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Superposition and interference of waves (QCE Physics Unit 2)

Syllabus dot point

“Describe the superposition of mechanical waves and explain constructive and destructive interference in terms of phase relationships”

8 min read

Quick answer

Mechanical waves obey superposition (the total displacement is the sum of individual displacements at a point), and two sources in phase produce constructive interference at points where the path-length difference is an integer wavelength and destructive interference at points where the difference is a half-integer wavelength.

Jump to a section
  1. What this dot point is asking
  2. The principle of superposition
  3. Constructive interference
  4. Destructive interference
  5. Path-length difference rule
  6. Where this leads next
  7. Examples in context
  8. Try this
  9. Exam-style questions

What this dot point is asking

QCAA wants you to state the principle of superposition for mechanical waves and apply it to constructive and destructive interference. The link between path-length difference and phase is the key idea: integer wavelengths of path difference (for sources in phase) give a maximum, half-integer wavelengths give a minimum.

The principle of superposition

When two or more waves meet at a point, the total displacement of the medium at that point is the sum of the displacements that each wave would produce on its own.

This is a linear principle: it applies to mechanical waves with small amplitudes (water, sound, strings) and to electromagnetic waves in vacuum.

After superposing, the waves continue past one another unchanged. They do not interact, only their displacements add at the moment of overlap.

Constructive interference

When two waves meet in phase (peaks line up with peaks, troughs with troughs), the resultant amplitude is the sum of the individual amplitudes. The waves reinforce.

For two identical waves of amplitude AA, the resultant amplitude is 2A2A at points of full constructive interference. Wave energy depends on A2A^2, so the energy at a constructive maximum is 44 times the energy of either wave alone (not 22 times, because energy scales with the square of amplitude).

Destructive interference

When two waves meet exactly out of phase (peaks line up with troughs), the resultant amplitude is the difference of the individual amplitudes. For identical waves, the resultant is zero: total destructive interference.

Energy is not destroyed. It is redistributed to the constructive maxima elsewhere in the interference pattern.

Path-length difference rule

For two sources oscillating in phase at the same frequency, the type of interference at a point depends on the path-length difference Δd\Delta d:

  • Constructive: Δd=nλ\Delta d = n \lambda, where n=0,1,2,3,…n = 0, 1, 2, 3, \ldots
  • Destructive: Δd=(n+12)λ\Delta d = (n + \tfrac{1}{2}) \lambda, where n=0,1,2,3,…n = 0, 1, 2, 3, \ldots

If the sources are out of phase by half a cycle, swap the two rules.

Worked example

Two speakers 1.01.0 m apart emit 1.01.0 kHz sound in phase. A listener stands 3.03.0 m directly in front of one speaker. Find the path-length difference and predict the interference at the listener's position. Use v=343v = 343 m s−1^{-1}.

Wavelength: λ=343/1000=0.343\lambda = 343 / 1000 = 0.343 m.

Distance from far speaker: 3.02+1.02=10=3.162\sqrt{3.0^2 + 1.0^2} = \sqrt{10} = 3.162 m.

Path difference: Δd=3.162−3.0=0.162\Delta d = 3.162 - 3.0 = 0.162 m.

Ratio: 0.162/0.343=0.4720.162 / 0.343 = 0.472.

This is close to a half wavelength, so the listener is near a destructive minimum but not exactly on one. Sound at this location is much quieter than at points where Δd\Delta d is an integer or half-integer multiple of λ\lambda.

Common traps
Treating interference as energy disappearing
Energy is conserved. Nodes (zero amplitude) exist alongside antinodes (large amplitude). Average over a full interference pattern recovers the input energy.
Confusing in-phase with same-amplitude
Constructive interference requires the waves to be in phase. Same-amplitude does not by itself guarantee constructive.
Forgetting that the rule flips for out-of-phase sources
If the sources oscillate 180°180° out of phase (one peak corresponds to the other's trough), integer path differences give minima.
Treating a small-amplitude resultant as no wave
Two waves of slightly different frequencies superpose to give beats (a slowly modulating amplitude). The destructive moments are real instants of zero displacement, not silence.

