Superposition · H2 9749 Topic 12
Principle of Superposition Lab
y_res = y₁ + y₂ · Δφ = 2πΔx/λ · R = 2A cos(Δφ/2)
Try first:hit “set equal & opposite”, then “jump to full overlap”.
Station 1 — Pulses pass through each other
resultant = y₁ + y₂— pulse 1— pulse 2— resultant (what the string does)
time t0.000 s (centres meet at t = 2.000 s)
Pulse 1 — travelling +x
Pulse 2 — travelling −x
largest |y| on string
0.3000 m
peak of the sum
largest |∂y/∂t|
0.536 m s⁻¹
peaks when the string is flat → energy is all kinetic
pulse 1 peak, measured off the resultant
0.300 m
= A₁ = 0.30 m ✓ unchanged
pulse 2 peak, measured off the resultant
-0.300 m
= −A₂ = -0.30 m ✓ unchanged
Space–time map of the resultantx → across (0…10 m) · t ↓ down (0…4 s) · pink +y · blue −y
ⓘ more
Two straight tracks with opposite gradients: each pulse keeps a constant speed and a constant shape right through the crossing. Where the tracks meet, the colour is the SUM — for equal-and-opposite pulses that row goes black across the whole width, which is the flat string. Above and below that row the two tracks continue with their original colour and width, unbent and undimmed.
Station 2 — Two sine waves, same frequency
R = 2A cos(Δφ/2)measured peak of y₁+y₂
0.600 m
brute-force scan of the summed wave
2A cos(Δφ/2)
0.600 m
✓ agrees with the scan
√(A₁²+A₂²+2A₁A₂cosΔφ)
0.600 m
✓ agrees with the scan
Δφ ↔ path differenceΔφ = 2πΔx/λ ⇒ Δx = (Δφ/2π)λ = 0.000 λ
0 = nλ¼λ½λ = (n+½)λ¾λλ = nλ
CONSTRUCTIVE — Δx = 0λ = nλ
ⓘ more
Constructive interference needs the two waves to arrive IN PHASE: Δφ = 2πn, i.e. path difference Δx = nλ (n = 0, 1, 2 …). Destructive needs antiphase: Δφ = (2n+1)π, i.e. Δx = (n + ½)λ. Those conditions assume the two sources themselves start in phase — if the sources are in antiphase the two conditions swap over.
Station 3 — Coherence
why the sources must be coherentphase difference now
0.00 rad
constant — that IS coherence
resultant amplitude now
2.000 A
2A|cos(Δφ/2)| at the centre
time-averaged ⟨I⟩ at centre
4.00 I₀
0 frames averaged
fringe visibility of the average
1.000
(I_max − I_min)/(I_max + I_min)
running average of the central intensity4.00 I₀ / 4I₀ max
02I₀ — no interference4I₀ — full constructive
ⓘ more
Incoherent sources still produce fringes at every single instant — but the pattern jumps to a new position every few hundred milliseconds, so an eye or a detector, which averages over time, records the smooth 2I₀ line: exactly I₁ + I₂, the plain sum of the two separate intensities with no interference term left. That is why observable two-source interference requires a constant phase difference. Same reason two separate lamps never produce fringes, while one laser through a double slit does.
What this confirms
- Principle of superposition — LO (a). At every x the bold curve is the arithmetic sum of the two faint ones, computed point by point. Invert one pulse and the sum goes negative where the other is positive: it is a signed (vector) sum of displacements, not of amplitudes or intensities.
- Total destructive interference is real — LO (a), (b). Equal-and-opposite pulses at full overlap give “largest |y| on string” < 0.5 mm: the string is momentarily flat everywhere. On the space–time map that is a black band straight across.
- A flat string still holds the energy. At that same instant “largest |∂y/∂t|” is at its maximum — every particle is moving fastest as it passes through zero displacement. The energy is entirely kinetic, so nothing has been destroyed; a moment later the pulses reappear.
- Waves pass through one another unchanged. After the crossing, each pulse's peak measured off the resultant returns to exactly its launch amplitude, and its half-width and speed are unaltered. Two pulses do not collide, scatter or exchange anything — superposition applies only while they overlap.
- Resultant amplitude of two sine waves — LO (a), (b). With A₁ = A₂ the brute-force scan of y₁ + y₂ matches 2A cos(Δφ/2) to 3 d.p.; with unequal amplitudes it matches the general phasor result √(A₁² + A₂² + 2A₁A₂cos Δφ). Δφ = 0 → 2A; Δφ = π → 0.
- Phase difference ↔ path difference — LO (b). The translator strip is Δφ = 2πΔx/λ made visible. Constructive interference is flagged at Δx = nλ, destructive at Δx = (n + ½)λ — the standard conditions for two sources that are themselves in phase (RI notes §11.4).
- Interference needs coherence — LO (b), (g), (h). Locked phase: the amplitude trace is a flat line, the averaged pattern keeps visibility ≈ 1, fringes stay put. Random phase: the trace becomes a staircase and the averaged pattern collapses onto the 2I₀ line with visibility → 0. Coherent means constant phase difference, not necessarily zero — drag the locked-phase slider and the fringes shift but stay sharp.
- Why 2I₀ is the give-away number. Averaged over random phases, ⟨cos Δφ⟩ = 0, so ⟨I⟩ = I₁ + I₂ = 2I₀ — intensities simply add, exactly as if there were no interference at all. Coherent and in phase gives 4I₀ instead: twice the amplitude, four times the intensity.