Questions: Acoustic Resonance in Pipes and Air Columns

Question 1

A closed-open pipe and an open-open pipe have the same length L. What is the ratio of their fundamental frequencies (closed : open)?

• A. 2 : 1 — the closed pipe vibrates twice as fast
• B. 1 : 2 — the closed pipe has half the fundamental frequency of the open pipe
• C. 1 : 1 — same length means same fundamental frequency
• D. 1 : 4 — the closed pipe vibrates at one-quarter the frequency

Answer: (B)

Open-open: f₁ = v/(2L). Closed-open: f₁ = v/(4L). Ratio = (v/4L)/(v/2L) = 1/2. The closed pipe's fundamental frequency is half that of the open pipe — one octave lower — because the node at the closed end forces the standing wave to fit a quarter-wavelength across L rather than a half-wavelength. The boundary conditions, not the length alone, determine the pitch.

Question 2

A clarinet (closed-open pipe) produces a fundamental frequency of 220 Hz. Which harmonics does it produce?

• A. 220 Hz, 440 Hz, 660 Hz, 880 Hz (all integer multiples)
• B. 220 Hz, 660 Hz, 1100 Hz (odd multiples only)
• C. 440 Hz, 880 Hz, 1320 Hz (even multiples only)
• D. 220 Hz, 330 Hz, 440 Hz (multiples of 1.5)

Answer: (B)

A closed-open pipe supports only odd harmonics: f_n = (2n−1)v/(4L), giving frequencies at 1×, 3×, 5×, … the fundamental. At 220 Hz: 220, 660, 1100, 1540 Hz, etc. The even harmonics (440, 880 Hz) are absent because no standing wave satisfying 'node at closed end, antinode at open end' can fit an even number of quarter-wavelengths across the pipe. This restricted harmonic series produces the clarinet's characteristically hollow timbre.

Question 3

An open end of a resonating pipe is always a displacement antinode.

• T. True
• F. False

Answer: True

At an open end, air molecules are free to move — there is no wall to constrain them. This freedom means displacement is at maximum, making the open end an antinode by physical necessity. At a closed end, the wall prevents air displacement, forcing a node. These boundary conditions are imposed by the physics of the situation, and they determine which wavelengths can fit in the pipe and therefore which frequencies resonate.

Question 4

Two pipes of the same length but different end conditions (one open-open, one closed-open) will resonate at the same fundamental frequency.

• T. True
• F. False

Answer: False

End conditions change everything. An open-open pipe of length L fits a half-wavelength as its fundamental: f₁ = v/(2L). A closed-open pipe fits only a quarter-wavelength: f₁ = v/(4L). The closed-open pipe sounds one full octave lower despite being the same physical length. This is why instrument designers treat end conditions as a design variable — the same tube can produce different pitches depending on how the ends are configured.

Question 5

Why does a closed-open pipe produce only odd harmonics, and what is the physical reason no even harmonics can form?

Model answer: A closed-open pipe requires a node at the closed end and an antinode at the open end. The simplest fitting pattern is one quarter-wavelength. The next allowed pattern requires three quarter-wavelengths, then five, then seven — always an odd number of quarter-wavelengths. An even number would put both ends in the same condition (both antinodes or both nodes), violating the boundary conditions. Since frequency is proportional to the number of quarter-wavelengths that fit, only odd multiples of the fundamental are supported.

The answer reveals that odd-harmonic production is not an arbitrary fact to memorize — it is the direct logical consequence of the boundary conditions. Once you know what each end must do, the allowed wavelengths follow automatically. This same reasoning applies to all resonance problems: identify the boundary conditions first, then derive the allowed modes.