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What happens when two waves meet?

The principle of superposition, constructive and destructive interference, coherence and path difference, and the formation of stationary waves with nodes and antinodes on strings and in pipes.

A focused answer to the Edexcel 9PH0 superposition content, covering the principle of superposition, constructive and destructive interference, coherence and path difference, and stationary waves with nodes and antinodes on strings and in pipes.

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What this dot point is asking

Edexcel wants you to apply the principle of superposition, explain constructive and destructive interference in terms of path difference and coherence, and describe how stationary waves form, with nodes and antinodes, on strings and in pipes.

The answer

The principle of superposition

After overlapping, the waves continue unchanged. Superposition is the basis of all interference and stationary-wave effects, and applies to every type of wave.

Constructive and destructive interference

For a stable, observable pattern the sources must be coherent (constant phase difference and the same frequency) and ideally of similar amplitude. Light from two separate lamps is incoherent, which is why interference is demonstrated by splitting one source, as in Young's double slit, where the fringe spacing is w=λDsw = \frac{\lambda D}{s}.

Coherence and path difference

Two sources are coherent if they maintain a fixed phase relationship over time. The path difference at a point is the extra distance one wave travels compared with the other; it decides whether the waves arrive in or out of phase. As you move along a screen, the path difference changes smoothly, producing alternating bright and dark fringes.

Stationary waves

A stationary (standing) wave forms when two waves of the same frequency and amplitude travel in opposite directions and superpose, typically a wave and its reflection. The result does not move along: it has nodes (points of permanent zero displacement) and antinodes (points of maximum displacement), spaced half a wavelength apart.

A string fixed at both ends has nodes at the ends; its fundamental fits half a wavelength into the length. A closed pipe has a node at the closed end and an antinode at the open end, giving only odd harmonics. Unlike a progressive wave, a stationary wave transfers no net energy and all points between adjacent nodes move in phase.

Examples in context

Noise-cancelling headphones generate a wave in antiphase with incoming sound, using destructive interference to silence it. Musical instruments produce notes from stationary waves: a guitar string and an organ pipe resonate at frequencies set by their length, and the harmonics give each instrument its timbre. Young's double-slit experiment was historic evidence for the wave nature of light. Microwave ovens can leave cold spots at the nodes of a stationary wave, which is why the food is rotated.

Try this

Q1. State the principle of superposition. [1 mark]

  • Cue. When waves overlap, the resultant displacement is the vector sum of the individual displacements.

Q2. Two coherent sources have a path difference of 1.51.5 wavelengths at a point. State whether interference there is constructive or destructive. [1 mark]

  • Cue. Destructive, since 1.5λ1.5\lambda is an odd number of half wavelengths.

Q3. A string fixed at both ends is 0.500.50 m long with a wave speed of 300300 m per second. Find its fundamental frequency. [2 marks]

  • Cue. f1=v2L=3002×0.50=300f_1 = \frac{v}{2L} = \frac{300}{2 \times 0.50} = 300 Hz.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20194 marksIn a double-slit experiment, slits 0.400.40 mm apart are illuminated by light of wavelength 600600 nm. The screen is 2.02.0 m away. Calculate the fringe spacing.
Show worked answer →

Fringe spacing: w=λDsw = \frac{\lambda D}{s}, where ss is the slit separation and DD the slit-to-screen distance.

w=600×10−9×2.00.40×10−3=1.2×10−64.0×10−4=3.0×10−3w = \frac{600 \times 10^{-9} \times 2.0}{0.40 \times 10^{-3}} = \frac{1.2 \times 10^{-6}}{4.0 \times 10^{-4}} = 3.0 \times 10^{-3} m, that is 3.03.0 mm.

Markers reward the formula w=λDsw = \frac{\lambda D}{s}, consistent SI conversion, and the value 3.03.0 mm.

Edexcel 20225 marksA stretched string of length 0.600.60 m fixed at both ends vibrates in its fundamental mode at 200200 Hz. Determine the speed of waves on the string and the frequency of the third harmonic, and explain how a stationary wave forms.
Show worked answer →

Fundamental wavelength: a string fixed at both ends has λ=2L=2×0.60=1.2\lambda = 2L = 2 \times 0.60 = 1.2 m in the fundamental.

Wave speed: v=fλ=200×1.2=240v = f\lambda = 200 \times 1.2 = 240 m per second.

Third harmonic: f3=3f1=3×200=600f_3 = 3 f_1 = 3 \times 200 = 600 Hz.

Formation: the wave reflects off the fixed ends and the incident and reflected waves (same frequency, travelling in opposite directions) superpose, producing fixed nodes and antinodes when the length fits a whole number of half wavelengths.

Markers reward λ=2L\lambda = 2L, v=fλ=240v = f\lambda = 240 m per second, f3=600f_3 = 600 Hz, and the superposition-of-reflected-waves explanation.

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