# 14.2: Two Coupled Linear Oscillators

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

Consider the two-coupled linear oscillator, shown in Figure $$\PageIndex{1}$$, which comprises two identical masses each connected to fixed locations by identical springs having a force constant $$\kappa$$. A spring with force constant $$\kappa^{\prime}$$ couples the two oscillators. The equilibrium lengths of the outer two springs are $$l$$ while that of the coupling spring is $$l^{\prime}$$. The problem is simplified by restricting the motion to be along the line connecting the masses and assuming fixed endpoints. The small displacements of $$m_1$$ and $$m_2$$ are taken to be $$x_1$$ and $$x_2$$ with respect to the equilibrium positions $$l$$ and $$l + l^{\prime}$$ respectively. The restoring force on $$m_1$$ is $$−\kappa x_1−\kappa^{\prime} (x_1 − x_2)$$ while the restoring force on $$m_2$$ is $$−\kappa x_2 − \kappa^{\prime} (x_2 − x_1)$$. This coupled double-oscillator system exhibits basic features of coupled linear oscillator systems.

Assuming $$m_1 = m_2 = m$$, then the equations of motion are

\begin{align} m\ddot{x}_1 + (\kappa + \kappa^{\prime} ) x_1 − \kappa^{\prime} x_2 = 0 \label{14.1} \\ \notag m\ddot{x}_2 + (\kappa + \kappa^{\prime} ) x_2 − \kappa^{\prime} x_1 = 0 \notag\end{align}

Assume that the motion for these coupled equations is oscillatory with a solution of the form

\begin{align} x_1 = B_1 e^{i\omega t} \label{14.2}\\ x_2 = B_2e^{i\omega t} \notag\notag\end{align}

where the constants $$B$$ may be complex to take into account both the magnitude and phase. Substituting these possible solutions into the equations of motion gives

\begin{align} −m\omega^2 B_1 e^{i\omega t} + (\kappa + \kappa^{\prime} ) B_1e^{i\omega t} − \kappa^{\prime} B_2e^{i\omega t} = 0 \label{14.3}\\ −m\omega^2B_2 e^{i\omega t} + (\kappa + \kappa^{\prime} ) B_2e^{i\omega t} − \kappa^{\prime} B_1 e^{i\omega t} = 0 \notag\end{align}

Collecting terms, and cancelling the common exponential factor, gives

\begin{align} (\kappa + \kappa^{\prime} − m\omega^2) B_1 − \kappa^{\prime} B_2 = 0 \label{14.4} \\ ( \kappa + \kappa^{\prime} − m\omega^2 ) B_2 − \kappa^{\prime} B_1 = 0 \notag\end{align}

The existence of a non-trivial solution of these two simultaneous equations requires that the determinant of the coefficients of $$B_1$$ and $$B_2$$ must vanish, that is

$\begin{vmatrix} \kappa + \kappa^{\prime} − m\omega^2 & −\kappa^{\prime} \\ −\kappa^{\prime} & \kappa + \kappa^{\prime} − m\omega^2 \end{vmatrix} = 0 \label{14.5}$

The expansion of this secular determinant yields

$( \kappa + \kappa^{\prime} − m\omega^2 )^2 − \kappa^{\prime 2} = 0 \label{14.6}$

Solving for $$\omega$$ gives

$\omega = \sqrt{\frac{\kappa + \kappa^{\prime} \pm \kappa^{\prime}}{m}} \label{14.7}$

That is, there are two characteristic frequencies (or eigenfrequencies) for the system

$\omega_1 = \sqrt{\frac{\kappa + 2\kappa^{\prime} }{m}} \label{14.8}$

$\omega_2 = \sqrt{\frac{\kappa}{m}} \label{14.9}$

Since superposition applies for these linear equations, then the general solution can be written as a sum of the terms that account for the two possible values of $$\omega$$.

Figure $$\PageIndex{2}$$ shows the solutions for a case where $$\kappa = 4$$ and $$\kappa^{\prime} = 1$$, in arbitrary units, with the initial condition that $$x_2 = D$$, and $$x_1 = \dot{x}_1 = \dot{x}_2 = 0$$. The two characteristic frequencies are $$\omega_1 = \sqrt{\frac{6}{m}}$$ and $$\omega_2 = \sqrt{\frac{4}{m}}$$. The characteristic beats phenomenon is exhibited where the envelope over one complete cycle of the low frequency encompasses several higher frequency oscillations. That is, the solution is

$x_{2}(t)=\frac{D}{4}\left[e^{i \omega_{1} t}+e^{-i \omega_{1} t}+e^{i \omega_{2} t}+e^{-i \omega_{2} t}\right]=D \cos \left[\left(\frac{\omega_{1}+\omega_{2}}{2}\right) t\right] \cos \left[\left(\frac{\omega_{1}-\omega_{2}}{2}\right) t\right]$

while

$x_{1}(t)=\frac{D}{4}\left[e^{i \omega_{1} t}+e^{-i \omega_{1} t}-e^{i \omega_{2} t}-e^{-i \omega_{2} t}\right]=D \sin \left[\left(\frac{\omega_{1}+\omega_{2}}{2}\right) t\right] \sin \left[\left(\frac{\omega_{1}-\omega_{2}}{2}\right) t\right]$

The energy in the two-coupled oscillators flows back and forth between the coupled oscillators as illustrated in Figure $$\PageIndex{2}$$.

A better understanding of the energy flow occurring between the two coupled oscillators is given by using a $$(x_1, x_2)$$ configuration-space plot, shown in Figure $$14.3.1$$. The flow of energy occurring between the two coupled oscillators can be represented by choosing normal-mode coordinates $$\eta_1$$ and $$\eta_2$$ that are rotated by $$45^{\circ}$$ with respect to the spatial coordinates $$(x_1, x_2)$$. These normal-mode coordinates $$(\eta_1, \eta_2)$$ correspond to the two normal modes of the coupled double-oscillator system.

This page titled 14.2: Two Coupled Linear Oscillators is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Douglas Cline via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.