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Small Oscillations and Normal Modes

At a stable equilibrium, VV has a local minimum. Expanding around equilibrium (qj=qj0+ηjq_j = q_j^0 + \eta_j):

L12j,kTjkη˙jη˙k12j,kVjkηjηkL \approx \frac{1}{2}\sum_{j,k} T_{jk}\dot{\eta}_j\dot{\eta}_k - \frac{1}{2}\sum_{j,k} V_{jk}\eta_j\eta_k

Where Tjk=2Tq˙jq˙k0T_{jk} = \frac{\partial^2 T}{\partial \dot{q}_j \partial \dot{q}_k}\bigg|_0 is the (constant) Mass matrix and Vjk=2Vqjqk0V_{jk} = \frac{\partial^2 V}{\partial q_j \partial q_k}\bigg|_0 is the (constant) Stiffness matrix.

Both TjkT_{jk} and VjkV_{jk} are symmetric matrices. TT is positive definite (kinetic energy is always positive), and VV is positive definite at a stable equilibrium.

Writing the Lagrangian in matrix form:

L=12η˙TTη˙12ηTVηL = \frac{1}{2}\dot{\boldsymbol{\eta}}^T \mathbf{T}\, \dot{\boldsymbol{\eta}} - \frac{1}{2}\boldsymbol{\eta}^T \mathbf{V}\, \boldsymbol{\eta}

The Euler-Lagrange equations become:

Tη¨+Vη=0\mathbf{T}\, \ddot{\boldsymbol{\eta}} + \mathbf{V}\, \boldsymbol{\eta} = \mathbf{0}

Assuming solutions of the form ηj=ajeiωt\eta_j = a_j e^{i\omega t}The eigenvalue problem is:

(Vω2T)a=0(\mathbf{V} - \omega^2 \mathbf{T})\mathbf{a} = \mathbf{0}

The secular equation (characteristic equation) is:

det(Vω2T)=0\det(\mathbf{V} - \omega^2 \mathbf{T}) = 0

This is a polynomial of degree nn in ω2\omega^2 whose roots are the squared normal mode frequencies ωα2\omega_\alpha^2.

Theorem 7.1. For a stable system, all normal mode frequencies are real and positive. The normal Modes are orthogonal with respect to both T\mathbf{T} and V\mathbf{V}.

Proof. Since T\mathbf{T} is positive definite, we can write T=LLT\mathbf{T} = \mathbf{L}\mathbf{L}^T (Cholesky decomposition). Defining ξ=LTa\boldsymbol{\xi} = \mathbf{L}^T\mathbf{a} and W=L1VLT\mathbf{W} = \mathbf{L}^{-1}\mathbf{V}\mathbf{L}^{-T}The eigenvalue problem becomes Wξ=ω2ξ\mathbf{W}\boldsymbol{\xi} = \omega^2\boldsymbol{\xi}. Since W\mathbf{W} is symmetric and V\mathbf{V} is positive definite, all eigenvalues ω2\omega^2 are real and positive. Orthogonality follows from the symmetry of W\mathbf{W}. \blacksquare

Theorem 7.2. The normal mode vectors a(α)\mathbf{a}^{(\alpha)} satisfy:

j,kaj(α)Tjkak(β)=Tαδαβ,j,kaj(α)Vjkak(β)=ωα2Tαδαβ\sum_{j,k} a_j^{(\alpha)} T_{jk} a_k^{(\beta)} = T_\alpha\, \delta_{\alpha\beta}, \quad \sum_{j,k} a_j^{(\alpha)} V_{jk} a_k^{(\beta)} = \omega_\alpha^2 T_\alpha\, \delta_{\alpha\beta}

Where TαT_\alpha is a normalisation constant.

This allows us to expand any motion as a superposition of normal modes.

Problem. Two identical simple pendulums of length ll and mass mm are coupled by a spring of constant kk connecting the bobs. Find the normal modes and frequencies.

Solution

Let θ1\theta_1 and θ2\theta_2 be the small angles from vertical. The separation between bobs (to first order) is approximately l(θ2θ1)l(\theta_2 - \theta_1) So the spring potential energy is 12kl2(θ2θ1)2\frac{1}{2}k l^2(\theta_2 - \theta_1)^2.

