University Physics I · Optional extension · Institutional Extension · Waves & Sound · WS.5
Superposition, Reflection & Interference
Two waves crossing one patch of string do not collide. They add, displacement by displacement, and separate unchanged. That one rule delivers interference patterns, and — with a boundary attached — the inverted echo from a fixed end.
Build the model
Connect the measurement to the mechanism.
The wave equation is linear, so if y₁ and y₂ are solutions, y₁ + y₂ is one too: overlapping waves add displacement by displacement and emerge unchanged. Everything else follows from that arithmetic. Two identical waves meeting with phase difference Δφ sum to amplitude 2A|cos(Δφ/2)| — full strength when Δφ = 0, nothing when Δφ = π — and for two in-phase sources Δφ is set by geometry alone, Δφ = 2πΔr/λ, so the pattern is a map of path difference.
Reflection is the same sum read at a boundary. Displacement and transverse force must stay continuous across a junction, which fixes the reflected amplitude at r = (v₂ − v₁)/(v₂ + v₁) and the transmitted one at t = 1 + r. A slower medium ahead gives r < 0, an inverted echo, and a fixed end is that limit taken to v₂ = 0, r = −1.
No energy is created or destroyed anywhere: interference moves power around, and r² + (Z₂/Z₁)t² = 1 at every junction.
- Simple definition
- Superposition: where two waves overlap, the medium's displacement is the sum of the displacements each wave would produce alone, and both waves continue afterwards as if the other had never been there.
- Example
- Two in-phase 343 Hz speakers, λ = 1.00 m. Stand 4.00 m from one and 5.00 m from the other: the 1.00 m path difference is one whole wavelength, the crests arrive together, and the sound is loud.
Each wave keeps its own shape and speed; the medium shows the sum.
Holds while the medium responds linearly to displacement
Δr = λ gives Δφ = 2π — one full wavelength of extra travel.
Δφ in rad; Δr and λ in m; Δφ₀ is any source phase offset
Swap the two lines if the sources are driven π out of phase.
m = 0, ±1, ±2, …; in-phase sources only (Δφ₀ = 0)
A phasor sum. Intensity peaks at 4I₁ but averages 2I₁ across the pattern.
Equal amplitudes: A = 2A₁|cos(Δφ/2)|, and I = 4I₁cos²(Δφ/2)
Fixed end is v₂ = 0, r = −1; free end is v₂ → ∞, r = +1.
Amplitude ratios; v₂ < v₁ gives r < 0, an inverted pulse
Impedance mismatch, not density alone, decides how much comes back.
Z in kg s⁻¹ for a string; R and T are power fractions summing to 1
Overlap is addition, not collision
Two pulses travelling towards each other on a string do not scatter or bounce. While they overlap, every particle of the string is displaced by the sum of what each pulse alone would demand: y(x, t) = y₁(x, t) + y₂(x, t). Afterwards each pulse continues with its original shape, amplitude, and speed, as if nothing had happened. This is a property of the wave equation, not a convenient assumption. ∂²y/∂x² = (1/v²)∂²y/∂t² is linear — no y², no |y| — so any sum of solutions is itself a solution. The physical requirement behind that linearity is that the restoring force stays proportional to displacement, which for a string means small slopes and a tension that does not change measurably as the wave passes. Break it and superposition goes with it: a loudspeaker driven past its linear range emits harmonics that were not in the input, and a sound wave loud enough to change the local air density carries its crests faster than its troughs, which is how a shock front forms.
What the sum looks like depends only on phase
Add two sinusoidal waves of equal amplitude travelling the same way, one lagging the other by a constant phase Δφ: y = A sin(kx − ωt) + A sin(kx − ωt + Δφ). The identity sin P + sin Q = 2 sin((P+Q)/2) cos((P−Q)/2) collapses this to y = 2A cos(Δφ/2) sin(kx − ωt + Δφ/2). The result is still a travelling wave of the same frequency and wavelength; only the amplitude has changed, to 2A|cos(Δφ/2)|. At Δφ = 0 that is 2A, the crests coincide, and the interference is constructive. At Δφ = π it is zero: crest sits on trough and the string stays flat. Every value between is allowed — at Δφ = π/2 the amplitude is 2A cos(π/4) = 1.41A, neither maximum nor null. Unequal amplitudes need the phasor sum, A² = A₁² + A₂² + 2A₁A₂ cos Δφ, which is the cosine rule applied to two rotating arrows; it reduces to (A₁ + A₂)² at Δφ = 0 and (A₁ − A₂)² at Δφ = π.
