OCR Specification focus:
‘Adjacent node spacing equals λ/2; identify fundamental and higher harmonics.’
In stationary waves, wavelength and harmonics describe how patterns of vibration form within a confined medium such as a string or air column. Understanding their relationships explains how musical notes and resonant frequencies arise across systems.
Understanding Wavelength in Stationary Waves
A stationary wave (also called a standing wave) is produced when two progressive waves of identical frequency and amplitude travel in opposite directions and interfere. This interference leads to fixed points of no displacement and points of maximum displacement.
Node: A point on a stationary wave that remains at zero displacement at all times due to complete destructive interference.
Antinode: A point on a stationary wave that experiences maximum displacement due to constructive interference.
In a stationary wave, the distance between two adjacent nodes (or two adjacent antinodes) is half of a wavelength (λ/2). This relationship is fundamental for interpreting and measuring wave patterns in strings, air columns, and microwave experiments.
Diagram of a stationary wave showing nodes (zero displacement) and antinodes (maximum displacement), with node–node and antinode–antinode separations labelled as λ/2. This directly supports using node spacing to infer the full wavelength. (The lower-right legend includes a brief resonance note, which is extra but does not exceed the syllabus scope.) Source.
This means that if the length between nodes is known, the full wavelength can be easily determined:
Node to adjacent node = λ/2
Antinode to adjacent antinode = λ/2
EQUATION
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Wavelength from Node Spacing (λ) = 2 × Distance between adjacent nodes
λ = Wavelength (metre, m)
—-----------------------------------------------------------------
Harmonics and Their Significance
A harmonic represents a resonant mode of vibration in a stationary wave system. Each harmonic corresponds to a frequency at which a stationary wave pattern fits exactly within the boundaries of the medium.
Harmonic: A vibration mode where an integer number of half-wavelengths fit exactly into the medium’s length.
Harmonics determine the frequency pattern of the system, with higher harmonics corresponding to higher frequencies and shorter wavelengths. The fundamental or first harmonic produces the lowest possible frequency for a given system length, while successive harmonics are integer multiples of this frequency.
EQUATION
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Harmonic Frequency (fₙ) = n × f₁
fₙ = Frequency of the nth harmonic (hertz, Hz)
n = Harmonic number (integer: 1, 2, 3, …)
f₁ = Fundamental frequency (hertz, Hz)
—-----------------------------------------------------------------
This relationship is consistent across all stationary wave systems, though the boundary conditions—such as whether the ends are fixed or open—affect the harmonic pattern.
Harmonics in Strings
A string fixed at both ends can only support stationary waves that satisfy the condition that both ends are nodes. The string length must therefore accommodate a whole number of half-wavelengths.
Fundamental (First Harmonic)
The simplest stationary wave pattern.
One-half of a wavelength fits along the string length.
There is one antinode at the midpoint and two nodes at the ends.
EQUATION
—-----------------------------------------------------------------
Length–Wavelength Relationship (Strings): L = ½λ₁
L = Length of the string (m)
λ₁ = Wavelength of the first harmonic (m)
—-----------------------------------------------------------------
The fundamental frequency is the lowest natural frequency of vibration and forms the basis for higher harmonics.
Second and Higher Harmonics
For the second harmonic, the string supports two half-wavelengths across its length (L = λ₂).
For the third harmonic, three half-wavelengths fit (L = 3λ₃/2), and so on.
The general relationship is:
EQUATION
—-----------------------------------------------------------------
General Harmonic Relationship (Strings): L = n(λₙ/2)
L = Length of the string (m)
λₙ = Wavelength of the nth harmonic (m)
n = Harmonic number
—-----------------------------------------------------------------
Each harmonic increases the number of nodes and antinodes:
First harmonic: 2 nodes, 1 antinode
Second harmonic: 3 nodes, 2 antinodes
Third harmonic: 4 nodes, 3 antinodes
This progression explains the overtones heard in musical instruments.
Harmonics in Air Columns
Air columns, such as those in open and closed tubes, follow similar principles but differ due to their boundary conditions.
Open Tube
Both ends are antinodes (pressure varies most).
Fundamental harmonic fits half a wavelength in the tube.
All harmonics are possible.
EQUATION
—-----------------------------------------------------------------
Open Tube Relationship: L = n(λₙ/2)
L = Length of the air column (m)
λₙ = Wavelength of the nth harmonic (m)
n = 1, 2, 3, …
—-----------------------------------------------------------------
Closed Tube
One end is a node, the other an antinode.
The fundamental harmonic fits a quarter of a wavelength.
Only odd harmonics occur, as even ones violate boundary conditions.
Standing waves for an open–open tube (left) and an open–closed tube (right), showing displacement patterns for the fundamental and higher harmonics. Open ends are displacement antinodes; closed ends are nodes, so only odd harmonics fit in a closed tube. The schematic matches the length relations L = nλ/2 (open) and L = nλ/4 with odd n (closed). (The figure focuses on displacement shapes and does not include pressure plots, which are beyond the syllabus requirement.) Source.