Where this leads next

Superposition is the foundation for standing waves (next dot point), for diffraction patterns (which you will meet in Year 12), and for double-slit interference of light (Unit 4 quantum context). The same path-difference rule applies in all three.

Examples in context

Example 1. A QPAC sound technician sets up two stage speakers 3.0 m3.0 \text{ m} apart playing the same 680 Hz680 \text{ Hz} test tone (λ=0.50 m\lambda = 0.50 \text{ m}). At a point 4.0 m4.0 \text{ m} from one speaker and 4.25 m4.25 \text{ m} from the other, path difference is 0.25 m=λ/20.25 \text{ m} = \lambda/2, so destructive interference produces a quiet patch. Moving 0.25 m0.25 \text{ m} sideways shifts the path difference to 0.50 m=λ0.50 \text{ m} = \lambda for constructive (loud). QCAA EA Unit 2 thematic items use this two-source setup.

Example 2. A Sunshine Coast tidal study models wave interaction between an ocean swell and a tide-driven counter-flow at a river mouth. When the two surface-wave trains meet 180∘180^\circ out of phase at 1.2 Hz1.2 \text{ Hz}, amplitudes near-cancel; in phase they double, creating standing crests visible to drone surveys. The same superposition principle underpinning QCAA Unit 2 IA1 data-test stems works at any wavelength.

Try this

Q1. State the principle of superposition. [2 marks]

  • Cue. When two or more waves overlap, the resultant displacement is the algebraic sum of the individual displacements.

Q2. Two speakers emit 1700 Hz1700 \text{ Hz} tones in phase. Calculate the wavelength using v=340 m s−1v = 340 \text{ m s}^{-1} and the path-difference required for destructive interference. [3 marks]

  • Cue. λ=0.20 m\lambda = 0.20 \text{ m}; destructive at (n+12)λ=0.10,0.30,0.50 m(n+\tfrac{1}{2})\lambda = 0.10, 0.30, 0.50 \text{ m}.

Q3. A QPAC two-speaker setup at 680 Hz680 \text{ Hz} (v=340 m s−1v = 340 \text{ m s}^{-1}) is placed 3.0 m3.0 \text{ m} apart. (a) Calculate the wavelength. (b) Determine the path-difference for the first constructive maximum off-axis. (c) Explain why such interference is rarely heard outdoors and discuss one design consideration. [2+3+3 marks; ISMG: Analysis and interpretation, Evaluation]

  • Cue. (a) 0.50 m0.50 \text{ m}; (b) Δ=λ=0.50 m\Delta = \lambda = 0.50 \text{ m}; (c) reflections off ground/walls scramble phase, design uses directional baffles.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Year 11 SAC-style4 marks
Two loudspeakers emit identical sound of frequency 686686 Hz in phase. A listener sits at a point that is 3.503.50 m from one speaker and 4.004.00 m from the other. Take the speed of sound as 343343 m s−1^{-1}. Determine whether the listener experiences constructive or destructive interference.
Show worked answer →

Wavelength: λ=v/f=343/686=0.500\lambda = v / f = 343 / 686 = 0.500 m.

Path-length difference: Δd=4.00−3.50=0.50\Delta d = 4.00 - 3.50 = 0.50 m.

Ratio: Δd/λ=0.50/0.50=1\Delta d / \lambda = 0.50 / 0.50 = 1, an integer.

For sources in phase, integer wavelengths of path difference give constructive interference.

The listener experiences constructive interference (a maximum).

Markers reward computing λ\lambda from v=fλv = f \lambda, the explicit calculation of path-length difference, and the conclusion stated against the integer rule.

Practise this

Sources & how we know this

ExamExplained