Add wave displacements and interpret cancellation
Compare signed component displacements with their sum and test what cancellation does and does not imply.
Learning outcome: Apply linear superposition at the same position and time, distinguish displacement from velocity, and identify limits of the prescribed-wave model.
Starting recap
Displacement and transverse velocity can be positive, negative or zero. A sum must compare values at the same position and instant.
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Try
Two contributions are +0.030 m and −0.020 m at one point. Is the resultant 0.050 m or 0.010 m?
Open the experimentPredict
With zero phase offset, predict the resultant at x = 0.5 m at times 0 and 0.25 s. Does a zero displacement imply zero velocity?
Experiment
Capture zero phase at 0 and 0.25 s. Compare the two contributions and their sum on identical axes, then compare the velocity entries. Change the phase to 90 degrees at time zero and explain the new sum.
Add two counterpropagating waves
Worked teaching model · synthetic values · Linear superposition of two prescribed transverse sinusoidal waves, each amplitude 0.050 m, wavelength 2 m and frequency 1 Hz; speeds are +2 and −2 m/s. Right wave is A sin(kx − omega t); left wave is A sin(kx + omega t + phi). The phase control changes phi for a separate steady pattern, not a mid-run source adjustment. Fixed viewing window 0–4 m, 161 samples, tracked material point x = 0.5 m. No boundary, reflection process, source startup, damping, nonlinear coupling or energy flux is computed. Nodes describe positions with zero resultant displacement at all times, not every zero of a snapshot.
Reset keeps your prediction, captures and practice history.
The three panels decompose one physical displacement into two contributions and their sum. The filled markers share x = 0.5 m. In the resultant panel, the dashed envelope gives maximum absolute displacement over a full cycle, not another traveling wave. Open circles mark persistent nodes. Node positions stay fixed when time changes; changing phase selects a different pattern.
| Quantity | Value | Interpretation |
|---|---|---|
| Right contribution | 0.050000 m | A sin(kx − omega t) |
| Left contribution | 0.050000 m | A sin(kx + omega t + phi) |
| Resultant displacement | 0.100000 m | Signed sum, not a sum of magnitudes |
| Right velocity contribution | 0.000000 m/s | Time rate of the right contribution |
| Left velocity contribution | 0.000000 m/s | Time rate of the left contribution |
| Resultant transverse velocity | 0.000000 m/s | Sum of both velocity contributions |
| Local amplitude envelope | 0.100000 m | Maximum absolute resultant displacement at x = 0.5 m |
| Persistent nodes | 0.000, 1.000, 2.000, 3.000, 4.000 m | Zero displacement throughout the cycle |
| Antinodes | 0.500, 1.500, 2.500, 3.500 m | Local amplitude 0.100 m |
At x = 0.5 m: 0.050000 m + 0.050000 m = 0.100000 m.
Resultant transverse velocity 0.000000 m/s; local amplitude envelope 0.100000 m. Nodes in this window: 0.000, 1.000, 2.000, 3.000, 4.000 m. A flat instantaneous displacement curve need not imply rest.
Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.
Understand
Apply
Public practice checks a specific claim. It does not establish independent mastery.
Use a capture to distinguish signed displacement cancellation from rest. Explain why this model cannot establish energy loss or calculate a transported energy flux.
Continue
Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.
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