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A stationary wave (also called a standing wave) is formed by the superposition of two progressive waves of the same frequency, wavelength, and amplitude travelling in opposite directions. Unlike a progressive wave, a stationary wave does not transfer energy along its length.
Spec mapping: This lesson sits under AQA 7408 section 3.3.1 and develops stationary waves: their formation by superposition of two oppositely travelling progressive waves, the structure of nodes (zero displacement) and antinodes (maximum displacement), the comparison with progressive waves, the harmonic series on strings fixed at both ends (fₙ = nv/2L), the harmonic series in pipes (open–open: all harmonics; closed–open: odd harmonics only), and the wave-speed relation v = √(T/μ) for transverse waves on a stretched string. This material underpins Required Practical 3 (resonance in standing waves). (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- Required Practical 3: the melde-string and resonance-tube experiments are the explicit practical here; students must understand the apparatus, the variables, and the error analysis.
- Section 3.3.2 (superposition): the principle of superposition is the conceptual prerequisite — stationary waves are superposition applied to two oppositely directed progressive waves.
- Section 3.2.1 (mechanics — SHM): every point on a stationary wave executes simple harmonic motion with a position-dependent amplitude, linking this lesson to the oscillations and damping topic on Paper 2.
- Section 3.7 (fields): the quantised wavelengths λₙ = 2L/n that emerge from boundary conditions on a string are conceptually analogous to the quantised energy levels of an electron in a one-dimensional potential well — a synoptic bridge to quantum mechanics that some examiners exploit.
Stationary waves are typically formed by the superposition of an incident wave and its reflection. For example:
At certain frequencies, the pattern of constructive and destructive interference creates a stable, fixed pattern with positions that always have zero displacement (nodes) and positions that oscillate with maximum displacement (antinodes).
A node is a point on a stationary wave that has zero displacement at all times. Nodes are caused by persistent destructive interference at that point. Adjacent nodes are separated by λ/2.
An antinode is a point on a stationary wave that oscillates with maximum amplitude. Antinodes are caused by persistent constructive interference. They are located midway between adjacent nodes.
| Property | Progressive Wave | Stationary Wave |
|---|---|---|
| Energy transfer | Transfers energy along the wave | No net energy transfer |
| Amplitude | All points oscillate with the same amplitude | Amplitude varies: zero at nodes, maximum at antinodes |
| Phase | Phase varies continuously along the wave | All points between two adjacent nodes are in phase; points in adjacent segments are in antiphase |
| Wavelength | Distance for one full cycle | Distance between alternate nodes = λ |
| Frequency | All points oscillate at the same frequency | All points oscillate at the same frequency |
Exam Tip: A key distinguishing feature of stationary waves is that the amplitude varies from point to point. In a progressive wave, all points have the same amplitude. This is a common exam question.
When a string is fixed at both ends (e.g., a guitar string), standing waves can be set up. Both ends must be nodes because the string cannot move at a fixed end.
The different modes of vibration are called harmonics. The lowest frequency is the fundamental (1st harmonic).
1st Harmonic (Fundamental):
2nd Harmonic:
3rd Harmonic:
General pattern for the nth harmonic:
The speed of a transverse wave on a string depends on the tension T and the mass per unit length μ:
v = √(T/μ)
where T is the tension (N) and μ = m/L is the mass per unit length (kg m⁻¹).
Worked Example 1 — A guitar string has a length of 0.65 m, mass per unit length 3.5 × 10⁻³ kg m⁻¹, and is under a tension of 80 N. Calculate the fundamental frequency.
v = √(T/μ) = √(80/(3.5 × 10⁻³)) = √(22857) = 151.2 m s⁻¹
f₁ = v/(2L) = 151.2/(2 × 0.65) = 151.2/1.30 = 116 Hz
Worked Example 2 — The same string vibrates in its 3rd harmonic. What is the frequency and wavelength?
f₃ = 3f₁ = 3 × 116 = 349 Hz
λ₃ = 2L/3 = 2 × 0.65/3 = 1.30/3 = 0.433 m
Worked Example 3 — A string of length 1.20 m vibrates at its fundamental frequency of 200 Hz. Calculate the speed of the wave on the string and the tension if the string has a mass of 4.8 × 10⁻³ kg.
f₁ = v/(2L), so v = 2Lf₁ = 2 × 1.20 × 200 = 480 m s⁻¹
μ = m/L = (4.8 × 10⁻³)/1.20 = 4.0 × 10⁻³ kg m⁻¹
v = √(T/μ), so T = μv² = (4.0 × 10⁻³) × 480² = (4.0 × 10⁻³) × 230400 = 922 N
Sound waves can also form stationary waves in pipes. The boundary conditions depend on whether the ends are open or closed.
Both ends are antinodes.
1st Harmonic: L = λ₁/2, so λ₁ = 2L, f₁ = v/(2L) 2nd Harmonic: L = λ₂, so λ₂ = L, f₂ = 2f₁ nth Harmonic: λₙ = 2L/n, fₙ = nv/(2L) = nf₁
All harmonics (n = 1, 2, 3, ...) are present.
One end is a node (closed), the other is an antinode (open).
