University Physics III · Oscillations · 1.7
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.
Build the model
Connect the measurement to the mechanism.
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.
- Simple definition
- 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 quadratic term leads; higher powers measure anharmonicity.
U′(x₀)=0 at equilibrium.
Potential curvature is the effective spring constant.
Requires U″₀>0 and sufficiently small displacement.
The cubic correction to force grows linearly relative to the harmonic force.
Compare force corrections, not only potential terms.
Find 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.
Curvature 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.
Amplitude 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.
Stable 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.
Change one variable at a time
Make the relationship visible.
Vary mass or curvature to verify the period law. Then vary cubic strength and excursion until the anharmonicity readout is no longer small.
ANGULAR FREQUENCY1.41 rad/s
PERIOD4.44 s
CUBIC FORCE RATIO0.032
Live interpretationANGULAR FREQUENCY: 1.41 rad/s. PERIOD: 4.44 s. CUBIC FORCE RATIO: 0.032
Catch the common trap
Explain before calculating.
At equilibrium U′=0, U″=0, and U⁗>0. What follows?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
Worked calculationA 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.
- 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.
Answerω=5.66 rad s⁻¹, T=1.11 s, and the cubic force correction is 1.875% at the stated amplitude.
DerivationFor U(x)=a(x²−b²)² with a, b>0, classify the equilibria and find the small-oscillation frequency about either minimum.
- 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).
Answerx=0 is unstable; x=±b are stable; ω=√(8ab²/m) about either minimum.