Stationary waves in air columns
Section: Superposition | Syllabus: Cambridge AS Level Physics (9702)
Stationary Sound Waves in Pipes
Sound waves travelling inside a tube (pipe) reflect from the ends, creating stationary waves through superposition. The pressure variations (compressions and rarefactions) of the longitudinal sound wave form patterns of nodes and antinodes.
Longitudinal Wave RepresentationDiagrams of stationary waves in pipes show displacement amplitude, not the actual compression/rarefaction pattern. The curves represent the maximum displacement of air molecules along the pipe axis-they are not transverse waves.
Boundary Conditions
- Closed end: Air molecules cannot move → displacement node (zero amplitude)
- Open end: Air molecules free to oscillate → displacement antinode (maximum amplitude)
Closed Pipes (One End Closed)
A closed pipe has a node at the closed end and an antinode at the open end.
FIG 8.8: Closed Pipe HarmonicsShow three diagrams for a closed pipe of length L: (a) 1st harmonic with N at closed end, A at open end (¼λ fits, λ = 4L), (b) 3rd harmonic with pattern N-A-N-A (¾λ fits, λ = 4L/3), (c) 5th harmonic with pattern N-A-N-A-N-A (5/4λ fits, λ = 4L/5). Label nodes N and antinodes A. Show yellow envelope curves representing amplitude variation.
| Harmonic | Pattern | λ in terms of L | Frequency |
|---|---|---|---|
| 1st (fundamental) | N-A (¼λ fits) | λ = 4L | f_1 = (v/4L) |
| 3rd | N-A-N-A (¾λ fits) | λ = (4L/3) | f_3 = 3f_1 |
| 5th | N-A-N-A-N-A | λ = (4L/5) | f_5 = 5f_1 |
| nth (odd only) | - | λ = (4L/n) | f_n = nf_1 |
Only odd harmonics (1st, 3rd, 5th, ...) exist in closed pipes. Even harmonics are impossible because they would require either two nodes or two antinodes at the ends, violating the boundary conditions.
Open Pipes (Both Ends Open)
An open pipe has antinodes at both ends.
FIG 8.9: Open Pipe HarmonicsShow three diagrams for an open pipe of length L: (a) 1st harmonic with A-N-A pattern (½λ fits, λ = 2L), (b) 2nd harmonic with A-N-A-N-A (1λ fits, λ = L), (c) 3rd harmonic with A-N-A-N-A-N-A (3/2λ fits, λ = 2L/3). Label nodes N and antinodes A.
| Harmonic | Pattern | λ in terms of L | Frequency |
|---|---|---|---|
| 1st (fundamental) | A-N-A (½λ fits) | λ = 2L | f_1 = (v/2L) |
| 2nd | A-N-A-N-A (1λ fits) | λ = L | f_2 = 2f_1 |
| 3rd | A-N-A-N-A-N-A | λ = (2L/3) | f_3 = 3f_1 |
| nth | - | λ = (2L/n) | f_n = nf_1 |
All harmonics (1st, 2nd, 3rd, ...) exist in open pipes. This gives open pipes a richer harmonic content than closed pipes of the same length.
Finding Wavelength Using Resonance
The wavelength of sound can be determined using resonance in air columns.
FIG 8.10: Resonance Tube ExperimentShow: open-ended tube partially submerged in water (forming a closed pipe), tuning fork held at open end. Label: tube length L (adjustable by raising/lowering tube), water level, tuning fork. Indicate that resonance (loud sound) occurs when L = λ/4, 3λ/4, etc.
Method (Conceptual)
- Hold a vibrating tuning fork (known frequency f) over the open end of a tube
- Adjust the tube length until resonance occurs (sudden increase in loudness)
- The first resonance occurs when L = λ/4 (fundamental of closed pipe)
- Calculate: λ = 4L, then verify with v = fλ
The antinode at an open end forms slightly outside the tube. This "end correction" is assumed negligible at AS Level. No knowledge of end corrections is required.
Comparing Closed and Open Pipes
Two instruments of the same length but different configurations produce different notes:
| Property | Closed Pipe | Open Pipe |
|---|---|---|
| Ends | N at closed, A at open | A at both ends |
| Fundamental λ | λ = 4L | λ = 2L |
| Fundamental f | f_1 = (v/4L) | f_1 = (v/2L) |
| Harmonics present | Odd only (1, 3, 5...) | All (1, 2, 3...) |
| Lowest note | Lower (longer λ) | Higher (shorter λ) |
A closed pipe produces a fundamental with wavelength 4L; an open pipe of the same length produces a fundamental with wavelength 2L. Therefore, the closed pipe produces a lower pitch (lower frequency) for the same length.
Worked Examples
Worked Example 1: Closed Pipe FundamentalQuestion: A closed pipe of length 0.75 m produces its fundamental note. The speed of sound is 340 m s⁻¹. Calculate: (a) the wavelength, (b) the frequency.
Solution(a) For fundamental in closed pipe: λ = 4L = 4 × 0.75 = 3.0 m
(b) f = (v/λ) = (340/3.0) = 113 Hz
Worked Example 2: Next Harmonic in Closed PipeQuestion: A closed pipe has a fundamental frequency of 125 Hz. What is the frequency of the next possible harmonic?
SolutionClosed pipes support only odd harmonics: 1st, 3rd, 5th...
Next harmonic after fundamental = 3rd harmonic
f_3 = 3f_1 = 3 × 125 = 375 Hz
Worked Example 3: Open Pipe HarmonicsQuestion: An open pipe has length 31 cm. The first resonance is heard at 520 Hz. Calculate: (a) the wavelength, (b) the speed of sound.
Solution(a) For fundamental in open pipe: λ = 2L = 2 × 0.31 = 0.62 m
(b) v = fλ = 520 × 0.62 = 322 m s^-1
Worked Example 4: Comparing PipesQuestion: Two pipes have the same length L. Pipe A is closed at one end; Pipe B is open at both ends. Which produces the lower fundamental note? By what factor?
SolutionClosed pipe A: _A = 4L, so f_A = (v/4L)
Open pipe B: _B = 2L, so f_B = (v/2L)
Ratio: (f_B/f_A) = (v/2L/v/4L) = 2
Pipe A (closed) produces the lower note, with frequency half that of Pipe B.
Common Exam Mistakes
Avoid These Errors- Mixing up boundary conditions: Closed end = node; Open end = antinode. Never the reverse.
- Including even harmonics in closed pipes: Only odd harmonics (1st, 3rd, 5th...) exist in closed pipes.
- Wrong fundamental formula: Closed pipe: λ = 4L. Open pipe: λ = 2L. Don't confuse them.
- Forgetting that diagrams show displacement: The curves represent displacement amplitude, not the actual sound wave shape.
- Temperature effects: Sound speed increases with temperature, affecting the frequencies produced.
Concept Check Questions
Interactive revision notes, videos and practice questions load below.