Coupled Oscillators
Connected Masses ($m(2)$ and $k(3)$)
Two masses, $m_1$ and $m_2$ connected via three springs.
Two masses can move horizontally. They are connected to each other and the walls by 3 springs as shown.
consider the forces on mass 1:
$$ \begin{aligned} F_\textrm{on mass 1} & = -k_1 x_1 + k_2 \left(x_2-x_1 \right) \\ & = -\left(k_1+k_2 \right)x_1 + k_2 x_2 \end{aligned} $$
consider the forces on mass 2:
$$ \begin{aligned} F_\textrm{on mass 2} & = -k_3 x_2 - k_2 \left(x_2-x_1 \right) \\ & = k_2 x_1 - \left(k_2+k_3 \right)x_2 \end{aligned} $$
This leads to 2 eqs. of motion.
$$ \begin{aligned} m_1 \ddot{x}_1 & = -\left(k_1+k_2 \right)x_1 + k_2 x_2 \\ m_2 \ddot{x}_2 & = k_2 x_1 - \left(k_2+k_3 \right)x_2 \end{aligned} $$
Matrix Form
With the following definitions: \begin{equation} \mathbf {x} = \begin{bmatrix}x_1 \\ x_2 \end{bmatrix} \end{equation} \begin{equation} \mathbf {M} = \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix} \end{equation} \begin{equation} \mathbf {K} = \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 + k_3 \end{bmatrix} \end{equation}
We can write: \begin{equation} \mathbf{M} \ddot {\mathbf{x}} = - \mathbf{K} \mathbf{x} \end{equation}
How to solve this?
Assume a complex solution: $$ \begin{aligned} z_1 & = x_1(t) + i y_1 (t) = \alpha_1 e^{i \left(\omega t - \delta_1 \right)} \\ & = \alpha_1 e^{-i \delta_1} e^{i \omega t} = a_1 e^{i \omega t} \end{aligned} $$ $$ \begin{aligned} z_2 & = \ldots = a_2 e^{i \omega t} \end{aligned} $$
\begin{equation} \mathbf{z}(t) = \begin{bmatrix}z_1(t) \\ z_2(t) \end{bmatrix} = \begin{bmatrix}a_1 \\ a_2 \end{bmatrix}e^{i\omega t} = \mathbf{a} e^{i \omega t} \end{equation}
note: $ \mathbf{x}(t) = \textrm{Re}\; \mathbf{z}(t)$
Start with the equation of motion: \begin{equation} \mathbf{M} \ddot {\mathbf{x}} = - \mathbf{K} \mathbf{x} \end{equation}
Replace $\mathbf{x}$ with our complex $\mathbf{z}$ and we obtain: \begin{equation} - \omega^2 \mathbf{M} \; \mathbf{a}\; e^{i \omega t} = - \mathbf{K}\; \mathbf{a} \; e^{i \omega t} \end{equation}
Canceling the exponential terms and re-arranging: \begin{equation} \left(\mathbf{K} - \omega^2 \mathbf{M} \right) \mathbf{a} = 0 \end{equation}
Now we ask, what do solutions to this equation look like?
Clearly, $\mathbf{a}$ could be zero, then we have the so-called trivial solution, i.e. nothing moves. Boring.
But if : \begin{equation} \textrm{det}\left(\mathbf{K - \omega^2 \mathbf{M}} \right) =0 \end{equation} then we might have more interesting solutions.
