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 Representation Diagrams 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 Harmonics Show 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 = fraction 3rd N-A-N-A (¾λ fits) λ = fraction f_3 = 3f_1 5th N-A-N-A-N-A λ = fraction f_5 = 5f_1 nth (odd only) - λ = fraction f_n = nf_1 Critical Point 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 Harmonics Show 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 = fraction 2nd A-N-A-N-A (1λ fits) λ = L f_2 = 2f_1 3rd A-N-A-N-A-N-A λ = fraction f_3 = 3f_1 nth - λ = fraction f_n = nf_1 Key Difference 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 Experiment Show: 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 Resonance An increase in amplitude that occurs when an object is made to vibrate at its natural frequency.
In air columns, resonance produces a noticeably louder sound. End Corrections 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 = fraction f_1 = fraction Harmonics present Odd only (1, 3, 5...) All (1, 2, 3...) Lowest note Lower (longer λ) Higher (shorter λ) Exam Insight 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 Fundamental Question: 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 = fraction = fraction = 113 Hz Worked Example 2: Next Harmonic in Closed Pipe Question: A closed pipe has a fundamental frequency of 125 Hz.
What is the frequency of the next possible harmonic? Solution Closed 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 Harmonics Question: 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 Pipes Question: Two pipes have the same length L.
Pipe A is closed at…
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