Superposition, reflection, and interference

Write the wave equation as ∂²y/∂x² = (1/v²) ∂²y/∂t². Both sides are built from derivatives, and the derivative of a sum is the sum of the derivatives, so if y₁ and y₂ each satisfy the equation then c₁y₁ + c₂y₂ satisfies it too for any constants c₁ and c₂. That is the whole justification for superposition: it is a property of the equation, not an extra physical postulate. Two consequences matter. First, waves pass through one another and emerge unchanged — a 40 Hz and a 400 Hz sound crossing a room do not scramble each other. Second, you add displacements and never intensities, because intensity goes as A² and (A₁ + A₂)² is not A₁² + A₂². Linearity is itself an approximation. The exact transverse equation for a string carries the curvature y″/(1 + y′²)³⁄², which reduces to the linear form only while the slope is small. At a slope of 0.10 the correction factor is (1.01)³⁄² = 1.015, a 1.5% error; at a slope of 0.50 it is 1.40, and the waves genuinely act on one another.

Two sources are coherent when they hold the same frequency and a fixed phase relation. Give them a source offset Δφ₀ and let the two paths to a listening point differ by ΔL. One extra wavelength of path is one extra full cycle of lag, so the arriving phase difference is Δφ = 2πΔL/λ + Δφ₀. Now add two equal cosines with cos α + cos β = 2 cos((α − β)/2) cos((α + β)/2). The first factor holds still in time and is therefore the resultant amplitude, 2A|cos(Δφ/2)|; the second is a wave of the original frequency carrying the mean phase. Take 686 Hz sound at 343 m s⁻¹, so λ = 0.500 m. With the sources in phase and ΔL = 1.00 m, ΔL/λ = 2 and Δφ = 4π, so Aᵣₑₛ = 2A and the intensity is four times that of one source. Step to where ΔL = 1.25 m: now ΔL/λ = 2.5, Δφ = 5π and Aᵣₑₛ = 0. Nothing about the sources changed — only the geometry did.

When the two amplitudes differ, the cosine identity no longer collapses so neatly, and the tidy route is a phasor: draw each wave as an arrow whose length is its amplitude and whose angle is its phase. Every arrow rotates at the same ω, so the angle between them is frozen and the sum is ordinary vector addition. The cosine rule then gives Aᵣₑₛ² = A₁² + A₂² + 2A₁A₂ cos Δφ. The result is trapped between A₁ + A₂ and |A₁ − A₂|, which is why unequal sources never cancel completely. With A₁ = 3.0 mm and A₂ = 2.0 mm at Δφ = 60°, Aᵣₑₛ² = 9.0 + 4.0 + 2(3.0)(2.0)(0.500) = 19.0 mm², so Aᵣₑₛ = 4.4 mm; at Δφ = 180° the same pair still leaves 1.0 mm. Setting A₁ = A₂ = A recovers Aᵣₑₛ² = 2A²(1 + cos Δφ) = 4A²cos²(Δφ/2), the equal-amplitude result.

Join string 1 to string 2 at x = 0 under one tension F. Write the left side as an incident plus a reflected wave, y = A cos(k₁x − ωt) + rA cos(k₁x + ωt), and the right side as y = tA cos(k₂x − ωt). The string cannot part, so y is continuous at x = 0; setting x = 0 and matching the cos ωt terms gives 1 + r = t. The junction carries no mass, so the transverse forces F ∂y/∂x on the two sides must balance, and with equal tension that makes the slope continuous; matching the sin ωt terms gives k₁(1 − r) = k₂t, and since k = ω/v, (1 − r)/v₁ = t/v₂. Substitute t = 1 + r and rearrange: r = (v₂ − v₁)/(v₂ + v₁). Check the limits. v₂ = v₁ gives r = 0 and t = 1, as it must when there is no junction at all. v₂ → 0, a string too heavy to move, gives r = −1 and t = 0: total inverted reflection, because yᵢ + yᵣ must vanish at a fixed end at every instant. A free end, v₂ → ∞, gives r = +1 and t = 2.

A wave on a string carries P = ½ μ v ω² A², and with μ = F/v² this is P = ½ (F/v) ω² A². Frequency and tension are the same on both sides of the junction, so power goes as A²/v. The reflected wave runs back through medium 1, so the reflected power fraction is simply R = r². The transmitted wave runs in medium 2, so T = (v₁/v₂)t². Substitute r and t and the two fractions sum to exactly 1 at every speed ratio — the boundary conditions were never free to break energy conservation. Numbers make it concrete. With v₁ = 40 m s⁻¹ and v₂ = 10 m s⁻¹, a string sixteen times heavier, r = −0.60 and t = 0.40, so R = 0.36 and T = (40/10)(0.40)² = 0.64, and 0.36 + 0.64 = 1.00. Run it the other way with v₂ = 3v₁: now t = 1.5, a transmitted pulse half again as tall as the incident one, yet T = (1/3)(1.5)² = 0.75 and R = 0.25. A taller pulse in a faster, lighter string is not extra energy.

A stationary interference pattern needs Δφ₀ to hold still. Two independent sources at nominally the same frequency drift in phase, and the time average of the cross term vanishes: ⟨I⟩ = I₁ + I₂ + 2√(I₁I₂)⟨cos Δφ⟩ collapses to I₁ + I₂, the incoherent result. That is why two loudspeakers on one oscillator interfere while two violins do not, and why a second coherent source in phase multiplies the intensity by four where a second incoherent source only doubles it. Amplitude sets the other limit. Superposition in air needs the acoustic pressure to stay a small perturbation on the ambient 1.01 × 10⁵ Pa. A painfully loud 120 dB tone carries I = 1.0 W m⁻², so its pressure amplitude is p = √(2ρvI) = √(2 × 1.2 × 343 × 1.0) = 29 Pa, a fraction 2.8 × 10⁻⁴ of ambient — still deeply linear. Blast waves and sonic booms are not, and there superposition stops holding.