Equal $m$ and $k$
Let's try a simple case, where the two masses are equal and the spring constants are all the same: \begin{equation} \mathbf {M} = \begin{bmatrix} m & 0 \\ 0 & m \end{bmatrix} \;\;\textrm{and}\;\; \mathbf {K} = \begin{bmatrix} 2k & -k \\ -k & 2k \end{bmatrix} \end{equation}
Our Matrix is thus: \begin{equation} \left(\mathbf{K - \omega^2 \mathbf{M}} \right) = \begin{bmatrix} 2 k - m \omega^2 & -k \\ -k & 2 k - m \omega^2 \end{bmatrix} \end{equation}
Taking the determinant of this yields: $$ \begin{align} \textrm{det}\left(\mathbf{K - \omega^2 \mathbf{M}} \right) & = \left(2 k - m \omega^2 \right)^2 - k^2\\ & = \left(k - m \omega^2 \right)\left(3 k - m \omega^2 \right) \end{align} $$
There will be two frequencies that create a 0 \begin{equation} \left(k - m \omega^2 \right)\left(3 k - m \omega^2 \right) \end{equation}
\begin{equation} \omega = \sqrt{\frac{k}{m}} = \omega_1 \;\; \textrm{and}\;\; \omega = \sqrt{\frac{3k}{m}} = \omega_2 \end{equation}
Thus we have two non-trivial solutions. Our next step is to find the motions that occur at these frequencies
Normal modes
First Normal Mode
\begin{equation} \left(\mathbf{K} - \omega^2 \mathbf{M} \right)\mathbf{a} = 0 \end{equation}
Start with $\omega_1 = \sqrt{k/m}$
\begin{equation} \left(\mathbf{K} - \omega_1^2 \mathbf{M} \right) = \begin{bmatrix} k & -k \\ -k & k \end{bmatrix} \end{equation}
\begin{equation} \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \end{bmatrix}= 0 \end{equation} which leads to two equations: $$ \begin{aligned} a_1 - a_2 & = 0 \\ -a_1 +a_2 & = 0 \end{aligned} $$ that both imply $a_1 = a_2$
if we say: $$a_1 = a_2 = A e^{-i \delta}$$ then we can write our complex solution $\mathbf{z}(t)$ as: \begin{equation} \mathbf{z}(t) = \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} e^{i \omega_1 t} = \begin{bmatrix} A \\ A \end{bmatrix} e^{i \left(\omega_1 t - \delta \right)} \end{equation} The actual motion of the masses is then described by the real part of the above: \begin{equation} \mathbf{x}(t) = \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} = \begin{bmatrix} A \\ A \end{bmatrix} \cos{\left(\omega_1 t - \delta \right)} \end{equation}
$$ \begin{aligned} x_1(t) & = A \cos \left(\omega_1 t - \delta \right) \\ x_2(t) & = A \cos \left(\omega_1 t - \delta \right) \end{aligned} $$
Second Normal Mode
Now Choose $\omega_2 = \sqrt{3k/m}$
\begin{equation} \left(\mathbf{K} - \omega_2^2 \mathbf{M} \right) = \begin{bmatrix} -k & -k \\ -k & -k \end{bmatrix} \end{equation}
\begin{equation} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \end{bmatrix}= 0 \end{equation} which leads to two equations: $$ \begin{aligned} a_1 + a_2 & = 0 \\ \end{aligned} $$ which implies $a_1 = - a_2$
if we say: $$a_1 = - a_2 = A e^{-i \delta}$$ then we can write our complex solution $\mathbf{z}(t)$ as: \begin{equation} \mathbf{z}(t) = \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} e^{i \omega_2 t} = \begin{bmatrix} A \\ -A \end{bmatrix} e^{i \left(\omega_2 t - \delta \right)} \end{equation} The motion of the masses is then described by the real part of the above: \begin{equation} \mathbf{x}(t) = \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix} = \begin{bmatrix} A \\ - A \end{bmatrix} \cos{\left(\omega_2 t - \delta \right)} \end{equation}
$$ \begin{aligned} x_1(t) & = A \cos \left(\omega_2 t - \delta \right) \\ x_2(t) & = -A \cos \left(\omega_2 t - \delta \right) \end{aligned} $$
Two Normal Modes
General Solution
$$ \begin{aligned} \mathbf{x}(t) & = A_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} \cos{\left(\omega_1 t - \delta_1 \right)} \\ \mathbf{x}(t) & = A_2 \begin{bmatrix} 1 \\ -1 \end{bmatrix} \cos{\left(\omega_2 t - \delta_2 \right)} \\ \end{aligned} $$
An arbitrary plot for the general solution.
Physical Example
A Carbon Dioxide Molecule
Larger systems
More masses?