Geometry supplies the phase
For two sources driven in phase, the only thing that can put them out of step at a listening point is the distance each wave has travelled. One extra wavelength of path is one extra cycle of phase, so Δφ = 2πΔr/λ, with Δr = r₂ − r₁. Constructive interference needs Δr = mλ and destructive needs Δr = (m + ½)λ, with m any integer. Put two in-phase speakers 3.00 m apart and stand 4.00 m directly in front of one: the 3-4-5 triangle makes r₁ = 4.00 m, r₂ = 5.00 m, Δr = 1.00 m. At 343 Hz in air at 343 m s⁻¹, λ = 1.00 m, so Δr = λ, Δφ = 2π, and the point is a maximum. Halve the frequency to 171.5 Hz and λ = 2.00 m, so Δr = λ/2, Δφ = π, and the same point is a null. Nothing moved; only the wavelength did. That frequency dependence is why one fixed listening position cannot be the good seat for every note, and why interference nulls in a room shift as you sweep a tone.
Interference moves energy; it does not create or destroy it
Intensity goes as amplitude squared, so for two equal coherent sources I = 4I₁cos²(Δφ/2), where I₁ is what one source alone delivers there. At a maximum that is 4I₁ — twice the 2I₁ you would get by simply adding intensities — and at a null it is zero. Neither result breaks conservation, because cos² averages to ½ over the pattern: integrate the intensity across a plane and you recover 2I₁, exactly the power the two sources emitted. Interference is bookkeeping about where the energy lands. The nulls are also shallower than the algebra suggests, because amplitude falls as 1/r from a point source. In the speaker example the far speaker arrives with amplitude 4/5 of the near one, so at Δφ = π the sum is 0.20A₁ rather than zero, and the intensity is (0.20)² = 0.040 of the near speaker alone — a drop of 10 log₁₀(0.040) = −14.0 dB. Deep, audible, and not silence.
A boundary sends part of the wave back
At a junction between two strings under the same tension, two conditions hold at every instant. The string does not break, so the displacement is continuous: A(i) + A(r) = A(t). And a massless junction point cannot support a net transverse force, so tension times slope is continuous too, which gives (A(i) − A(r))/v₁ = A(t)/v₂. Solve the pair: r = A(r)/A(i) = (v₂ − v₁)/(v₂ + v₁) and t = A(t)/A(i) = 1 + r = 2v₂/(v₁ + v₂). The sign of r is the whole story. Send a pulse from a light string into one four times heavier at the same tension: v = √(T/μ), so v₂ = v₁/2, giving r = −1/3 and t = 2/3. The echo comes back inverted at a third the amplitude; the transmitted pulse stays upright at two thirds, travels at half the speed, and so is compressed to half the length. Push μ₂ to infinity and you have a fixed end, v₂ = 0, r = −1: total inverted reflection. Let μ₂ go to zero and you have a free end, r = +1, upright.
Follow the power, and note what the model assumes
Amplitude ratios are not energy ratios. A string carries average power P = ½Zω²A² with impedance Z = μv = √(FT μ), FT being the tension, so the reflected fraction of the power is R = r² and the transmitted fraction is T = (Z₂/Z₁)t². For the four-times-heavier string, Z₂/Z₁ = √(μ₂/μ₁) = 2, so R = (1/3)² = 1/9 = 0.111 and T = 2 × (2/3)² = 8/9 = 0.889. They sum to 1, as they must. The same impedance argument works for sound, where Z = ρv. Air has Z ≈ 1.2 × 343 = 412 kg m⁻² s⁻¹ and water about 1.48 × 10⁶, a mismatch of 3600, so the power crossing from air into water is 1 − ((Z₂ − Z₁)/(Z₂ + Z₁))² = 0.0011 — about one part in 900. Two assumptions carry all of this: linearity, and coherence, meaning sources that hold a fixed phase relationship. Two independent sources drifting in frequency give no fixed pattern at all; they give beats, taken up later in this extension. And when a reflected wave meets the outgoing one on the same string, the sum is no longer a travelling wave — that is the standing-wave problem.
