14.2: Two Coupled Linear Oscillators

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.

Figure $\PageIndex{1}$: Two coupled linear oscillators. The equilibrium spring-lengths are $l$ for the outer springs and $l^{\prime}$ for the coupling spring. The displacement from the stable locations are given by $x_1$ and $x_2$. The separation between the two masses is $r$ and the location of the center-of-mass is $R_{cm}$.

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 $\PageIndex{2}$: Displacement of each of two coupled linear harmonic oscillators with $\kappa = 4$ and $\kappa^{\prime} = 1$ in relative units.

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.