Air columns can resonate just like strings, but the ends matter. Open pipes allow antinodes at both ends, while closed pipes force a node at the closed end and an antinode at the open end. That difference changes the harmonics the pipe can support.
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Why can a didgeridoo or bottle produce a deep note from a relatively short air column, while an open flute supports a different set of harmonics? Predict using the idea of what happens at the ends of the pipe.
Type your prediction below. You will revisit it at the end.
Write your prediction in your book. You will revisit it at the end.
Wrong: Vectors and scalars are just different ways of writing the same thing.
Right: Vectors have magnitude and direction; scalars have magnitude only. They follow different mathematical rules.
📚 Core Content
A pipe’s ends set the standing-wave boundary conditions, and those conditions decide exactly which patterns of vibration can survive inside the air column.
At an open end, the air is in direct contact with the atmosphere, so it can move most freely. That makes the open end a displacement antinode — a point of maximum air oscillation. At a closed end, the air is blocked by a solid surface and cannot move back and forth, so the closed end is a displacement node — a point of zero air oscillation. It is important to remember that pressure nodes and displacement nodes are swapped: where displacement is maximum (antinode), pressure variation is minimum (node), and vice versa.
These boundary conditions are not suggestions — they are strict mathematical requirements. Only standing-wave patterns that satisfy both ends simultaneously are allowed. This is why a didgeridoo, which is closed at the mouth end and open at the other, can only support a very specific set of resonant frequencies.
Open pipes have displacement antinodes at both ends and support the full harmonic series.
The fundamental in an open pipe fits exactly half a wavelength between the two open ends, so $L = \lambda/2$. Higher modes add extra half-wavelength sections, giving $L = n\lambda/2$ for $n = 1, 2, 3, ...$. Because every half-wavelength addition places an antinode at each end, all integer harmonics are permitted. The second harmonic has one full wavelength inside the pipe, the third harmonic has one and a half wavelengths, and so on.
Once the wavelength is known, the resonant frequency follows from $v = f\lambda$. For the fundamental, $f_1 = v/(2L)$. For the $n$th harmonic, $f_n = n \times f_1$. This linear spacing of harmonics gives open-pipe instruments their bright, rich timbre.
Closed pipes have a displacement node at the closed end, an antinode at the open end, and allow only odd harmonics.
The fundamental in a closed pipe fits one quarter of a wavelength inside the pipe, so $L = \lambda/4$. The next possible pattern must still have a node at the closed end and an antinode at the open end, which requires adding another half-wavelength segment. This gives $L = 3\lambda/4$, then $5\lambda/4$, and so on. The general condition is $L = n\lambda/4$ where $n$ is restricted to odd integers ($1, 3, 5, ...$).
Mathematically, an even value of $n$ would require either two nodes or two antinodes at the ends, violating the closed-open boundary condition. This absence of even harmonics gives closed-pipe instruments their mellower, more hollow timbre compared with open-pipe instruments of similar length.
A resonance tube uses a changing air-column length to find where standing-wave resonance occurs, turning a simple tube into a precise tool for measuring the speed of sound.
In the standard school practical, a hollow tube is placed over a water reservoir. A tuning fork of known frequency is struck and held above the open end of the tube. The tube is slowly raised or lowered, changing the length of the air column above the water. When the air-column length matches an allowed standing-wave mode for a closed pipe, the sound becomes noticeably louder — this is resonance.
The water surface acts as a closed end (displacement node), while the top of the tube is open (displacement antinode). The first resonance occurs when the air-column length equals $\lambda/4$. The next resonance occurs at $3\lambda/4$, so the distance between the first and second resonance positions is exactly $\lambda/2$. By measuring this distance and knowing the tuning-fork frequency, students can calculate the wavelength and then determine the speed of sound using $v = f\lambda$.
Different harmonic sets help give instruments their distinctive sounds, or timbres, even when they play the same fundamental note.
Timbre is essentially the harmonic fingerprint of an instrument. An open pipe can support all harmonics, while a closed pipe supports only odd harmonics. That changes which resonant frequencies are emphasised and contributes to the different tonal colours of instruments. A closed pipe of a given length has a fundamental frequency one octave lower than an open pipe of the same length, and its sound is mellower because the even harmonics that add brightness are missing.
Instrument makers exploit these differences deliberately. A flute is designed as an open pipe to achieve a bright, penetrating tone. A stopped organ pipe (closed at one end) produces a softer, more flute-like tone because the even harmonics are absent. The didgeridoo’s deep, droning sound comes from the strong fundamental and the sparse odd-harmonic series of its closed-pipe geometry.
