Physics 2Waves, Sound and Physical Optics

Superposition and Interference

Overlapping waves add point by point: crest on crest builds up, crest on trough cancels.

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When two waves meet in the same medium, what happens? With the wave graphs you can follow it point by point: the displacements simply add.

Key idea

Superposition: where waves overlap, the total displacement is the sum of the separate displacements. Crest on crest builds up (constructive interference); crest on trough cancels (destructive interference). Afterward, each wave goes on unchanged.
Two pulses (dashed) and their sum (solid). They pass through each other and come out as they went in.

Why it works

Why add? For small displacements, the force restoring each bit of string is proportional to how far it is displaced, so the equation of motion is linear: if each wave alone is a solution, their sum is a solution too. Nothing in the medium "remembers" which wave a bit of motion came from. Noise-canceling headphones use this: a microphone picks up the noise, and the speaker adds a copy turned upside down, so crest meets trough at your ear.

Common mistake

Thinking two waves that cancel are destroyed. At the instant two opposite pulses overlap exactly, the string is flat, but it is moving fast: the energy is all kinetic for that moment. A moment later both pulses re-emerge, unchanged. Check: energy cannot vanish, so the waves must still be there.

Worked example: Adding pulses

A pulse $3.0\ \text{cm}$ high meets a pulse $2.0\ \text{cm}$ deep (upside down) from the other direction. What is the displacement where they overlap exactly? What are they like afterward?
  1. Add with signs: $3.0 + (-2.0) = 1.0\ \text{cm}$.
  2. Afterward each continues unchanged: $3.0\ \text{cm}$ up and $2.0\ \text{cm}$ down, moving apart.

Answer: $1.0\ \text{cm}$ while they overlap; unchanged afterward

Worked example: Two speakers

Two speakers play the same $343\ \text{Hz}$ tone in step. You stand $4.0\ \text{m}$ from one and $5.5\ \text{m}$ from the other. Is the sound loud or quiet? ($v = 343\ \text{m/s}$.)
  1. Wavelength: $\lambda = \dfrac{343}{343} = 1.00\ \text{m}$.
  2. Path difference: $5.5 - 4.0 = 1.5\ \text{m} = 1.5\lambda$.
  3. A half-number of wavelengths means crest meets trough: destructive.
  4. Whole numbers of wavelengths ($0, \lambda, 2\lambda, \dots$) would be constructive.

Answer: Quiet (destructive interference)

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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Sources and further reading

  • OpenStax College Physics 2e, Section 16.10: Superposition and Interference
  • OpenStax University Physics Volume 1, Section 16.5: Interference of Waves
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