Questions: Resonance in Air Columns and Pipes

Question 1

A pipe of length L is open at one end and closed at the other (like a clarinet). Which harmonics are present in its resonant spectrum?

• A. All integer harmonics: f_n = nv/(2L) for n = 1, 2, 3, ...
• B. Only odd harmonics: f_n = (2n−1)v/(4L) for n = 1, 2, 3, ...
• C. All integer harmonics, but starting at f = v/(4L) rather than v/(2L)
• D. Only even harmonics, because the open-closed geometry requires an even number of quarter-wavelengths

Answer: (B)

A closed end requires a displacement node; an open end requires a displacement antinode. The simplest wave fitting these constraints is a quarter-wavelength (L = λ/4). The next requires L = 3λ/4, then L = 5λ/4 — only odd multiples of the quarter-wavelength fit. This yields f_n = (2n−1)v/(4L): only odd harmonics. This is the boundary-condition derivation of the formula — you don't need to memorize it if you understand why nodes occur at closed ends and antinodes at open ends.

Question 2

At an open end of a pipe, what boundary condition applies and why?

• A. Displacement node — air molecules must stop moving as they exit the pipe into open space
• B. Pressure node (displacement antinode) — air pressure must equal atmospheric pressure at the opening, so pressure variation goes to zero
• C. Pressure antinode (displacement node) — air compresses at the opening because of the sudden change in area
• D. No boundary condition applies — open ends are physically unconstrained

Answer: (B)

At an open end, the air pressure must match the atmosphere outside, so the excess pressure (deviation from atmospheric) must be zero there — a pressure node. Since pressure variation and displacement variation are 90° out of phase in a longitudinal sound wave, a pressure node corresponds to maximum displacement — a displacement antinode. This is the physically correct boundary condition, and it is what distinguishes open-open pipes (antinodes at both ends, all harmonics present) from closed-open pipes (node at closed end, antinode at open end, only odd harmonics).

Question 3

A clarinet (effectively a closed-open pipe) playing the same fundamental frequency as a flute (effectively an open-open pipe) will have a different timbre because the clarinet's resonant spectrum lacks even harmonics.

• T. True
• F. False

Answer: True

True. Both instruments can play the same fundamental pitch, but their overtone spectra differ. The flute (open-open) supports all harmonics (f, 2f, 3f, 4f, ...), adding brightness from even-numbered partials. The clarinet (closed-open) supports only odd harmonics (f, 3f, 5f, 7f, ...), giving it a darker, more hollow sound. This is a direct consequence of boundary conditions, not just tube length or mouthpiece design. The geometry of the air column is acoustic destiny.

Question 4

A pipe closed at both ends supports the same resonant frequencies as a pipe open at both ends of the same length.

• T. True
• F. False

Answer: True

True — and this surprises many students. A closed-closed pipe requires displacement nodes at both ends; an open-open pipe requires displacement antinodes at both ends. In both cases, the simplest mode fits one half-wavelength in the pipe (L = λ/2), and all integer multiples of the half-wavelength also fit. Both give f_n = nv/(2L) for all integers n = 1, 2, 3, ... The harmonic series is identical. The difference between open-open and closed-open (where one end has a node and the other has an antinode) is what breaks the symmetry and eliminates even harmonics.

Question 5

Explain, starting from physical boundary conditions (not just formulas), why a closed-open pipe supports only odd harmonics while an open-open pipe of the same length supports all harmonics.

Model answer: Boundary conditions: a closed end requires a displacement node (molecules cannot move through the wall); an open end requires a displacement antinode (pressure there equals atmospheric, so pressure variation is zero, meaning displacement variation is maximum). For an open-open pipe, the simplest standing wave has antinodes at both ends, fitting one half-wavelength (L = λ/2). Higher modes fit 2, 3, 4, ... half-wavelengths: L = nλ/2, giving f_n = nv/(2L) — all integers present. For a closed-open pipe, the simplest wave has a node at one end and an antinode at the other. This requires L = λ/4. The next mode must again have node at closed, antinode at open — the next fitting pattern has L = 3λ/4, then 5λ/4. Only odd multiples of λ/4 fit: L = (2n−1)λ/4, giving f_n = (2n−1)v/(4L) — only odd harmonics. The missing even harmonics are not a mathematical accident but a direct geometric consequence of the mixed boundary conditions.

The key is that mixed boundary conditions (node on one side, antinode on the other) break the symmetry that allows half-integer as well as odd-quarter-integer wavelengths. With symmetric boundary conditions (node-node or antinode-antinode), the fitting condition gives all harmonics. With asymmetric boundary conditions (node-antinode), only odd harmonics fit. This is why the clarinet — a closed-open cylindrical tube — sounds darker and more 'reedy' than a flute of comparable length: the even harmonics that would add brightness are simply absent.