Most oscillators are coupled, meaning they have multiple natural frequencies, example:

🪟 Scenarios:

1. Displace each pendulum by the same amount in the same direction
   • Frequencies are the same $\omega_1=\sqrt{\frac{g}{l}}$
   • in phase
2. Displace each pendulum by the same amount in opposite directions
   • out of phase
   • with the same frequency $\omega_2$, which is different from $\omega_1$

1. If $x_B>x_A$ then the tension in the spring is $T=k(x_B-x_A)$. The restoring forces are:

   $$ \begin{aligned} F_\mathrm{B} &= - \frac{mg x_\mathrm{B}}{l} - k (x_\mathrm{B} -x_\mathrm{A}) \\ F_\mathrm{A} &= - \frac{mg x_\mathrm{A}}{l} + k (x_\mathrm{B} -x_\mathrm{A}) \end{aligned} $$

   • Thus the equations of motion are

   $$ \begin{aligned} \ddot{x}\mathrm{A} + \frac{g}{l} x\mathrm{A} - \frac{k}{m} ( x_\mathrm{B}- x_\mathrm{A}) &=0\quad &(1)\\ \ddot{x}\mathrm{B} + \frac{g}{l} x\mathrm{B} + \frac{k}{m} ( x_\mathrm{B}- x_\mathrm{A}) &=0\quad &(2) \end{aligned} $$

   • The equations are coupled

   If we define $q_1=x_A+x_B$ and do $(1)+(2)$ we get:

   $$ \ddot{q}_1 + \frac{g}{l} q_1=0 $$

   which is an equation of SHM, we can also do $(1)-(2)$ we get:

   $$ \ddot{q}_2 + \left ( \frac{g}{l} + \frac{2k}{m} \right ) q_2 =0 $$

   which is also an equation of SHM

   • The variables $q_1, q_2$ are called the normal coordinates

   The normal frequencies are:

   $$ \omega_1^2 = \frac{g}{l} \quad \quad \mathrm{and} \quad \quad \omega_2^2= \frac{g}{l} + \frac{2k}{m} $$

   • Exciting the system in $q_1$ means $q_2=0$ i.e. $x_\mathrm{A}-x_\mathrm{B}=0 \quad \Rightarrow \quad x_\mathrm{A}=x_\mathrm{B}$
   • Exciting the system in $q_2$ means $q_1=0$ i.e. $x_\mathrm{A}+x_\mathrm{B}=0 \quad \Rightarrow \quad x_\mathrm{A}=-x_\mathrm{B}$
   • Any motion of the two pendula can be broken down into a linear combination of two independent harmonic oscillations

Eigenvectors

• Link to explanation

$n$ oscillators, $n$ nodes

Two identical masses, $A$ and $B$ connect by a spring of constant $k'$ each mass is is also connected to a side wall via a spring of spring constant $k$. The three springs are aligned horizontally as illustrated on the side. $x_A$ and $x_B$ are the displacements of the masses in the longitudinal $x$ direction.

• We can write the governing equations:

$$ \begin{aligned} m \ddot{x}\mathrm{A} &= - k x\mathrm{A} + k'(x_\mathrm{B}- x_\mathrm{A}) \\ m \ddot{x}\mathrm{B} &= - k x\mathrm{B} - k'(x_\mathrm{B}- x_\mathrm{A}) \end{aligned} $$

• These can be simplified using $\omega^2=k/m$ and ${\omega'}^2=k'/m'$ to:

$$ \begin{aligned} \ddot{x}\mathrm{A} + \omega^2 x\mathrm{A} + \omega'^2 (x_\mathrm{A}- x_\mathrm{B})&= 0 \\ \ddot{x}\mathrm{B} + \omega^2 x\mathrm{B} + \omega'^2 (x_\mathrm{B}- x_\mathrm{A})&= 0 \end{aligned} $$

• Assuming that both masses move at the same frequency $\Omega$ we can try the following solutions:

$$ x_\mathrm{A} = X_\mathrm{A} e^{i \Omega t}\quad ;\quad x_\mathrm{B} = X_\mathrm{B} e^{i \Omega t} $$

• Rewriting we get

$$ \begin{aligned} (\omega^2 + \omega'^2) X_\mathrm{A} - \omega'^2 X_\mathrm{B} &=\Omega^2 X_\mathrm{A} \\ -\omega'^2 X_\mathrm{A} + (\omega^2 + \omega'^2) X_\mathrm{B} &=\Omega^2 X_\mathrm{B} \end{aligned} $$

• Which can be written in matrix equation as