SCIENCE · INVESTIGATE AND EXPLAIN

Test a local harmonic approximation against a nonlinear force

Use a potential derivative to distinguish exact SHM from a local approximation.

Learning outcome: Compare lambda=0,4,8 at the same quadratic coefficient. Calculate the force-term ratio and exact center speed, then halve amplitude in a written changed case. State which conclusions are exact under the supplied potential and which require a small-displacement approximation.

Starting recap

Derive the restoring equation, retain initial conditions and distinguish exact harmonic results from approximation and calibration assumptions.

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Try

For m=2 kg,k=4 N/m,lambda=4 N/m³ and A=.5 m, force correction ratio is .25. Exact energy is .5625 J and exact center speed is .75 m/s; the quadratic prediction is sqrt(.5)≈.707107 m/s. The local angular frequency is sqrt(2), not a demonstrated finite-amplitude frequency.

Open the experiment

Predict

Does a stable nonlinear restoring force necessarily produce exact SHM?

Your prediction and reason

Experiment

Compare lambda=0,4,8 at the same quadratic coefficient. Calculate the force-term ratio and exact center speed, then halve amplitude in a written changed case. State which conclusions are exact under the supplied potential and which require a small-displacement approximation.

Test the local harmonic approximation to a nonlinear potential

Worked teaching model · synthetic values · Optional local-curvature enrichment: U=kx²/2+lambda*x⁴/4,m=2 kg. Compare force and energy at assigned amplitude .5 m. Positive coefficients give a stable origin; lambda>0 is not exactly SHM. No claimed exact nonlinear period.

Quadratic coefficient k: 4 N/m Quartic coefficient lambda: 4 N/m³

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Exact restoring force

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Exact restoring force: numerical coordinates

displacement (m)	force (N)
-0.5	2.5
-0.483333	2.38498
-0.466667	2.27319
-0.45	2.1645
-0.433333	2.05881
-0.416667	1.95602
-0.4	1.856
-0.383333	1.75865
-0.366667	1.66385
-0.35	1.5715
-0.333333	1.48148
-0.316667	1.39369
-0.3	1.308
-0.283333	1.22431
-0.266667	1.14252
-0.25	1.0625
-0.233333	0.984148
-0.216667	0.907352
-0.2	0.832
-0.183333	0.757981
-0.166667	0.685185
-0.15	0.6135
-0.133333	0.542815
-0.116667	0.473019
-0.1	0.404
-0.0833333	0.335648
-0.0666667	0.267852
-0.05	0.2005
-0.0333333	0.133481
-0.0166667	0.0666852
0	0
0.0166667	-0.0666852
0.0333333	-0.133481
0.05	-0.2005
0.0666667	-0.267852
0.0833333	-0.335648
0.1	-0.404
0.116667	-0.473019
0.133333	-0.542815
0.15	-0.6135
0.166667	-0.685185
0.183333	-0.757981
0.2	-0.832
0.216667	-0.907352
0.233333	-0.984148
0.25	-1.0625
0.266667	-1.14252
0.283333	-1.22431
0.3	-1.308
0.316667	-1.39369
0.333333	-1.48148
0.35	-1.5715
0.366667	-1.66385
0.383333	-1.75865
0.4	-1.856
0.416667	-1.95602
0.433333	-2.05881
0.45	-2.1645
0.466667	-2.27319
0.483333	-2.38498
0.5	-2.5

Local linear restoring force

Axes rescale independently. Curves join sampled coordinates; a joining segment is not additional measured evidence. Read physical units rather than screen angles or apparent slopes.

