Normal Modes of a Bounded System
Boundary conditions select the normal modes of an ideal uniform string or air column. Fixed-end strings and pipes open at both ends allow all whole-number harmonics. Pipes closed at one end allow odd harmonics when end corrections are neglected.
What a learner can do afterwards
- Finds allowed wavelengths and frequencies for an ideal uniform string fixed at both ends
- Compares ideal open-open and open-closed pipes using displacement end conditions, neglecting end corrections
- Explains how relative harmonic strengths contribute to an instrument's timbre
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A normal mode is a standing-wave pattern allowed by the boundaries. On an ideal uniform string, displacement nodes stay at zero. Antinodes lie midway between neighbouring nodes and have the greatest oscillation amplitude, but pass through zero each cycle. Fixed ends require nodes, so L = nλ_n/2, λ_n = 2L/n and f_n = nv/(2L) = n f_1. Here L is length, v is travelling-wave speed and n = 1, 2, 3, ... . The lowest mode frequency f_1 is the fundamental.
A string has fundamental frequency 120 Hz. Its second harmonic is 2 × 120 = 240 Hz, and its third is 3 × 120 = 360 Hz. A harmonic's number multiplies the fundamental frequency. The bars compare frequencies in Hz; their heights do not show loudness.
For an ideal uniform pipe, neglect end corrections. A rigid closed end is an air-displacement node; an open end is a displacement antinode. Pressure variation has the opposite end pattern. A pipe open at both ends fits whole numbers of half-wavelengths: λ_n = 2L/n and f_n = nv/(2L), for n = 1, 2, 3, ... . Its fundamental wavelength is 2L and frequency is v/(2L).
A pipe closed at one end needs a displacement node there and an antinode at its open end. Node-to-antinode spacing is λ/4. Only odd numbers of quarter-wavelengths join those endpoints: λ_n = 4L/n and f_n = nv/(4L), for n = 1, 3, 5, ... . An overtone is a mode above the fundamental. This pipe's first overtone is the third harmonic at 3f_1; the second harmonic is absent.
Starting at a node, odd quarter-wave counts reach an antinode. These are listed positions, not a drawing of the pipe.
Compare equal-length pipes at the same sound speed, ignoring end corrections. The open-open fundamental has wavelength 2L; the open-closed fundamental has wavelength 4L. Since f = v/λ, the open-closed frequency is half as large, one octave lower. These are natural-mode predictions. A driven system can still move at other frequencies.
The values compare wavelength in units of L. They do not compare pipe lengths or sound strengths.
Instruments can sustain the same fundamental while their harmonics have different relative strengths. That spectral mix contributes to timbre, the quality that helps us distinguish their sounds. How a sound begins and fades also affects timbre. Allowed mode frequencies alone do not tell us how strongly each mode is excited.
Ideal boundaries select normal modes. Fixed-end strings and open-open pipes allow all harmonics; open-closed pipes allow odd ones. Relative harmonic strengths contribute to timbre.
Normal-mode lesson, questions, worked solutions and flashcards written for this dataset.
Reading passages and flashcards written for this dataset. Standing waves in open tubes and Standing waves in closed tubes are linked Khan Academy videos.
More to read
From node spacing to mode frequency
An ideal string with fixed ends must have a displacement node at each endpoint. Each gap between neighbouring nodes is half a wavelength. If n such gaps fill length L, then L = nλ_n/2. Rearranging gives λ_n = 2L/n. Dividing the travelling-wave speed v by that wavelength gives f_n = nv/(2L).
For example, take a 0.75 m string with wave speed 120 m/s. The fundamental wavelength is 1.50 m and its frequency is 80 Hz. The third mode has wavelength 0.50 m and frequency 240 Hz. The sound produced in the surrounding air has the same frequency as the string's vibration, but its wavelength depends on the sound speed in air.
Mode frequencies describe natural standing patterns. They do not say that a driven string is motionless at every other frequency, or that all modes receive equal energy from a pluck.
Choose the quantity before marking nodes
A sound wave in a pipe can be described using particle displacement or pressure variation. Their nodes are different. At a rigid closed end, the air cannot move along the pipe, giving a displacement node and a pressure antinode. At an ideal open end, pressure variation has a node and displacement has an antinode. We neglect end corrections here.
An open-open pipe has displacement antinodes at both ends. Its lowest mode contains half a wavelength, so λ_1 = 2L. An open-closed pipe needs a displacement node at one end and an antinode at the other. Its lowest mode contains a quarter wavelength, so λ_1 = 4L.
At equal length and sound speed, the open-closed fundamental is half the open-open fundamental. The next open-closed mode is at 3f_1. It is called the first overtone because it follows the fundamental, and the third harmonic because its frequency is three times the fundamental.
A stationary pattern can contain moving particles
In a linear model, add signed wave displacements at the same position and time. Contributions of +4 mm and -1 mm give +3 mm. Complete cancellation requires equal opposite values, such as +4 mm and -4 mm. A crest and a trough need not cancel completely if their contributions have different sizes.
Two equal-amplitude sinusoidal waves of the same frequency travelling in opposite directions can form a stationary wave. The node locations stay fixed. Points between the nodes still move, and antinodes have the greatest oscillation amplitude. An antinode passes through zero displacement during each cycle without becoming a node.
A moment of zero combined displacement does not destroy the component waves or their energy. Boundaries select which stationary patterns are normal modes. Their positions stay fixed while the material between nodes oscillates.
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