EQUATION
—-----------------------------------------------------------------
Closed Tube Relationship: L = n(λₙ/4)
L = Length of the air column (m)
λₙ = Wavelength of the nth harmonic (m)
n = 1, 3, 5, … (odd integers)
—-----------------------------------------------------------------
This means that closed tubes produce deeper tones for the same tube length compared to open tubes.
Determining Wavelength and Harmonic Order Experimentally
In laboratory settings, the wavelength and harmonic order can be determined using resonance or wave interference patterns. The process involves:
Generating a wave (sound, microwave, or string vibration) of variable frequency.
Observing nodes and antinodes that appear due to interference.
Measuring the distance between adjacent nodes to find λ/2.
Using the relationship L = n(λ/2) (or L = n(λ/4) for closed tubes) to identify the harmonic.
EQUATION
—-----------------------------------------------------------------
Wave Speed Relation: v = fλ
v = Wave speed (metres per second, m/s)
f = Frequency (hertz, Hz)
λ = Wavelength (metre, m)
—-----------------------------------------------------------------
Using this, one can determine either the wave speed, wavelength, or frequency, depending on the measured and known quantities.
Summary of Key Relationships
Adjacent node spacing = λ/2
L = n(λ/2) for strings and open tubes
L = n(λ/4) for closed tubes (odd n only)
fₙ = n × f₁ defines harmonic frequencies
These relationships form the mathematical framework behind the formation of stationary wave patterns, the identification of harmonics, and their practical applications in physics and acoustics.
FAQ
Harmonics can be identified by observing the number of nodes and antinodes formed along the string.
The fundamental (first harmonic) has two nodes at the ends and one central antinode.
The second harmonic shows an extra node–antinode pair, giving three nodes in total.
Each successive harmonic adds one more node and antinode.
Using a strobe light or video analysis can help make these stationary patterns visible and allow accurate measurement of wavelength.
In a closed tube, the closed end must be a node (no air movement) and the open end must be an antinode (maximum air movement).
For even harmonics, this pattern cannot fit the tube’s length because an even number of quarter-wavelengths would require both ends to have the same boundary condition.
Therefore, only harmonics with odd values of n (1, 3, 5, …) satisfy the conditions, producing resonances at frequencies that are odd multiples of the fundamental.
The strength of harmonics depends on how energy is distributed among vibrational modes.
Material and tension of the string or air column affect resonance efficiency.
Method of excitation (e.g. plucking position or blowing technique) influences which harmonics dominate.
Damping and boundary shape determine how quickly higher harmonics fade.
These factors combine to produce a characteristic sound, or timbre, for each instrument, even when playing the same fundamental note.
Wavelength can be inferred from resonance positions rather than direct measurement.
In an air column, gradually change the tube length until resonance is heard — the distance between resonant positions equals half a wavelength.
On a string, vary the frequency of vibration and measure the frequency difference between successive resonances, then use v = fλ to find wavelength indirectly.
Both methods rely on identifying where constructive interference reinforces the wave pattern.
The energy stored in a stationary wave increases with harmonic number because higher harmonics involve shorter wavelengths and greater curvature of the medium.
The energy is proportional to both amplitude squared and frequency squared.
Higher harmonics oscillate faster and often require greater tension or energy input to sustain.
However, damping effects usually cause higher harmonics to lose energy more rapidly, which is why they are often less prominent in real systems.
Practice Questions
Question 1 (2 marks)
A stationary wave is formed on a stretched string fixed at both ends. The distance between two adjacent nodes is measured to be 0.25 m.
Calculate the wavelength of the wave.
Mark scheme:
Correctly uses the relationship between adjacent nodes and wavelength: λ = 2 × distance between nodes (1 mark)
Correctly calculates the wavelength: λ = 2 × 0.25 = 0.50 m (1 mark)
Question 2 (5 marks)
A tube open at one end and closed at the other resonates with a sound source at a fundamental frequency of 300 Hz. The speed of sound in air is 340 m s⁻¹.
(a) Calculate the wavelength of the fundamental stationary wave in the tube. (2 marks)
(b) Determine the length of the tube. (2 marks)
(c) State and explain what happens to the pattern of nodes and antinodes if the next higher resonant frequency is produced. (1 mark)
Mark scheme:
(a)
States or uses relationship v = fλ (1 mark)
Substitutes correctly: λ = 340 / 300 = 1.13 m (1 mark)
(b)
Recognises that for a closed tube, the fundamental has L = λ/4 (1 mark)
Calculates correctly: L = 1.13 / 4 = 0.283 m (1 mark)
(c)
States that a node remains at the closed end and an antinode at the open end, with an additional node–antinode pair appearing (1 mark)
