Superposition of waves: interference and beats
Where waves overlap, displacements add. Waves of equal frequency give interference; slightly different frequencies give beats, ν₁ − ν₂ per second.
The principle of superposition
Send two pulses towards each other along a stretched slinky. Where they overlap, the spring's displacement is the sum of the two. Two crests make a bigger crest. A crest and a trough partly cancel. After crossing, each pulse moves on with its own shape and direction, as if the other had never been there. This is the principle of superposition: the resultant displacement is the vector sum of the separate displacements. Because waves pass through each other unchanged, your radio can pick out one station although many stations broadcast at once. Superposition explains interference, beats and standing waves.
Derivation: resultant of two waves (interference)
Two waves of the same frequency travel in the same direction along the same line, with amplitudes a₁ and a₂ and a constant phase difference φ.
- 1y₁ = a₁ sin(ωt − kx) and y₂ = a₂ sin(ωt − kx + φ). Put θ = ωt − kx.
- 2By superposition, y = a₁ sin θ + a₂ sin(θ + φ) = (a₁ + a₂ cos φ) sin θ + (a₂ sin φ) cos θ.
- 3Let a₁ + a₂ cos φ = A cos α and a₂ sin φ = A sin α. Then y = A sin(θ + α): a wave of the same frequency.
- 4★Squaring and adding the two lines gives A² = a₁² + a₂² + 2a₁a₂ cos φ.
- 5Constructive: φ = 0, 2π, 4π ... (path difference 0, λ, 2λ ...). cos φ = 1, so A = a₁ + a₂, the largest value.
- 6Destructive: φ = π, 3π, 5π ... (path difference λ/2, 3λ/2 ...). cos φ = −1, so A = a₁ − a₂, the smallest value.
- 7So, Imax/Imin = (a₁ + a₂)²/(a₁ − a₂)². Interference only redistributes energy in space; no energy is lost.
★ key step
Beats
Sound two tuning forks of nearly equal frequency together. The loudness rises and falls regularly. One rise and fall is one beat. Let the forks have frequencies N and N + n. Every second, the second fork makes n more vibrations. Start with both in phase: the sound is loud. After 1/(2n) s, the second fork has gained half a vibration. The forks are now opposite in phase, and the sound is faint. After 1/n s, it has gained one full vibration. The forks are in phase again, and the sound is loud. So there is one beat every 1/n s, and n beats per second = (N + n) − N. The ear holds a sound for about 1/10 s, so more than about 10 beats per second cannot be heard separately.
Finding an unknown frequency with beats
| What you do to the unknown fork | Beats per second | So the unknown frequency was |
|---|---|---|
| Load a prong with wax (its frequency falls) | Decrease | Higher: ν = ν₀ + n |
| Load a prong with wax (its frequency falls) | Increase | Lower: ν = ν₀ − n |
| File a prong (its frequency rises) | Decrease | Lower: ν = ν₀ − n |
| File a prong (its frequency rises) | Increase | Higher: ν = ν₀ + n |
Worked example: interference and beats
(a) Two waves with intensities in the ratio 1 : 16 interfere. Find Imax/Imin. (b) A tuning fork A gives 6 beats per second with a 384 Hz fork B. When A is loaded with a little wax, the beats fall to 2 per second. Find the frequency of A.
- (a) I ∝ a², so the amplitudes are in the ratio √1 : √16 = 1 : 4.
- Imax/Imin = (1 + 4)² ÷ (4 − 1)² = 25 ÷ 9.
- (b) Before loading, A = 384 ± 6 Hz, so 390 Hz or 378 Hz.
- Wax lowers the frequency of A. The beats fell, so A moved closer to 384 Hz. That happens only if A was above 384 Hz.
- So A = 390 Hz. Check: after loading, about 386 Hz gives 386 − 384 = 2 beats per second.
(a) Imax : Imin = 25 : 9; (b) 390 Hz
How to draw the graph of beats
- Draw three horizontal time axes, one below the other, each 12 cm long for 1 second.
- On the top axis, draw 12 equal waves (12 Hz). On the middle axis, draw 10 equal waves (10 Hz). Start both at zero together.
- On the bottom axis, draw their sum: waves whose height swells and shrinks.
- Mark the points of largest height loud (maximum) and of smallest height faint (minimum).
- Mark one beat period 1/(12 − 10) = 0.5 s between two loud points, and write 2 beats per second.
A sitar or tanpura player sounds a string against a reference note and adjusts the tension until the slow wah-wah of beats disappears. Noise-cancelling earphones use destructive interference: they make a sound wave opposite in phase to the outside noise.
2 marks: “What are beats? Tuning forks of 256 Hz and 260 Hz are sounded together. How many beats are heard per second?” Beats are the regular rise and fall in loudness when two sounds of slightly different frequencies are heard together. Beats per second = 260 − 256 = 4.
2 marks: “In an interference pattern, the ratio of maximum to minimum intensity is 25 : 1. Find the ratio of the amplitudes of the two waves.” (a₁ + a₂)/(a₁ − a₂) = √25 = 5, so a₁ + a₂ = 5a₁ − 5a₂, giving 4a₁ = 6a₂. So a₁ : a₂ = 3 : 2.
3 marks: “Show that the number of beats heard per second equals the difference of the frequencies of the two sources.” Forks of N and N + n; the second gains one vibration on the first every 1/n s; at t = 0 in phase (loud), at 1/(2n) s opposite in phase (faint), at 1/n s in phase again (loud); one beat every 1/n s, so n = (N + n) − N beats per second.
Beats = difference. Wax makes a fork slower; a file makes it faster.