Local linear restoring force: numerical coordinates

displacement (m)	force (N)
-0.5	2
-0.483333	1.93333
-0.466667	1.86667
-0.45	1.8
-0.433333	1.73333
-0.416667	1.66667
-0.4	1.6
-0.383333	1.53333
-0.366667	1.46667
-0.35	1.4
-0.333333	1.33333
-0.316667	1.26667
-0.3	1.2
-0.283333	1.13333
-0.266667	1.06667
-0.25	1
-0.233333	0.933333
-0.216667	0.866667
-0.2	0.8
-0.183333	0.733333
-0.166667	0.666667
-0.15	0.6
-0.133333	0.533333
-0.116667	0.466667
-0.1	0.4
-0.0833333	0.333333
-0.0666667	0.266667
-0.05	0.2
-0.0333333	0.133333
-0.0166667	0.0666667
0	0
0.0166667	-0.0666667
0.0333333	-0.133333
0.05	-0.2
0.0666667	-0.266667
0.0833333	-0.333333
0.1	-0.4
0.116667	-0.466667
0.133333	-0.533333
0.15	-0.6
0.166667	-0.666667
0.183333	-0.733333
0.2	-0.8
0.216667	-0.866667
0.233333	-0.933333
0.25	-1
0.266667	-1.06667
0.283333	-1.13333
0.3	-1.2
0.316667	-1.26667
0.333333	-1.33333
0.35	-1.4
0.366667	-1.46667
0.383333	-1.53333
0.4	-1.6
0.416667	-1.66667
0.433333	-1.73333
0.45	-1.8
0.466667	-1.86667
0.483333	-1.93333
0.5	-2

Calculated values under the declared model
Quantity	Value	Unit
Local harmonic angular frequency	1.41421	rad/s
Quadratic-only predicted center speed	0.707107	m/s
Exact energy-based center speed	0.75	m/s
Cubic/linear restoring-force ratio at amplitude	0.25	ratio
Exact total energy from rest at amplitude	0.5625	J
Curvature at equilibrium	4	N/m

The stable nonlinear potential is only approximately harmonic near its minimum.

Optional enrichment: F=−dU/dx=−kx−lambda*x³ and U second derivative at0 is k. The local linear equation has omega=sqrt(k/m), but a nonzero cubic force at finite amplitude violates exact SHM. Comparing force error and exact energy does not determine an exact nonlinear period.

Derive before checking. Retain contrasting captures and explain one assumption that could fail; calculated evidence is not actual laboratory work.

The stable nonlinear potential is only approximately harmonic near its minimum.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Optional calculus enrichment: expand near equilibrium

This lesson extends the core linear oscillator with an explicitly supplied nonlinear potential. For U(x)=kx²/2+lambda x⁴/4, F=−dU/dx=−kx−lambda x³. At x=0 the force vanishes and U second derivative is k. Positive k gives a locally stable minimum and a local harmonic frequency sqrt(k/m). It does not make the entire nonlinear force exactly proportional to x.

Quantify the neglected term

At amplitude A, the magnitude ratio of cubic to linear restoring force is lambda A²/k. This is a declared diagnostic for the force approximation, not a general exact bound on period error. Halving amplitude quarters that ratio. If lambda=0 the model reduces to exact SHM; if lambda is positive the force stiffens with increasing displacement.

Use exact energy without claiming exact harmonic timing

Released from rest at A, total E=kA²/2+lambda A⁴/4. At the center U=0, so v_max=sqrt(2E/m). This energy result is exact for the stipulated conservative potential even when a sinusoidal time history is not. The quadratic-only prediction is A sqrt(k/m); comparing them diagnoses one consequence of neglecting the quartic term.

Separate different evidence claims

A stable periodic trajectory need not be sinusoidal. Measuring a restoring slope only very close to equilibrium estimates k, not the higher-order coefficient throughout the path. The workbench plots the exact force and its local tangent over the assigned range. It does not calculate or claim the exact nonlinear period, damping, resonant response or physical material behavior.

Worked original case

Investigation

Public numerical transfer practice

Try a new practice case

For U=4x²/2+4x⁴/4, what is U second derivative at x=0 in N/m?

Numerical result in the requested units

This public, unsaved check is not a secure assessment. Explain the derivative, integral or boundary in the retained course evidence workspace.

Sources and evidence boundary

Original derivations and assigned numerical cases checked 30 October 2026 against the official AP Physics C Mechanics framework and OpenStax harmonic motion and pendulums. These are supplied or generated records, not physical measurements or performed laboratory time. Do not construct rotating equipment, release loads or conduct launch experiments. Public feedback and retained self-reports do not certify independent mastery.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Why can the local frequency be correct while a sinusoidal prediction fails at a finite amplitude?

Your explanation and limitations

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Previous: Recover harmonic timing from an energy integral Next: Audit a pendulum small-angle approximation

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