Locate equilibrium and test the restoring-force law

1. Frame the question

An equilibrium position has zero net force. Stability asks what happens after a small displacement: a force back toward that position supports oscillation, while a force away from it destabilizes the equilibrium. Zero velocity alone does not identify equilibrium; a mass at a spring turning point is momentarily at rest under a nonzero restoring force.

2. Build the mechanism

Simple harmonic motion requires the stronger linear relation Fnet=−kx with positive constant k, where x is displacement from equilibrium. Newton’s second law gives a=−(k/m)x. The minus sign means acceleration points opposite displacement, not necessarily opposite velocity. A mass moving toward equilibrium speeds up even though its displacement magnitude shrinks.

For a vertical ideal spring, choose downward extension y from the unstretched length. The force is mg−ky, so equilibrium is yeq=mg/k. Define x=y−yeq and substitute: mg−k(x+mg/k)=−kx. Gravity changes the location of equilibrium, not the stiffness governing motion about it. The vertical spring potential alone is not the complete energy ledger; gravitational potential must also be included or combined into the shifted effective potential.

A force −20x−bx³ with b positive also points toward equilibrium. However, its force-to-displacement ratio changes with position when b is nonzero. It is a deliberate nonlinear contrast, not evidence that every restoring system has a sinusoidal motion or amplitude-independent period. At x=0 the ratio F/x is undefined; use the slope or nearby values instead.

3. Work an example

For m=2 kg, k=20 N/m and g=10 m/s², vertical equilibrium extension is 1 m. An additional downward displacement x=0.20 m gives net force −4 N and acceleration −2 m/s². With an optional cubic coefficient 100 N/m³, that displacement instead gives −4.8 N and −2.4 m/s²; the restoring direction is unchanged but the linear law fails.

4. Predict and investigate

Does a restoring force necessarily make the period independent of amplitude?

Compare x=0.1 and 0.2 m first with b=0, then b=100. Calculate F/x away from zero. Save contrasting captures and distinguish linearity from restoring direction; predict the force at negative displacement before checking.

5. Transfer to a new case

Try a new practice case

For m=4 kg and k=40 N/m, find signed acceleration at x=−0.30 m.

Your numerical result (m/s²) Check number

Reveal worked check (practice, not mastery)

This extra practice is not saved and does not award mastery. Record your first answer, explanation and whether you revealed help in the lesson notebook. Use the lesson’s saved question separately.

6. Explain the boundary

Why does adding gravity shift an ideal vertical oscillator’s equilibrium without changing k/m?

These are bounded virtual models and supplied numerical cases, not independent measurements or an observed physical investigation. Do not construct spring, pendulum or collision apparatus from these instructions. Practical evidence requires an appropriate supervised setting.

Sources

Original explanations, worked cases and numerical models. Framework checked 30 October 2026: AP Physics 1 Course and Exam Description, topics 7.1. Mechanism reference: OpenStax University Physics 1, simple harmonic motion and pendulums. No official assessment items reproduced.