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SCIENCE · INVESTIGATE AND EXPLAIN

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 experiment

Predict

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.

Rightward contribution (m) · t = 0.000 s-0.1200.1201234Position x (m) · same axes for contributions and sum
Leftward contribution (m) · t = 0.000 s-0.1200.1201234Position x (m) · same axes for contributions and sum
Resultant displacement (m) · t = 0.000 s-0.1200.1201234Position x (m) · same axes for contributions and sum

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.

Superposition at the tracked point
QuantityValueInterpretation
Right contribution0.050000 mA sin(kx − omega t)
Left contribution0.050000 mA sin(kx + omega t + phi)
Resultant displacement0.100000 mSigned sum, not a sum of magnitudes
Right velocity contribution0.000000 m/sTime rate of the right contribution
Left velocity contribution0.000000 m/sTime rate of the left contribution
Resultant transverse velocity0.000000 m/sSum of both velocity contributions
Local amplitude envelope0.100000 mMaximum absolute resultant displacement at x = 0.5 m
Persistent nodes0.000, 1.000, 2.000, 3.000, 4.000 mZero displacement throughout the cycle
Antinodes0.500, 1.500, 2.500, 3.500 mLocal 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

Add contributions at one place and time

When a medium responds linearly, two overlapping wave displacements add algebraically. Resultant displacement = right-wave contribution + left-wave contribution, evaluated at the same position and time. If one contribution is +0.030 m and the other is −0.020 m, the result is +0.010 m. Adding their magnitudes would lose the direction information.

The panels show two mathematical contributions and their resultant on identical scales. They are not three separate strings. The tracked material coordinate is fixed at x = 0.5 m. Both contributions have amplitude 0.050 m, wavelength 2 m and frequency 1 Hz, but travel in opposite directions at 2 m/s. Their sum produces a standing pattern.

Compare reinforcement with a flat instant

Start at phase offset 0° and time 0 s. At the tracked point, both contributions are +0.050 m, giving +0.100 m. This is constructive interference at that point and instant. The resultant may exceed either component’s amplitude without violating the addition rule.

Move time to 0.25 s. The resultant graph is now flat across the viewing window. At the tracked point both contributions happen to cross zero, but at many other positions they are equal and opposite. For example, at x = 0 the right contribution is −0.050 m and the left contribution is +0.050 m. Their sum is zero.

A flat displacement snapshot does not mean the medium is at rest. At x = 0.5 m and t = 0.25 s, the velocity contributions are each approximately −0.314159 m/s, so the resultant transverse velocity is −0.628319 m/s. The point is crossing equilibrium at its greatest speed. At t = 0.50 s, its displacement is −0.100 m and its velocity is zero again.

Use the ledger to distinguish adding displacements from adding velocities. Both operations are valid here, but one cannot infer the second result merely from the first. A zero value and a zero rate of change are different claims.

Change phase without inventing a collision

The phase control shifts the left wave’s cycle relative to its specified formula. It selects a different steady pattern; it does not simulate changing a source while a wave is already in flight. At 90° and time zero, the tracked point has a right contribution of +0.050 m and a left contribution of zero, giving +0.050 m. Its resultant velocity is approximately −0.314159 m/s.

The phase label is not a universal statement that the two waves are 90° apart at every time: because they propagate oppositely, their relative phase at a point also evolves with time. In this model the right contribution uses sin(kx − omega t), while the left uses sin(kx + omega t + phi). The control changes phi only.

These prescribed waves pass through one another according to linear superposition. They do not bounce off each other or permanently erase one another when the sum vanishes. Strongly nonlinear media need a different model; adding two arbitrary disturbances is not always a physically accurate rule.

Cancellation is not an energy measurement

Zero displacement at an instant does not establish that energy has disappeared. In the zero-phase example, points away from nodes are moving when the resultant passes through its flat configuration. Their kinetic energy would require material mass information to calculate. The browser supplies no material density, tension, source work or energy flux, so it does not calculate an energy budget.

Likewise, do not add component intensities as though they were signed displacements. Energy-related quantities generally depend on squares and can include interference terms. This lesson verifies displacement and velocity addition, not an energy-loss claim.

Transfer to an unfamiliar overlap

At one point two contributions are +0.040 m and −0.040 m, with transverse velocity contributions +0.10 m/s and +0.20 m/s. Predict the resultant displacement and velocity, then decide whether the point is a persistent node.

Check the superposition transfer

The displacement is zero and the velocity is +0.30 m/s. The point is crossing equilibrium, so this observation cannot identify it as a persistent node. A node must retain zero resultant displacement throughout the motion, not merely at one instant.

OpenStax interference of waves introduces linear superposition. The browser’s component formulas and numerical cases are original prescribed examples, not measurements of a particular medium.

Apply

At the same position and time, two small linear-wave contributions are +0.030 m and −0.020 m. Their resultant displacement is:

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.

Skipping retains experiments and awards no mastery.

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