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Small Oscillations & Normal Modes (IPhO Mechanics)

IPhO mechanics lesson on linearising about equilibrium, small oscillations, and normal modes in coupled systems.

  • International Physics Olympiad preparation

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Near a stable equilibrium, a nonlinear system becomes a linear one. That is why many olympiad mechanics problems reduce to “find equilibrium, expand to quadratic order, read off frequencies and mode shapes”.

Learning objectives

By the end, you should be able to:

  • IPHO-M-OSC-01: locate the actual equilibrium, test stability and retain the leading restoring term;
  • IPHO-M-OSC-02: find the frequencies and amplitude ratios of a two-coordinate linear system; and
  • interpret symmetry, zero-frequency and decoupling limits physically.

The official syllabus names stable and unstable equilibria and the single harmonic oscillator. Coupled normal modes are a Mini Physics training extension that connects those official ideas to superposition, beats and linear systems.

Symmetric and antisymmetric normal modes of two coupled massesTwo identical masses between three springs move together in the lower-frequency symmetric mode, leaving the middle spring unchanged. They move oppositely in the higher-frequency antisymmetric mode, strongly changing the middle spring.Symmetric mode (1, 1)Antisymmetric mode (1, −1)middle spring unchangedmiddle spring changes mostω₁ = √(k/m)ω₂ = √(3k/m)
Scroll diagram horizontally to read all labels.
Symmetry gives the mode shapes before any determinant: (1, 1) leaves the coupling spring unchanged; (1, −1) stretches or compresses it.
Prerequisites (already covered elsewhere)

1. Definitions

  • Equilibrium: configuration q = q₀ where net generalized force is zero. For conservative 1D systems: V'(q₀) = 0.
  • Stable equilibrium: small displacements produce restoring forces. For conservative 1D systems: V''(q₀) is positive.
  • Small oscillations / linearisation: write q(t) = q₀ + η(t) and keep only lowest nonzero order terms (equivalently: keep T and V to quadratic order in η and η dot).
  • Normal mode: motion where all coordinates oscillate sinusoidally at one frequency with fixed amplitude ratios.
  • Mass matrix M and stiffness matrix K (after shifting equilibrium to q = 0):
    T = 1/2 q dot ^T M q dot, V = 1/2 q^T K q.
  • Eigenfrequency condition: det(K-ω² M) = 0.
  • Beats: slow envelope due to superposition of two close frequencies (often energy exchange in weakly coupled oscillators).

2. Key ideas

  • Find equilibrium first, then expand. If you expand about the wrong point, you will get the wrong frequency.
  • For one coordinate:
    ω = square root of (k/M), k = V''(q₀),
    with T ≈ 1/2 Mη dot ² and V ≈ 1/2 kη².
  • For coupled systems, symmetry often gives the mode shapes immediately (in-phase and out-of-phase for identical parts).
  • If there is a cyclic coordinate, it typically produces a zero-frequency mode (a free rigid motion).
Quadratic-form recipe (multi-DOF)
  1. Choose small coordinates q about equilibrium.
  2. Expand T and V to quadratic order.
  3. Read off M and K.
  4. Solve det(K-ω² M) = 0 for ω.
  5. Solve (K-ω² M)A = 0 for the mode shape A.

3. Detailed explanation

3.1 Why everything becomes SHM (one DOF)

Taylor expand the potential about equilibrium:

V(q) = V(q₀) + V'(q₀)η + 1/2 V''(q₀)η² + …

At equilibrium V'(q₀) = 0, so the leading restoring term is quadratic. If near equilibrium the kinetic energy is T ≈ 1/2 Mη dot ², then:

Mη double dot + V''(q₀)η = 0.

3.2 Normal modes as an eigenvalue problem

For small oscillations in multiple coordinates:

Mq double dot + Kq = 0.

Try q = A cos(ω t):

(K-ω² M)A = 0.

Nontrivial A requires det(K-ω² M) = 0.

3.3 Symmetry shortcuts

If the system is invariant under swapping labels 1 and 2, then the symmetric vector (1,1) and antisymmetric vector (1,-1) are guaranteed mode shapes. Plug those in to find ω without determinants.

4. Common mistakes

  • Expanding about the wrong equilibrium.
  • Dropping the quadratic term you actually need (for a stable equilibrium, the linear term is zero).
  • Missing factors of 2 in spring extensions (especially for antisymmetric motion).
  • Assuming M is the identity when your coordinates are angles or coupled distances.

5. Problem-solving tips

  1. Write the equilibrium condition in one line before expanding.
  2. State the symmetry guess for mode shapes before heavy algebra.
  3. Check a decoupling limit (coupling constant to zero should recover independent oscillators).
  4. A zero-frequency result is usually a symmetry, not a mistake.

6. Worked examples

1) Physical pendulum: uniform rod pivoted at one end

A uniform rod of length L and mass m is pivoted at one end and oscillates in a vertical plane with small angle θ about the lowest position.

About the pivot, small-angle torque is τ ≈ -mgd θ where d = L/2 is the centre-of-mass distance from the pivot.

Rotational equation:

Iₚᵢᵥₒₜθ double dot + mgd θ = 0.

For a uniform rod about one end:

Iₚᵢᵥₒₜ = 1/3 mL².

Therefore

ω² = (mg(L/2))/((1/3)mL²) = 3g/2L, T = 2π square root of (2L/3g) .
2) Two masses with three springs: normal modes by symmetry

Two equal masses m move on a frictionless line. They are connected as: wall - spring k - mass 1 - spring k - mass 2 - spring k - wall.

Let displacements from equilibrium be x₁,x₂.

Forces give:

mx double dot ₁ = -2kx₁ + kx₂, mx double dot ₂ = kx₁ - 2kx₂.

Symmetric mode (x₁ = x₂ = x):

mx double dot = -kx ⇒ ω₁ = square root of (k/m) .

(The middle spring is not stretched.)

Antisymmetric mode (x₁ = -x₂ = x):

mx double dot = -3kx ⇒ ω₂ = square root of (3k/m) .

(The middle spring stretches strongly and increases the restoring force.)

7. Extension

1) Weak coupling and beats: estimate the energy-swap time

Two equal masses m are each attached to a wall by a spring k, and the masses are coupled by a weak spring k_c.

Symmetry gives the mode frequencies:

  • In-phase: coupling spring not stretched, so ω₁ = square root of (k/m) = ω₀.
  • Out-of-phase: effective stiffness is k + 2k_c, so ω₂ = square root of ((k + 2k_c)/m).

For k_c much smaller than k:

ω₂ = ω₀ square root of (1 + 2k_c/k) ≈ ω₀(1 + k_c/k).

So the splitting is

Δω ≈ ω₂-ω₁ ≈ ω₀k_c/k.

A displacement that excites both modes produces an amplitude beat period

T_beat ≈ 2π/Δω ≈ (2π/ω₀)k/k_c.

Starting with one oscillator displaced and the other at rest, the first nearly complete energy transfer occurs after approximately π/Δω, half of this beat period. Very weak coupling still transfers energy, but on a long timescale.

8. Practice and evidence

Practice
  • Do one problem where equilibrium is nontrivial (find it, then linearise).
  • Do one 2-DOF normal-mode problem and solve it twice: symmetry guess, then det(K-ω² M) = 0.
  • Treat IPHO-M-OSC-01/02 as passed only if you state the equilibrium, mode shapes and one limiting-case check.
  • Then continue via the IPhO Mechanics Hub.
Syllabus and review details

No official syllabus alignment is listed for this lesson.

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