Physics III course sequence · Interactive lesson
Small oscillations in a potential well
Turn any smooth stable equilibrium into a local harmonic oscillator. Read the frequency from curvature, then test how small ‘small’ must be.
UPIII-01.07 · Lesson 7 of 9 in this unit. Position is not a completion score.Build the model
Curvature sets the local clock; higher derivatives expose its limits.
The key idea
Near a smooth stable equilibrium x₀ with U′(x₀)=0 and U″(x₀)>0, the potential is locally quadratic and a mass m oscillates with ω=√(U″(x₀)/m).
Example. A marble near the bottom of a smooth bowl needs no literal spring: the bowl's local curvature supplies keff=U″(x₀).
The complete physical model and assumptions
At equilibrium the first derivative of the potential vanishes. If the second derivative there is positive, the leading change in potential is quadratic, so the force is linear in displacement and the motion is simple harmonic. The local angular frequency is fixed by curvature divided by mass. Cubic and quartic terms eventually shift the symmetry and make the period amplitude-dependent. If the curvature vanishes, the harmonic conclusion cannot be drawn from stability alone.
1Find equilibrium before expanding
Solve U′(x₀)=0 and test the point. Positive U″ gives a stable minimum; negative U″ gives an unstable maximum. Expanding in ξ=x−x₀ prevents a hidden constant-force term.
2Curvature becomes stiffness
Taylor's theorem gives U≈U₀+½U″₀ξ². Therefore F≈−U″₀ξ and ξ̈+(U″₀/m)ξ=0. The dimensions agree: curvature is energy per length squared, or force per length.
3Amplitude reveals anharmonicity
The cubic force correction is −½U‴₀ξ², so its magnitude relative to the harmonic force is |U‴₀ξ|/(2U″₀). Quartic terms can stiffen or soften the well symmetrically.
4Stable does not always mean harmonic
For U=aξ⁴ with a>0, the origin is stable but U″(0)=0. The restoring force is cubic and the period depends on amplitude even close to equilibrium.
Key equations
Read the units, assumptions and limits before substituting.
Local expansion
U(x₀+ξ) = U₀ + ½U″₀ξ² + ⅙U‴₀ξ³ + 1/24 U⁗₀ξ⁴ + …
The quadratic term leads; higher powers measure anharmonicity.
U′(x₀)=0 at equilibrium.Equation and frequency
mξ̈ = −U″₀ξω = √(U″₀/m)T = 2π√(m/U″₀)
Potential curvature is the effective spring constant.
Requires U″₀>0 and sufficiently small displacement.First boundary check
|U‴₀ξ|/(2U″₀) ≪ 1
The cubic correction to force grows linearly relative to the harmonic force.
Compare force corrections, not only potential terms.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Vary mass or curvature to verify the period law. Then vary cubic strength and excursion until the anharmonicity readout is no longer small.
Live results
- ANGULAR FREQUENCY
- 1.41 rad/s
- PERIOD
- 4.44 s
- CUBIC FORCE RATIO
- 0.032
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Mass m: 1.0 kg; Curvature U″₀: 2.0 N/m; Cubic strength: 0.03 relative; Excursion A: 0.7 relative
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Stable equilibrium is not automatically SHM. The local result also needs non-zero positive curvature and an excursion small enough that higher derivatives do not compete.
Practice & worked examples
2 authored examples. Make an attempt, then reveal the reasoning one step at a time.
Worked calculationExample 1
A 0.500 kg particle moves in U(x)=3.00+8.00x²+2.00x³ J near x=0. Find the small-oscillation frequency and the cubic force correction at amplitude 0.050 m.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 3 steps revealed.
- U′=16x+6x², so x₀=0; U″(0)=16 N m⁻¹>0.
- ω=√(16/0.500)=5.66 rad s⁻¹ and T=2π/ω=1.11 s.
- U‴=12 J m⁻³, so the force-correction ratio is (12)(0.050)/(2×16)=0.01875.
Answer. ω=5.66 rad s⁻¹, T=1.11 s, and the cubic force correction is 1.875% at the stated amplitude.
DerivationExample 2
For U(x)=a(x²−b²)² with a, b>0, classify the equilibria and find the small-oscillation frequency about either minimum.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 3 steps revealed.
- U′=4ax(x²−b²), so equilibria occur at x=0 and x=±b.
- U″=4a(3x²−b²): x=0 is unstable and x=±b are stable.
- At either minimum keff=U″=8ab², so ω=√(8ab²/m).
Answer. x=0 is unstable; x=±b are stable; ω=√(8ab²/m) about either minimum.
Quick check
Select an answer and explain it before checking.
Review & continue
Near a smooth stable equilibrium x₀ with U′(x₀)=0 and U″(x₀)>0, the potential is locally quadratic and a mass m oscillates with ω=√(U″(x₀)/m).
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.