1st Harmonic: L = λ₁/4, so λ₁ = 4L, f₁ = v/(4L) 3rd Harmonic (next mode): L = 3λ₃/4, so λ₃ = 4L/3, f₃ = 3v/(4L) = 3f₁ 5th Harmonic: L = 5λ₅/4, so λ₅ = 4L/5, f₅ = 5f₁
Only odd harmonics (n = 1, 3, 5, ...) are present in a pipe closed at one end.
Worked Example 4 — A pipe open at both ends has a length of 0.85 m. Calculate the fundamental frequency and the frequency of the first three harmonics. (Speed of sound = 340 m s⁻¹.)
f₁ = v/(2L) = 340/(2 × 0.85) = 340/1.70 = 200 Hz
f₂ = 2f₁ = 400 Hz
f₃ = 3f₁ = 600 Hz
Worked Example 5 — A pipe closed at one end has a length of 0.50 m. What are the frequencies of the first three modes of vibration? (Speed of sound = 340 m s⁻¹.)
f₁ = v/(4L) = 340/(4 × 0.50) = 340/2.00 = 170 Hz (1st harmonic)
f₃ = 3f₁ = 510 Hz (3rd harmonic — next available mode)
f₅ = 5f₁ = 850 Hz (5th harmonic)
Note: there is no 2nd or 4th harmonic for a pipe closed at one end.
Exam Tip: A very common error is to state that a closed pipe has "1st, 2nd, 3rd harmonics." Remember: only odd harmonics are present when one end is closed.
Resonance occurs when a system is driven at one of its natural frequencies of vibration. At resonance:
For a string or pipe, resonance occurs at the harmonic frequencies. Tuning forks, musical instruments, and microwave ovens all exploit resonance.
A string is attached to a vibration generator (connected to a signal generator) at one end and passes over a pulley with a hanging mass at the other. By adjusting the frequency of the signal generator, standing wave patterns are observed at the resonant frequencies. Adding or removing mass changes the tension and hence the wave speed and resonant frequencies.
A stationary wave can be written as the superposition of two oppositely directed progressive waves of equal amplitude and frequency:
y₁(x, t) = A sin(ωt − kx) (rightward-travelling) y₂(x, t) = A sin(ωt + kx) (leftward-travelling)
Using the sum-to-product trig identity sin P + sin Q = 2 sin((P+Q)/2) cos((P−Q)/2):
y(x, t) = y₁ + y₂ = 2A sin(ωt) cos(kx)
This is the standard form of a stationary wave. Notice the separation of variables: the time-dependence sin(ωt) is the same at every point in space, while the spatial pattern cos(kx) has fixed nodes (where cos(kx) = 0) and fixed antinodes (where cos(kx) = ±1).
cos(kx) = 0 gives kx = π/2, 3π/2, 5π/2, ...
Using k = 2π/λ: x_node = λ/4, 3λ/4, 5λ/4, ...
Adjacent nodes are therefore separated by λ/2, confirming the earlier qualitative result.
cos(kx) = ±1 gives kx = 0, π, 2π, 3π, ...
x_antinode = 0, λ/2, λ, 3λ/2, ...
Adjacent antinodes are also λ/2 apart, with antinodes interleaved between nodes at quarter-wavelength intervals.
(a) Maximum amplitude is 2A = 0.012 m at antinodes (where cos(4πx) = ±1). So A_progressive = 0.006 m and A_antinode = 0.012 m = 12 mm.
(b) ω = 120π → f = ω/(2π) = 60 Hz
(c) k = 4π → λ = 2π/k = 0.50 m
(d) v = fλ = 60 × 0.50 = 30 m s⁻¹
This is the speed at which the underlying progressive waves travel — the stationary wave itself does not move.
Unlike a progressive wave, a stationary wave does not transport energy along its length. Instead, energy oscillates locally between kinetic and potential forms within each segment between adjacent nodes:
The total mechanical energy between any two adjacent nodes is conserved over each oscillation cycle, even though it sloshes between KE and PE. There is no net energy transfer across a node — this is the rigorous content of "stationary waves do not transfer energy".
In any real resonance system — a guitar string, an organ pipe, a tuning fork — energy is gradually lost to friction, air viscosity, and radiation. The quality factor Q is defined as 2π × (energy stored / energy lost per cycle), and is a measure of how sharply a resonance is tuned to its natural frequency.
| System | Approximate Q |
|---|---|
| Guitar string (steel-core) | ~1000 |
| Tuning fork | ~10⁴ |
| Quartz crystal oscillator | ~10⁶ |
| Atomic clock (caesium hyperfine transition) | ~10¹⁰ |
High Q means a sharp resonance peak (a narrow range of frequencies excites a large response) and slow decay after the driver is switched off. Atomic clocks exploit extremely high Q to keep time with parts-per-10¹⁵ accuracy — the foundation of GPS and the modern definition of the second.
Number of oscillations in the decay time: N = f × τ = 220 × 2.0 = 440 cycles.
Q ≈ πN = π × 440 = 1380 — consistent with the typical Q for a metal guitar string.
This means the string oscillates for ~440 cycles before its amplitude falls significantly — explaining why a plucked guitar string sustains its tone for a noticeable interval before fading.
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