Modes of oscillation

The equation of motion of bead $j$ is

$m\ddot x_j= -k(x_j-x_{j-1}-x_{j+1})$

We look for the solution in the form of

$x_j= A e^{ i(\kappa j-\omega t)}$

and we get

$-m \omega^2 = -k (2-e^{i\kappa}-e^{-i\kappa})=-2k(1-\cos \kappa)$

Here $A$ and $\kappa$ are free parameters. The parameter $\kappa$ is constrained by the fact that if we go around the circle, we need to get the same displacement: $x_j=x_{j+N}$ . Therefore $e^{i\kappa N}=1$ and $\kappa=2\pi n/N$ , where $n$ is an integer. The lowest frequency is with $n=1$

$\omega= \sqrt{\frac{2k}{m}\left(1-\cos \frac{2\pi}{N}\right)}= 6.18 sec^{-1}$

$f=\omega/2 \pi= 0.984$ Hz

We get the highest frequency with $n=5$ . For $n=0$ we get zero frequency; that corresponds to the uniform sliding of all beads in the same direction.

Suppose that $N = 2M$ is even, and consider the $N \times N$ symmetric matrix $U(N)$ , where $U(N)_{uv} \; = \; \left\{ \begin{array}{lll} 1 & \hspace{1cm} & |u-v| \equiv 1 \; \mathrm{mod}\; N \\ 0 & & \mbox{otherwise} \end{array} \right.$ Then $\mathbf{y}$ will be an eigenvector of $U(N)$ with eigenvalue $\lambda$ provided that $y_{u-1} + y_{u+1} \;= \; \lambda y_u \hspace{2cm} 2 \le u \le N-1$ and in addition provided that $y_N + y_2 \; = \; \lambda y_1 \hspace{3cm} y_{N-1} + y_1 \; = \; \lambda y_N$ If we write $\lambda = 2\cos\theta$ then the main recurrence relation is satisfied by $y_j \; = \; e^{ij\theta} \hspace{2cm} 1 \le j \le N$ and the additional two equations will also be true provided that $e^{iN\theta} = 1$ , and so provided that $\theta = \tfrac{2\alpha\pi}{N} = \tfrac{\alpha\pi}{M}$ for some integer $\alpha$ . Since $U(N)$ is a symmetric matrix, it has a basis of eigenvectors:

• The eigenvalue $\lambda_0 = 2$ has associated eigenvector $\mathbf{y}(0)$ , where $\mathbf{y}(0)_j \; = \; 1 \hspace{2cm} 1 \le j \le N$
• For any $1 \le \alpha \le M-1$ , the eigenvalue $\lambda_\alpha = 2\cos\big(\tfrac{\alpha\pi}{M}\big)$ has two associated eigenvectors $\mathbf{y}(\alpha,1)$ and $\mathbf{y}(\alpha,2)$ , where $\mathbf{y}(\alpha,1)_j \; = \; \cos\big(\tfrac{\pi i\alpha j}{M}\big) \hspace{1cm} \mathbf{y}(\alpha,2)_j \; = \; \sin\big(\tfrac{\pi i\alpha j}{M}\big) \hspace{2cm} 1 \le j \le M-1$
• The eigenvalue $\lambda_M = -2$ has associated eigenvector $\mathbf{y}(M)$ , where $\mathbf{y}(M)_j \; = \; (-1)^j \hspace{2cm} 1 \le j \le N$

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Assuming that there is a spring joining the first and $N$ th particles, the kinetic energy of the system is $T \; = \; \tfrac12m\sum_{j=1}^N \dot{x}_j^2$ and the kinetic energy of the system is $V \; = \; \tfrac12k\sum_{j=1}^{N-1} (x_j + x_{j+1})^2 + \tfrac12k(x_N - x_1)^2$ Lagrange's equations for this system can be written as $\begin{aligned} m\ddot{x}_1 + k(-x_N + 2x_1 - x_2) & = \; 0 \\ m\ddot{x}_2 + k(-x_1 + 2x_2 - x_3) & = \; 0 \\ m\ddot{x}_3 + k(-x_2 + 2x_3 - x_4) & = \; 0\\ \cdots & \hspace{0.5cm} \cdots \\ m\ddot{x}_{N-1} + k(-x_{N-2} + 2x_{N-1} + x_N)& = \; 0 \\ m\ddot{x}_N + k(-x_{N-1} + 2x_N - x_1) & = \; 0 \end{aligned}$ or, more compactly, as $m \ddot{\mathbf{x}} + 2k\mathbf{x} - kU(N)\mathbf{x} \; = \; \mathbf{0}$ We look for normal modes of oscillation of the form $\mathbf{x} \,= \, e^{i\omega t}\mathbf{y}$ for a constant vector $\mathbf{y}$ , which means we require that $-m\omega^2\mathbf{y} + 2k\mathbf{y} - kU(N)\mathbf{y} \; = \; \mathbf{0}$ or, equivalently, $U(N)\mathbf{y} \; = \; \left(2 - \tfrac{m\omega^2}{k}\right)\mathbf{y}$ so that $\mathbf{y}$ is an eigenvector of $U(N)$ with associated eigenvalue $2 - \tfrac{m\omega^2}{k}$ . Thus the possible values of $\omega$ are given by the equation $2 - \tfrac{m\omega^2}{k} = \lambda_\alpha = 2\cos\big(\tfrac{\alpha\pi}{M}\big)$ for $0 \le \alpha \le M$ , so that $\omega \; = \; 2\sin\big(\tfrac{\alpha\pi}{2M}\big)\sqrt{\tfrac{k}{m}} \; = \; 2\sin\big(\tfrac{\alpha\pi}{N}\big)\sqrt{\tfrac{k}{m}} \hspace{2cm} 0 \le \alpha \le M$ The largest possible value of $\omega$ is given by $\alpha = M$ , and is $2\sqrt{\tfrac{k}{m}}$ . The value $\omega=0$ can be ignored, since it corresponds to $y_j=1$ for all $j$ , which happens when no oscillations occur at all, and the particles just neatly rotate around the circle.

Thus the smallest possible value of $\omega$ is $2\sin\big(\tfrac{\pi}{N}\big)\sqrt{\tfrac{k}{m}}$ .

In this case, $N=10$ , $k=0.1$ , $m=0.001$ , this leads to the smallest value of $\omega$ being $20\sin\tfrac{\pi}{10} = 6.180339887$ , and the smallest possible frequency being $\boxed{0.9836316431}$ Hz.