Superposition of progressive waves, nodes, antinodes, harmonics and resonance
AQA A-Level Physics 3
🌊Explain how stationary waves form by superposition of two progressive waves
📍Identify nodes and antinodes and explain their properties
📏Relate the wavelength to the length of a string or pipe for each harmonic
🎵Calculate resonant frequencies for strings and pipes
⚡Compare stationary and progressive waves
🔬Carry out the required practical: stationary waves on a string
Formation of Stationary Waves
Stationary (standing) wave: A wave pattern formed by the superposition of two progressive waves of equal frequency and amplitude travelling in opposite directions along the same line.
When two such waves overlap, constructive and destructive interference create a pattern of fixed points:
Node: A point of permanent destructive interference — zero amplitude. Nodes are always at rest.
Antinode: A point of permanent constructive interference — maximum amplitude. Antinodes oscillate with the greatest displacement.
Adjacent nodes are separated by λ/2. Adjacent antinodes are also separated by λ/2. A node and adjacent antinode are separated by λ/4. The pattern does not move along the medium — hence "stationary."
A stationary wave does not transfer energy (on average). This contrasts with a progressive wave, which continuously transfers energy in the direction of travel.
Harmonics on a Stretched String
A string fixed at both ends can only support stationary waves with nodes at both ends. Allowed wavelengths depend on the length L of the string:
Speed of waves on string: v = √(T/μ)
T = tension (N), μ = mass per unit length (kg m⁻¹)
Harmonic
Nodes
Antinodes
Wavelength
Frequency
1st (fundamental)
2
1
2L
f₁
2nd
3
2
L
2f₁
3rd
4
3
2L/3
3f₁
All harmonics (n = 1, 2, 3, …) are possible on a string fixed at both ends. The fundamental frequency f₁ = v/(2L). Higher harmonics are integer multiples: f_n = nf₁.
Stationary Waves in Pipes
Sound waves in pipes form stationary waves. Boundary conditions differ depending on whether each end is open or closed:
Open end: Antinode (air is free to move — displacement antinode, pressure node)
Pipe open at both ends: supports all harmonics
f_n = nv/(2L), n = 1, 2, 3, …
Pipe closed at one end: supports only odd harmonics
f_n = nv/(4L), n = 1, 3, 5, …
Fundamental: λ = 4L, f₁ = v/(4L)
A pipe closed at one end has a fundamental frequency half that of a pipe of the same length open at both ends. It also only produces odd harmonics (1st, 3rd, 5th), giving it a distinctive timbre.
Comparing Stationary and Progressive Waves
Property
Progressive wave
Stationary wave
Energy transfer
Yes, in direction of travel
No net energy transfer
Amplitude
Same for all points
Varies (0 at nodes, max at antinodes)
Phase
Changes continuously along wave
All points between two nodes in phase; points either side of a node in antiphase
Wavelength
Distance between same-phase points
Twice the node-to-node distance
Wave pattern
Moves through medium
Fixed pattern, does not move
A string of length 0.80 m is fixed at both ends. The wave speed on the string is 240 m s⁻¹. Calculate the frequencies of the first three harmonics.
1st resonance = 131 Hz; 2nd resonance = 392 Hz (no 2nd harmonic for closed pipe)
On a stationary wave pattern, the distance between the 1st and 5th nodes is 32 cm. What is the wavelength of the original progressive waves?
1From node 1 to node 5 spans 4 gaps between adjacent nodes
2Each gap between adjacent nodes = λ/2
34 × (λ/2) = 32 cm → 2λ = 32 cm → λ = 16 cm
λ = 16 cm = 0.16 m
1. What is the separation between adjacent nodes in a stationary wave of wavelength 0.40 m?
Adjacent nodes are separated by λ/2 = 0.40/2 = 0.20 m.
2. A stationary wave forms on a 1.2 m string vibrating in its 3rd harmonic. How many antinodes are present?
The nth harmonic has n antinodes. The 3rd harmonic has 3 antinodes (and 4 nodes, including the two fixed ends).
3. How does increasing the tension of a stretched string affect the fundamental frequency of stationary waves on it?
v = √(T/μ). Increasing T increases v. Since f₁ = v/(2L) and L is fixed, increasing v increases f₁. Pitch rises as a string is tightened.
