Small Oscillations and Normal Modes

7.1 Equilibrium and Small Oscillations

At a stable equilibrium, $V$ has a local minimum. Expanding around equilibrium ( $q_j = q_j^0 + \eta_j$ ):

$L \approx \frac{1}{2}\sum_{j,k} T_{jk}\dot{\eta}_j\dot{\eta}_k - \frac{1}{2}\sum_{j,k} V_{jk}\eta_j\eta_k$

Where $T_{jk} = \frac{\partial^2 T}{\partial \dot{q}_j \partial \dot{q}_k}\bigg|_0$ is the (constant) Mass matrix and $V_{jk} = \frac{\partial^2 V}{\partial q_j \partial q_k}\bigg|_0$ is the (constant) Stiffness matrix.

Both $T_{jk}$ and $V_{jk}$ are symmetric matrices. $T$ is positive definite (kinetic energy is always positive), and $V$ is positive definite at a stable equilibrium.

7.2 Matrix Formulation

Writing the Lagrangian in matrix form:

$L = \frac{1}{2}\dot{\boldsymbol{\eta}}^T \mathbf{T}\, \dot{\boldsymbol{\eta}} - \frac{1}{2}\boldsymbol{\eta}^T \mathbf{V}\, \boldsymbol{\eta}$

The Euler-Lagrange equations become:

$\mathbf{T}\, \ddot{\boldsymbol{\eta}} + \mathbf{V}\, \boldsymbol{\eta} = \mathbf{0}$

7.3 Normal Modes and the Secular Equation

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

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

The secular equation (characteristic equation) is:

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

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

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

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

7.4 Orthogonality of Normal Modes

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

$\sum_{j,k} a_j^{(\alpha)} T_{jk} a_k^{(\beta)} = T_\alpha\, \delta_{\alpha\beta}, \quad \sum_{j,k} a_j^{(\alpha)} V_{jk} a_k^{(\beta)} = \omega_\alpha^2 T_\alpha\, \delta_{\alpha\beta}$

Where $T_\alpha$ is a normalisation constant.

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

7.5 Worked Example: Coupled Pendulums

Problem. Two identical simple pendulums of length $l$ and mass $m$ are coupled by a spring of constant $k$ connecting the bobs. Find the normal modes and frequencies.

Solution

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

Kinetic energy:

$T = \frac{1}{2}ml^2\dot{\theta}_1^2 + \frac{1}{2}ml^2\dot{\theta}_2^2$

Potential energy (gravitational + spring):

$V = \frac{1}{2}mgl\theta_1^2 + \frac{1}{2}mgl\theta_2^2 + \frac{1}{2}kl^2(\theta_2 - \theta_1)^2$

$= \frac{1}{2}(mgl + kl^2)\theta_1^2 + \frac{1}{2}(mgl + kl^2)\theta_2^2 - kl^2\theta_1\theta_2$

The mass matrix and stiffness matrix are:

$\mathbf{T} = ml^2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad \mathbf{V} = \begin{pmatrix} mgl + kl^2 & -kl^2 \\ -kl^2 & mgl + kl^2 \end{pmatrix}$

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

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

$mgl + kl^2 - ml^2\omega^2 = \pm kl^2$

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

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

The general solution is a superposition:

$\theta_1(t) = A\cos(\omega_1 t + \phi_1) + B\cos(\omega_2 t + \phi_2)$

$\theta_2(t) = A\cos(\omega_1 t + \phi_1) - B\cos(\omega_2 t + \phi_2)$

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

$\blacksquare$

7.6 Worked Example: Double Pendulum (Small Oscillations)

For two equal masses $m$ on massless rods of length $l$ :

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

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).

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