hearing is most sensitive at about 2000–5000 Hz. Suggest why. (d) In a young baby, the canal's resonance is near 8000 Hz. Estimate the length of the baby's ear canal.

Hint 1
The eardrum is a closed end, so the canal is a closed pipe: $f_{1} = \frac{v}{4 L}$ .

Hint 2
(b) A closed pipe has no 2nd harmonic. (d) Rearrange to $L = \frac{v}{4 f _{1}}$ .

Full answer
(a) $f_{1} = \frac{343}{4 \times 0.025} \approx 3430 Hz$ . (b) The next is the 3rd harmonic, $3 \times 3430 \approx 10300 Hz$ . (c) Sounds near 3–4 kHz set up a standing wave in the canal, so they arrive at the eardrum with a larger amplitude. Medical sources describe a boost of about 10–15 dB in this range (Weber and Khariwala). (d) $L = \frac{343}{4 \times 8000} \approx 0.011 m$ , about 1 cm. Check: a shorter pipe gives a higher fundamental. ✓ The real canal is curved and not a perfect cylinder, so these are estimates.

The bigger idea

Strings, open pipes and closed pipes all follow one method. Find what each end forces: a node or an antinode. Count how many quarter-wavelengths fit between the ends. The allowed frequencies follow from $v = f λ$ . Same ends give whole numbers of half-wavelengths and every harmonic. Different ends give odd numbers of quarter-wavelengths and only odd harmonics.

You can hear the difference. In its low register, a clarinet's odd harmonics are strong and its even harmonics are weak. Acousticians link this to its "hollow" sound (UNSW Music Acoustics). In higher registers, where the bore stops behaving like an ideal cylinder, the even harmonics come back and the sound becomes brighter. A flute has every harmonic from the start. The next lecture asks what happens when you drive a pipe, a string or a bridge at one of its allowed frequencies from outside. The answer is resonance.

Key concepts in this lesson

In a pipe, a closed end is a displacement node, because the air cannot move into the wall, and an open end is a displacement antinode.
A pipe open at both ends holds a whole number of half-wavelengths, so $f_{n} = \frac{n v}{2 L}$ for every whole number $n$ , the same list as a string.
A pipe closed at one end holds an odd number of quarter-wavelengths, so $f_{n} = \frac{n v}{4 L}$ for $n = 1, 3, 5, \dots$ only.
A closed pipe's fundamental is an octave below that of an open pipe of the same length, because its fundamental wavelength is 4L instead of 2L.
The antinode at an open end lies slightly beyond the end, by the end correction, which is about 0.6 times the radius of the pipe.
In a resonance tube, neighbouring loud lengths are half a wavelength apart, so their difference gives the wavelength without needing the end correction.