4. Which statement is TRUE about stationary waves compared to progressive waves?
Between two adjacent nodes, all points oscillate in phase (reach maximum and zero simultaneously). Points on opposite sides of a node are in antiphase. Stationary waves do not transfer net energy.
5. A pipe open at both ends has length 0.34 m. Speed of sound = 340 m s⁻¹. Calculate the fundamental frequency.
1. A guitar string of length 0.65 m and mass per unit length 3.2 × 10⁻³ kg m⁻¹ is tuned to 440 Hz (A4). Calculate the tension in the string. Then determine the frequency if the string length is shortened to 0.49 m by pressing a fret.
f₁ = v/(2L) → v = 2Lf₁ = 2 × 0.65 × 440 = 572 m s⁻¹. T = μv² = 3.2×10⁻³ × 572² = 3.2×10⁻³ × 327184 = 1047 N ≈ 1050 N. With the shorter length (0.49 m) at same tension: v is the same (tension and μ unchanged). f = v/(2L) = 572/(2 × 0.49) = 572/0.98 = 583.7 Hz ≈ 584 Hz. (This corresponds to D5, a major sixth above A4.)
2. Explain why a pipe closed at one end only produces odd harmonics (1st, 3rd, 5th…), whereas a pipe open at both ends produces all harmonics. Use diagrams if helpful.
A pipe closed at one end requires a node at the closed end (air cannot move) and an antinode at the open end. The possible patterns must satisfy: one node at one end, one antinode at the other. The simplest pattern fits ¼ wavelength into the pipe: λ₁ = 4L → f₁ = v/4L. The next pattern fits ¾λ: λ₃ = 4L/3 → f₃ = 3v/4L = 3f₁ (third harmonic). Then 5/4λ: f₅ = 5f₁. Even harmonics (2nd, 4th…) would require a pattern with an even number of quarter-wavelengths between a node and an antinode — but these would place a node at the open end or an antinode at the closed end, violating the boundary conditions. So only n = 1, 3, 5, … are permitted. A pipe open at both ends has antinodes at both ends. The fundamental fits ½λ into L → f₁ = v/2L. Each subsequent harmonic adds one more half-wavelength: all integer multiples are possible.
3. Microwaves of frequency 10.5 GHz are directed at a metal reflector and form a stationary wave with a detector placed between the source and reflector. The detector records maxima separated by 1.42 cm. Calculate the speed of the microwaves and compare to c.
Maxima in a stationary wave are at antinodes, separated by λ/2. So λ/2 = 1.42 cm → λ = 2.84 cm = 0.0284 m. v = fλ = 10.5 × 10⁹ × 0.0284 = 2.982 × 10⁸ m s⁻¹ ≈ 2.98 × 10⁸ m s⁻¹. This is very close to c = 3.00 × 10⁸ m s⁻¹ (0.7% difference, within experimental uncertainty), confirming that microwaves are electromagnetic waves travelling at the speed of light.
Required Practical Stationary Waves on a String
Objective: Investigate the factors affecting the frequency of stationary waves on a stretched string (Melde's experiment).
Equipment
Signal generator and vibration generator (mechanical oscillator)
String (known μ) clamped at one end, over a pulley
Hanging masses (to vary tension T)
Ruler (to measure length L)
Stroboscope (optional, to visualise the pattern)
Method
Set up the string between the vibrator and pulley. Hang a fixed mass (e.g. 200 g) to give tension T = mg.
Vary the frequency on the signal generator. Identify the fundamental frequency f₁ (one antinode, large amplitude). Record L and f₁.
Change the hanging mass and repeat to find how f₁ varies with T. Plot f₁ vs √T — expect a straight line (f₁ ∝ √T).
Change the length L and find f₁. Plot f₁ vs 1/L — expect a straight line (f₁ ∝ 1/L).
Analysis
From f₁ = (1/2L)√(T/μ), the gradient of f₁ vs √T gives 1/(2L√μ), allowing μ to be determined and compared with the manufacturer's value.
Safety
Use a tray or cushion under hanging masses in case the string breaks. Do not stand directly under the masses. Ensure the vibration generator is securely clamped.