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Topic 6 of 11

Small Oscillations

Any system near a stable equilibrium oscillates. For small displacements, the equations of motion linearize and the problem reduces to an eigenvalue problem: find the normal mode frequencies and the corresponding normal coordinates in which the modes decouple.

Key Concepts

Equilibrium & Stability

A configuration

q_0

is in equilibrium when

\partial V/\partial q_i=0

. It is stable if

V

has a local minimum (the Hessian

\partial^2V/\partial q_i\partial q_j

is positive definite). Near stable equilibrium, motion is oscillatory.

Normal Modes

Special solutions where all coordinates oscillate at the same frequency

\omega

. A system with

n

degrees of freedom has

n

normal modes with frequencies

\omega_1\le\omega_2\le\cdots\le\omega_n

. General motion is a superposition of all normal modes.

Eigenvalue Problem

Small oscillations reduce to

(K-\omega^2 M)\vec a=0

, where

K_{ij}=\partial^2 V/\partial q_i\partial q_j

is the stiffness matrix and

M_{ij}

is the mass matrix. Normal frequencies are the square roots of eigenvalues.

Normal Coordinates

A change of coordinates that diagonalizes both

K

and

M

simultaneously. In normal coordinates, each mode is independent:

\ddot{\eta}_i+\omega_i^2\eta_i=0

Key Equations

Linearized EOM

M\ddot{\vec{\eta}} + K\vec{\eta} = 0

Matrix equation for small displacements η from equilibrium; M = mass matrix, K = stiffness matrix.

Eigenvalue condition

\det(K - \omega^2 M) = 0

Secular equation; solutions give the normal mode frequencies ω².

Normal mode frequencies (double pendulum)

\omega_{\pm}^2 = \frac{g}{l}(2 \pm \sqrt{2})

Two equal-mass, equal-length pendulums coupled at the bobs (symmetric case).

Worked Example

Two Coupled Springs

Problem

Two equal masses $m=1.0$ kg are connected by three identical springs ( $k=10$ N/m): wall–mass–spring–mass–spring–mass–wall. Find the two normal mode frequencies.

Solution

Let $x_1,x_2$ be displacements. EOM: $m\ddot x_1=-2kx_1+kx_2$ , $m\ddot x_2=kx_1-2kx_2$ .

Stiffness matrix: $K=k\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$ . Mass matrix: $M=mI$ .

$\det(K-\omega^2 M)=0$ : $(2k-m\omega^2)^2-k^2=0$ .

\omega_\pm^2 = \frac{(2k\pm k)}{m} \implies \omega_- = \sqrt{k/m},\; \omega_+ = \sqrt{3k/m}

\omega_- = \sqrt{10} = 3.16\text{ rad/s},\quad \omega_+ = \sqrt{30} = 5.48\text{ rad/s}

Answer

\omega_-=3.16

rad/s (in-phase),

\omega_+=5.48

rad/s (out-of-phase).

Practice

Exercises

7 problems

1 of 7

Two equal masses ( $m=1.0$ kg) are connected by three equal springs ( $k=10$ N/m) between two walls. Find the lower normal mode frequency $\omega_-$ (in rad/s).

Your answer

rad/s

2 of 7

Same system as above. Find the higher normal mode frequency $\omega_+$ (in rad/s).

Your answer

rad/s

3 of 7

For a double pendulum with two equal masses ( $m$ ) and two equal lengths ( $l=0.50$ m), find the lower normal mode frequency $\omega_-$ (in rad/s). Use $\omega_-^2=(2-\sqrt{2})g/l$ .

Unlock Exercise 3