In a standing sound wave, displacement nodes and pressure nodes are never in the same place — they are always half a wavelength out of step.
At an open end, the air can move freely, so displacement is maximum (antinode). However, because the open end is exposed to the atmosphere, the pressure cannot vary much from atmospheric pressure, so pressure variation is minimum (node). At a closed end, the air cannot move, so displacement is zero (node), but the pressure can build up and vary strongly, making it a pressure antinode.
This swap between pressure and displacement is crucial in practical applications such as microphone placement and speaker design. It also appears in exam questions that ask specifically about pressure nodes rather than displacement nodes.
Visual Break — Decision Flowchart
✏️ Worked Examples
Scenario: An open organ pipe has length 0.80 m. Find the wavelength of the fundamental and the second harmonic. Then calculate the fundamental frequency if the speed of sound in air is 340 m/s.
If the pipe were closed at one end instead, the fundamental wavelength would double to $3.2\ \text{m}$ (since $L = \lambda/4$) and the fundamental frequency would drop to approximately 106 Hz. Explain why this makes closed-pipe instruments sound deeper for the same physical length.
Scenario: A closed pipe has length 0.60 m. Find the wavelength of the fundamental and the next allowed mode. Then find the fundamental frequency using $v = 340\ \text{m/s}$.
If you mistakenly used open-pipe rules, you would predict a non-existent second harmonic with wavelength $0.60\ \text{m}$ for the closed pipe. Explain why this prediction violates the boundary conditions.
Scenario: A tuning fork of frequency 512 Hz is held above a resonance tube filled with water. The first resonant length is measured as 16.5 cm and the second resonant length as 50.5 cm. Find the wavelength of the sound and the speed of sound in air.
The first resonance length were measured as 16.5 cm but the tuning fork frequency were unknown. If the speed of sound in air is taken as 340 m/s, how could you determine the frequency of the tuning fork?
🏃 Activities
For each statement below, write whether it describes an open pipe or a closed pipe.
A closed pipe has length 0.50 m. Find the wavelength of the fundamental.
Explain in one or two sentences why different harmonic patterns in air columns help produce different instrument timbres.
A student uses a 480 Hz tuning fork with a resonance tube. The first resonant length is 17.5 cm and the second resonant length is 53.5 cm.
Earlier you were asked why the ends of a pipe matter so much for the note it produces.
The full answer: the ends set the boundary conditions. Open ends behave like antinodes, while closed ends behave like nodes. That changes the wavelengths that fit in the air column and therefore changes the allowed harmonics and resonant frequencies.
Now revisit your prediction. How do the boundary conditions explain the different harmonic sets?
Annotate your prediction in your book with what you now understand differently.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
✅ Check Your Understanding
1. An open pipe has which boundary conditions?
2. A closed pipe supports:
3. For the fundamental of a closed pipe:
4. For an open pipe of length 1.0 m in the fundamental, the wavelength is:
5. The next allowed mode after the fundamental in a closed pipe corresponds to:
6. A resonance-tube practical is useful because it helps determine:
7. Explain the difference in boundary conditions between an open pipe and a closed pipe. 3 MARKS
8. An open pipe has length 0.75 m. Find the wavelength of the second harmonic. 3 MARKS
9. A closed pipe has length 0.90 m. Find the wavelength of the fundamental and explain why even harmonics are absent. 4 MARKS
1. A — open pipes have antinodes at both ends.
2. D — closed pipes allow odd harmonics only.
3. B — the fundamental of a closed pipe fits one quarter wavelength.
4. C — for the open-pipe fundamental, $\lambda = 2L = 2.0\ \text{m}$.
5. A — the next allowed mode is $L = 3\lambda/4$.
6. D — resonance length gives wavelength, which can then be used to find speed.
Q7 (3 marks): An open pipe has displacement antinodes at both ends because the air can move freely there. A closed pipe has a node at the closed end because the air cannot move there, and an antinode at the open end. These boundary conditions determine which standing-wave modes are allowed.
Q8 (3 marks): For an open pipe, $L = n\lambda/2$. For the second harmonic, $0.75 = 2\lambda/2$, so $\lambda = 0.75\ \text{m}$.
Q9 (4 marks): For the fundamental of a closed pipe, $L = \lambda/4$. So $\lambda = 4L = 3.6\ \text{m}$. Even harmonics are absent because a closed pipe must have a node at one end and an antinode at the other, so only odd quarter-wavelength patterns satisfy the boundary conditions.
Climb platforms using your knowledge of harmonics and standing waves in open and closed pipes. Pool: lessons 1–11.
Tick when you have finished the activities and checked the answers.