General Mechanics/Coupled Oscillators

We often encounter systems which contain multiple harmonic oscillators, such as this:

two identical masses, m, the first attached to a wall by a spring with constant k, and the second attached to the first by another, identical spring.

If the springs weren't linked they'd both vibrate at the same frequency, ω=√(k/m). Linking the springs changes this.

To find out how the linked system behaves, we will start with the Lagrangian, using the displacements of the masses, x₁ and x₂, as our coordinates.

A moment's inspection of the system shows

$T={\frac {1}{2}}m\left({\dot {x}}_{1}^{2}+{\dot {x}}_{2}^{2}\right)\quad V={\frac {1}{2}}k\left(x_{1}^{2}+(x_{2}-x_{1})^{2}\right)$

so, using ω²=k/m,

$L={\frac {1}{2}}m\left({\dot {x}}_{1}^{2}+{\dot {x}}_{2}^{2}\right)-{\frac {1}{2}}m\omega ^{2}\left(x_{1}^{2}+(x_{2}-x_{1})^{2}\right)$

The equations of motion immediately follow.

${\begin{matrix}{\ddot {x}}_{1}&=&-\omega ^{2}(2x_{1}-x_{2})\\{\ddot {x}}_{2}&=&-\omega ^{2}(x_{2}-x_{1})\end{matrix}}\quad (1)$

To solve these equations we try a solution in trig functions

$x_{1}=A_{1}\sin \Omega t\quad x_{2}=A_{2}\sin \Omega t$

Substituting this into (1) gives

${\begin{matrix}-\Omega ^{2}A_{1}&+&\omega ^{2}(2A_{1}-A_{2})&=&0\\-\Omega ^{2}A_{2}&+&\omega ^{2}(A_{2}-A_{1})&=&0\\\end{matrix}}$

We would get the same equations from any trig function solution of the same frequency.

Gathering the coefficients of A₁ and A₂ together lets rewrite the last equation as

${\begin{pmatrix}2\omega ^{2}-\Omega ^{2}&-\omega ^{2}\\-\omega ^{2}&\omega ^{2}-\Omega ^{2}\end{pmatrix}}{\begin{pmatrix}A_{1}\\A_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}\quad (2)$

We can only solve this equation if the determinant of the matrix is zero.

$\Omega ^{4}-3\omega ^{2}\Omega ^{2}+\omega ^{4}=0$

The solutions are

$\Omega _{\pm }^{2}={\frac {3\pm {\sqrt {5}}}{2}}\omega ^{2}$

so the combined system has two natural frequencies, one lower and one higher than the natural frequency of the individual springs. This is typical.

We can also calculate the ratio of A₁ and A₂ from (2). Dividing by A₂ gives

${\frac {A_{1}}{A_{2}}}=1-{\frac {\Omega _{\pm }^{2}}{\omega ^{2}}}$

• For the lower frequency, Ω_-, which is less than ω, the ratio is positive, so the two masses move in the same direction with different amplitudes. They are said to be in phase.

• For the higher frequency, Ω₊, which is more than ω, the ratio is negative, so the two masses move in the opposite direction with different amplitudes. They are said to be in antiphase.

This behaviour is typical when pairs of harmonic oscillators are coupled.

The same approach can also be used for systems with more than two particles.