〰️Wave Pulse Simulator - Reflection, Transmission, Superposition, Standing Waves

How to use Wave Pulse on a String: Reflection & Transmission

1. 1Pick a scenario, or switch between Send from the left and Colliding pulses
2. 2Set the waveform, amplitude, pulse width or frequency, tension and the two linear densities, and choose a fixed, free or absorbing right end
3. 3Press Play and use slow motion and Step to catch the moment a pulse reaches the junction or the end; the grey dashed curve is the exact solution
4. 4Once the pulses separate, compare the measured r, t, R and T with theory; with a continuous wave, look for the nodes in the envelope

FAQ

Why does a pulse reflect upside down going from a light to a heavy string?

Displacement and slope must both be continuous at the junction. The heavy string has a larger impedance Z = √(Tμ) and is hard to move, so it holds the junction back like a partly fixed end, and the reflection is inverted: r = (Z₁−Z₂)/(Z₁+Z₂) < 0. In the limit Z₂ → ∞ you get a fixed end with r = −1, the “half-wave loss” of optics. Going from heavy to light, r > 0 and the reflection is upright, with a free end (r = +1) as the limit.

The transmitted pulse can be taller than the incident one. Doesn’t that break energy conservation?

From a heavy to a light string t = 2Z₁/(Z₁+Z₂) exceeds 1, for example t = 4/3 when μ drops from 0.4 to 0.1. But the energy a wave carries is also proportional to the impedance: the transmitted share is (Z₂/Z₁)t², and the light string’s Z₂ is small, so no energy is created. The reflected share r² plus the transmitted share (Z₂/Z₁)t² always equals 1, and the tool lists both measured values next to theory.

How can two pulses pass straight through each other?

The small-amplitude wave equation is linear, so the sum of two solutions is also a solution: the superposition principle. Where the pulses overlap the displacements add point by point. Same-sign pulses build a taller peak; opposite-sign pulses briefly cancel and the string looks flat, but it is still moving, so all the energy is kinetic at that instant. Afterwards each pulse carries on unchanged.

How does a standing wave form, and where are the nodes?

A continuous sine wave reflects from the fixed end and travels back with the same frequency. The two counter-propagating waves add to 2A sin(kx) cos(ωt): some points never move (nodes) and some swing with amplitude 2A (antinodes). A fixed end is a node and neighbouring nodes are λ/2 = c/(2f) apart; a free end is an antinode. The tool marks the predicted nodes with purple triangles and shades each point’s amplitude envelope.

How does the simulation stay stable and accurate?

It uses a centred finite-difference scheme with Δt = 0.98Δx/c_max, which satisfies the CFL condition cΔt ≤ Δx; otherwise errors would grow exponentially. The slower section runs at a lower Courant number and has slight numerical dispersion, so sharp square edges ring a little, which is why the square pulse edges are smoothed. The absorbing end uses a first-order Mur boundary that leaves around 10⁻⁶ of the energy behind, and for Gaussian pulses the measured coefficients typically match theory within 0.5%.