Change one variable at a time
Make the relationship visible.
Hold λ at 1.00 m and drag Δr from 0 to 1.00 m — the sum swells to 2A₁, flattens to a dead string at half a wavelength, and rebuilds — then change λ alone and watch the same Δr turn from a maximum into a null.
PHASE Δφ1.57 rad
PATH DIFFERENCE0.25 λ
AMPLITUDE A1.41 A₁
INTENSITY I2.00 I₁
Live interpretationPHASE Δφ: 1.57 rad. PATH DIFFERENCE: 0.25 λ. AMPLITUDE A: 1.41 A₁. INTENSITY I: 2.00 I₁
Catch the common trap
Explain before calculating.
A transverse pulse travels along a light string and reaches a junction with a heavier string of four times the linear density, held at the same tension. What happens at the junction?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyTwo loudspeakers driven in phase emit 686 Hz sound into air at 343 m s⁻¹. A listener stands 6.40 m from one speaker and 7.15 m from the other. Is that point a maximum or a minimum?
- Wavelength first: λ = v/f = 343/686 = 0.500 m.
- Path difference: Δr = 7.15 − 6.40 = 0.75 m.
- In wavelengths: Δr/λ = 0.75/0.500 = 1.5, so Δr = (1 + ½)λ — a half-integer, not a whole one.
- As phase: Δφ = 2πΔr/λ = 2π(1.5) = 3π rad = 9.42 rad, which is π plus one full cycle, so crest lands on trough.
AnswerA minimum — destructive interference, with Δr = 1.5λ and Δφ = 3π rad (9.42 rad).
MediumTwo coherent sources of the same frequency reach a point with amplitudes A₁ = 4.0 mm and A₂ = 3.0 mm, arriving with a phase difference of 60°. Find the resultant amplitude, and the intensity there as a multiple of what source 1 delivers alone.
- Unequal amplitudes need the phasor sum: A² = A₁² + A₂² + 2A₁A₂ cos Δφ.
- With cos 60° = 0.500: A² = (4.0)² + (3.0)² + 2(4.0)(3.0)(0.500) = 16 + 9 + 12 = 37 mm².
- A = √37 = 6.1 mm — between the 7.0 mm of Δφ = 0 and the 1.0 mm of Δφ = π.
- Intensity follows amplitude squared, so I/I₁ = A²/A₁² = 37/16 = 2.3.
AnswerA = 6.1 mm; I = 2.3 I₁.
HardA string of linear density 2.5 g m⁻¹ is knotted to one of 40 g m⁻¹, both under tension 40 N. A 5.0 mm pulse arrives from the light side. Find the two speeds, the reflected and transmitted amplitudes, and the fraction of the incident power that returns.
- v = √(F/μ): v₁ = √(40/0.0025) = √16000 = 126 m s⁻¹ and v₂ = √(40/0.040) = √1000 = 31.6 m s⁻¹, so the heavy string is exactly four times slower.
- r = (v₂ − v₁)/(v₂ + v₁) = (1 − 4)/(1 + 4) = −0.60; the minus sign is the inversion, and the echo is 0.60 × 5.0 = 3.0 mm upside down.
- t = 1 + r = 0.40, so the transmitted pulse is upright at 0.40 × 5.0 = 2.0 mm — and 5.0 − 3.0 = 2.0 mm, which is displacement continuity at the knot.
- Power fractions: R = r² = (−0.60)² = 0.36 and T = (Z₂/Z₁)t² = √(μ₂/μ₁) × t² = 4 × 0.16 = 0.64, summing to 1.00.
Answerv₁ = 126 m s⁻¹, v₂ = 31.6 m s⁻¹; reflected 3.0 mm inverted, transmitted 2.0 mm upright; 36% of the power comes back and 64% goes on.