Kinetic energy:

T=12ml2θ˙12+12ml2θ˙22T = \frac{1}{2}ml^2\dot{\theta}_1^2 + \frac{1}{2}ml^2\dot{\theta}_2^2

Potential energy (gravitational + spring):

V=12mglθ12+12mglθ22+12kl2(θ2θ1)2V = \frac{1}{2}mgl\theta_1^2 + \frac{1}{2}mgl\theta_2^2 + \frac{1}{2}kl^2(\theta_2 - \theta_1)^2

=12(mgl+kl2)θ12+12(mgl+kl2)θ22kl2θ1θ2= \frac{1}{2}(mgl + kl^2)\theta_1^2 + \frac{1}{2}(mgl + kl^2)\theta_2^2 - kl^2\theta_1\theta_2

The mass matrix and stiffness matrix are:

T=ml2(1001),V=(mgl+kl2kl2kl2mgl+kl2)\mathbf{T} = ml^2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad \mathbf{V} = \begin{pmatrix} mgl + kl^2 & -kl^2 \\ -kl^2 & mgl + kl^2 \end{pmatrix}

The secular equation det(Vω2T)=0\det(\mathbf{V} - \omega^2\mathbf{T}) = 0:

(mgl+kl2ml2ω2)2(kl2)2=0(mgl + kl^2 - ml^2\omega^2)^2 - (kl^2)^2 = 0

mgl+kl2ml2ω2=±kl2mgl + kl^2 - ml^2\omega^2 = \pm kl^2

Mode 1 (++ sign): ω12=g/l\omega_1^2 = g/l. The eigenvector is (1,1)(1, 1): both pendulums swing in phase, the spring is not stretched.

Mode 2 (- sign): ω22=g/l+2k/m\omega_2^2 = g/l + 2k/m. The eigenvector is (1,1)(1, -1): the pendulums swing out of phase, the spring is maximally stretched.

The general solution is a superposition:

θ1(t)=Acos(ω1t+ϕ1)+Bcos(ω2t+ϕ2)\theta_1(t) = A\cos(\omega_1 t + \phi_1) + B\cos(\omega_2 t + \phi_2)

θ2(t)=Acos(ω1t+ϕ1)Bcos(ω2t+ϕ2)\theta_2(t) = A\cos(\omega_1 t + \phi_1) - B\cos(\omega_2 t + \phi_2)

If initially only θ1\theta_1 is displaced and θ2=0\theta_2 = 0 with zero velocities, energy slowly transfers between the two pendulums --- the classic beat phenomenon.

\blacksquare

7.6 Worked Example: Double Pendulum (Small Oscillations)

Section titled “7.6 Worked Example: Double Pendulum (Small Oscillations)”

For two equal masses mm on massless rods of length ll:

The kinetic and potential energy matrices (to second order) give the eigenvalue problem with solutions ω1=g/l(22)1/2\omega_1 = \sqrt{g/l}(2 - \sqrt{2})^{1/2} and ω2=g/l(2+2)1/2\omega_2 = \sqrt{g/l}(2 + \sqrt{2})^{1/2}.

The corresponding normal modes are:

  • Mode 1: both pendulums swing in the same direction (in phase).
  • Mode 2: the pendulums swing in opposite directions (out of phase).
  • Assuming the mass matrix is always diagonal; in generalised coordinates it can have off-diagonal terms.
  • Forgetting that both TT and VV must be positive definite for all eigenvalues to be real and positive.
  • Confusing the secular equation det(Vω2T)=0\det(\mathbf{V} - \omega^2\mathbf{T}) = 0 with det(VλI)=0\det(\mathbf{V} - \lambda\mathbf{I}) = 0; the latter only applies when T=IT = I.
  • Neglecting to normalise eigenvectors with respect to the mass matrix, which leads to incorrect mode superpositions.
  • Misidentifying degenerate modes: when ωα=ωβ\omega_\alpha = \omega_\beta, any linear combination of the corresponding eigenvectors is also a normal mode.
  • Assuming that small oscillation analysis captures the full dynamics; it only describes motion near equilibrium and breaks down for large amplitudes.
flowchart TD
A[7_Small Oscillations And Normal Modes] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]

Normal modes are like the natural hum of a system. When you pluck a guitar string, it vibrates at specific frequencies determined by its length and tension. Push a swing at just the right moment and it builds up; push at the wrong frequency and it barely moves. Small oscillation analysis decomposes any complex motion near equilibrium into these natural frequencies. Think of coupled pendulums: energy sloshes back and forth between them like water in connected tanks, creating the beat phenomenon. Each normal mode is a pattern where every part of the system moves in phase with the same frequency. The secular equation is like tuning a radio: it tells you exactly which frequencies the system can sustain. Any motion is just a superposition of these natural frequencies, like a chord on a piano is a superposition of individual notes.

ResultStatement
Secular equationdet(Vω2T)=0\det(\mathbf{V} - \omega^2\mathbf{T}) = 0 gives normal mode frequencies
Orthogonalitya(α)TTa(β)=Tαδαβ\mathbf{a}^{(\alpha)T}\mathbf{T}\mathbf{a}^{(\beta)} = T_\alpha\delta_{\alpha\beta}
Stability conditionTT and VV positive definite     \implies all ωα2>0\omega_\alpha^2 > 0
Beat phenomenonTwo pendulums with close frequencies exchange energy periodically
DegeneracyWhen ωα=ωβ\omega_\alpha = \omega_\beta, eigenvectors are not uniquely determined

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