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← all cardsTopic 11 · Wave Motion · Phase & phase difference

Trig method — no axis values

● VERBATIM — examined word-for-word

When x and λ can't be read: relabel the x-axis as a phase axis (1λ ↔ 360°). A particle at displacement y sits at angle given by sin θ = y/y₀ — at half amplitude, θ = 30°. Work out each particle's angle, subtract. The "hard" phase question is pure trig, not a new formula.

sin θ = y/y₀

What this actually means

Some graphs give no numbers on the axes, only displacements as fractions of the amplitude. The trick: relabel the x-axis as phase, with one wavelength spanning 360°.

A particle at displacement y sits at the angle θ where sin θ = y/y₀. Half amplitude means 30°, not 45°, because the sine curve is steep near zero.

Find each particle's angle, subtract, done. The 'hard' phase question is O-level trigonometry wearing a wave costume.

The trap

Assuming half amplitude means 45° of phase — sin 30° = ½, so it is 30°.

Prove it — watch it be true

  1. Open the phase protractor and drag the marker to y = y₀/2
  2. The angle reads 30°, not 45°
  3. Drag a second marker, subtract the two angles: that is Δφ
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