Now, let's try it with the Lagrangian formulation: \begin{equation} T = \frac{1}{2}m(\dot{x}_1^2 + \dot{x}_2^2 + \dot{x}_3^2) \end{equation} \begin{equation} U = \frac{1}{2}k x_1^2 + \frac{1}{2}k(x_2-x_1)^2 + \frac{1}{2}k(x_3-x_2)^2 + \frac{1}{2}k x_3^2 \end{equation} Thus our $L$ is: \begin{equation} L = \frac{1}{2}m(\dot{x}_1^2 + \dot{x}_2^2 + \dot{x}_2^2) - \frac{1}{2}k x_1^2 - \frac{1}{2}k(x_2-x_1)^2 - \frac{1}{2}k(x_3-x_2)^2 - \frac{1}{2}k x_3^2 \end{equation}
After solving the Euler-Lagrange for this system: $$ \begin{aligned} m \ddot{x}_1 & = -k x_1 + k (x_2 - x_1) \\ m \ddot{x}_2 & = -k(x_2 -x_1) + k (x_3 - x_2) \\ m \ddot{x}_3 & = -k x_3 - k (x_3 - x_2) \\ \end{aligned} $$
Adopt exponential solutions: $$ \begin{aligned} x_1 & = b_1 e^{i\omega t} \\ x_2 & = b_2 e^{i\omega t} \\ x_3 & = b_3 e^{i\omega t} \\ \end{aligned} $$
Which leads to a 3 x 3 matrix version:
\begin{equation} \begin{pmatrix} - m \omega^2 + 2k & -k & 0 \\ -k & -m \omega^2 + 2k & -k \\ 0 & -k & -m \omega^2 + 2k \end{pmatrix} \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix} = 0 \end{equation}Taking the determinant of this and setting equal to zero: \begin{equation} \begin{vmatrix} - m \omega^2 + 2k & -k & 0 \\ -k & -m \omega^2 + 2k & -k \\ 0 & -k & -m \omega^2 + 2k \end{vmatrix} = 0 \end{equation}
Will lead to a polynomial \begin{equation} (-m \omega^2 + 2k)[(-m\omega^2 + 2k)^2 - k^2] + k(-k(-m\omega^2 + 2k)) = 0 \end{equation} This can be factored: \begin{equation} (-m \omega^2 + 2k)[(-m \omega^2 + 2k)^2-2k^2] =0 \end{equation}
if \begin{equation} (-m \omega^2 + 2k) = 0 \end{equation} then we have a root: \begin{equation} \omega_1 = \sqrt{\frac{2k}{m}} \end{equation}
likewise: if \begin{equation} [(-m \omega^2 + 2k)^2-2k^2] = 0 \end{equation} then: \begin{equation} \omega_2 = \sqrt{\frac{(2-\sqrt{2})k}{m}} \; \textrm{and} \;\omega_3 = \sqrt{\frac{(2+\sqrt{2})k}{m}} \end{equation}
The continuum limit
\begin{equation} L = T - U = \sum_i \frac{1}{2}m \dot{x}_i^2 - \frac{1}{2}\sum_i k(x_{i+1}-x_i)^2 \end{equation}
The lagrangian for a mass $m$ is then: \begin{equation} m \ddot{x}_i - k (x_{i+1}-x_i) + k(x_{i}-x_{i-1}) = 0 \end{equation} or: \begin{equation} k (x_{i+1}-x_i) - k(x_{i}-x_{i-1}) = m \ddot{x}_i \end{equation}
Define a mass per unit length: \begin{equation} \mu = \frac{m}{a} \end{equation} as well as the Young's Modulus: \begin{equation} Y = \frac{\textrm{stress}}{\textrm{strain}} \end{equation} \begin{equation} \textrm{stress} = T = k(x_{i+1}-x) \end{equation} and \begin{equation} \textrm{strain} = \frac{x_{i+1}-x_i}{a} \end{equation}
Thus: \begin{equation} Y = \frac{k(x_{i+1}-x_i)}{(x_{i+1}-x_i)/a} = k a \end{equation}
Now, let $a \rightarrow 0$ \begin{equation} \mu \ddot{x}_i - \frac{Y}{a}[(x_{i+1}-x_i) - (x_i - x_{i-1})] = 0 \end{equation}
Call $\eta(x,t)$ the displacement from equilibrium \begin{equation} \ddot{x}_i \rightarrow \left. \frac{\partial ^2\eta}{\partial t^2} \right|_{x_i} \end{equation}
\begin{equation} \frac{(x_{i+1}-x_i)}{a} \rightarrow \left. \frac{\partial \eta}{\partial x}\right|_{x_i+a} \end{equation}
\begin{equation} \frac{(x_{i}-x_{i-1})}{a} \rightarrow \left. \frac{\partial \eta}{\partial x}\right|_{x_i} \end{equation}
From Taylor Series: \begin{equation} \left. \frac{\partial \eta}{\partial x} \right|_{x_i+a} = \left. \frac{\partial \eta}{\partial x} \right|_{x_i} + a \left. \frac{\partial^2 \eta}{\partial x^2} \right|_{x_i+a} \end{equation}
Wave Equation:
\begin{equation} \frac{\partial^2 \eta}{\partial x^2} - \frac{\mu}{Y} \frac{\partial^2 \eta}{\partial t^2} = 0 \end{equation}
Replace the $\mu/Y$ with the velocity, $v^2$.
\begin{equation} \frac{\partial^2 \eta}{\partial x^2} - \frac{1}{v^2} \frac{\partial^2 \eta}{\partial t^2} = 0 \end{equation}