10.6: Normal Modes

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Figure 10.4: Top view of a double mass, triple spring system.

We now consider an application of the eigenvector analysis to the coupled massspring system shown in Fig. 10.4. The position variables $x_{1}$ and $x_{2}$ are measured from the equilibrium positions of the masses. Hooke’s law states that the spring force is linearly proportional to the extension length of the spring, measured from equilibrium. By considering the extension of the spring and the sign of the force, we write Newton’s law $F=m a$ separately for each mass

$$\begin{aligned} &m \ddot{x}_{1}=-k x_{1}-K\left(x_{1}-x_{2}\right), \\ &m \ddot{x}_{2}=-k x_{2}-K\left(x_{2}-x_{1}\right) . \end{aligned} \nonumber$$

Further rewriting by collecting terms proportional to $x_{1}$ and $x_{2}$ yields

$$\begin{aligned} &m \ddot{x}_{1}=-(k+K) x_{1}+K x_{2}, \\ &m \ddot{x}_{2}=K x_{1}-(k+K) x_{2} . \end{aligned} \nonumber$$

The equations for the coupled mass-spring system form a system of two secondorder linear homogeneous odes. In matrix form, $m \ddot{x}=A x$, or explicitly,

$$m \frac{d^{2}}{d t^{2}}\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{cc} -(k+K) & K \\ K & -(k+K) \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right) . \nonumber$$

In analogy to a system of first-order equations, we try the ansatz $x=v e^{r t}$, and upon substitution into Equation $\ref{10.14}$ we obtain the eigenvalue problem Av $=\lambda \mathrm{v}$, with $\lambda=m r^{2}$. The eigenvalues are determined by solving the characteristic equation

$$\begin{aligned} 0 &=\operatorname{det}(\mathrm{A}-\lambda \mathrm{I}) \\ &=\left|\begin{array}{cc} -(k+K)-\lambda & K \\ K & -(k+K)-\lambda \end{array}\right| \\ &=(\lambda+k+K)^{2}-K^{2} \end{aligned} \nonumber$$

The solution for $\lambda$ is

$$\lambda=-k-K \pm K \nonumber$$

and the two eigenvalues are

$$\lambda_{1}=-k, \quad \lambda_{2}=-(k+2 K) . \nonumber$$

The corresponding values of $r$ in our ansatz $\mathrm{x}=\mathrm{v} e^{r t}$, with $r=\pm \sqrt{\lambda / m}$, are

$$r_{1}=i \sqrt{k / m}, \quad \bar{r}_{1}, \quad r_{2}=i \sqrt{(k+2 K) / m}, \quad \bar{r}_{2} . \nonumber$$

Since the values of $r$ are pure imaginary, we know that $x_{1}(t)$ and $x_{2}(t)$ will oscillate with angular frequencies $\omega_{1}=\operatorname{Im}\left\{r_{1}\right\}$ and $\omega_{2}=\operatorname{Im}\left\{r_{2}\right\}$, that is,

$$\omega_{1}=\sqrt{k / m}, \quad \omega_{2}=\sqrt{(k+2 K) / m} \nonumber$$

The positions of the oscillating masses in general contain time dependencies of the form $\sin \omega_{1} t, \cos \omega_{1} t$, and $\sin \omega_{2} t, \cos \omega_{2} t$.

It is of further interest to determine the eigenvectors, or so-called normal modes of oscillation, associated with the two distinct angular frequencies. With specific initial conditions proportional to an eigenvector, the mass will oscillate with a single frequency. The eigenvector with eigenvalue $\lambda_{1}$ satisfies

$$-K v_{11}+K v_{12}=0 \nonumber$$

so that $v_{11}=v_{12}$. The normal mode with frequency $\omega_{1}=\sqrt{k / m}$ thus follows a motion where $x_{1}=x_{2}$. Referring to Fig. 10.4, during this motion the center spring length does not change, which is why the frequency of oscillation is independent of $K$.

Next, we determine the eigenvector with eigenvalue $\lambda_{2}$ :

$$K v_{21}+K v_{22}=0, \nonumber$$

so that $v_{21}=-v_{22}$. The normal mode with frequency $\omega_{2}=\sqrt{(k+2 K) / m}$ thus follows a motion where $x_{1}=-x_{2}$. Again referring to Fig. 10.4, during this motion the two equal masses symmetrically push or pull against each side of the middle spring.

A general solution for $\mathbf{x}(t)$ can be constructed from the eigenvalues and eigenvectors. Our ansatz was $\mathrm{X}=\mathrm{V} e^{r t}$, and for each of two eigenvectors $\mathrm{V}$, we have a pair of complex conjugate values for $r$. Accordingly, we first apply the principle of superposition to obtain four real solutions, and then apply the principle again to obtain the general solution. With $\omega_{1}=\sqrt{k / m}$ and $\omega_{2}=\sqrt{(k+2 K) / m}$, the general solution is given by

$$\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{l} 1 \\ 1 \end{array}\right)\left(A \cos \omega_{1} t+B \sin \omega_{1} t\right)+\left(\begin{array}{r} 1 \\ -1 \end{array}\right)\left(C \cos \omega_{2} t+D \sin \omega_{2} t\right) \nonumber$$

where the now real constants $A, B, C$, and $D$ can be determined from the four independent initial conditions, $x_{1}(0), x_{2}(0), \dot{x}_{1}(0)$, and $\dot{x}_{2}(0)$.