6.6 Coupled Oscillations and Normal Modes

A coupled oscillator is a system of more than one oscillators that have some way of transferring energy to one another. Generally what we see is that the kinetic energy is passed between each oscillator. A normal mode is a collective excitation of the whole system where all parts move with the same frequency. Normal modes are the generalisation of the resonant frequency for a single oscillator, and once the system is in a normal mode it does not decay or change its motion into another mode (unless there is damping). If we take a system of two pendula coupled by a spring as an example, the system has two normal modes. One where the pendula are swinging in-phase, which has a frequency $\omega_{1}=\sqrt{\frac{g}{L}}$ , and one where they are swinging out-of-phase, which has a frequency $\omega_{2}=\sqrt{\frac{g}{L}+2\frac{k}{m}}$ .

The principle of superposition states that any motion of the system can be expressed as a superposition (sum) of motion due to normal modes. Since the frequencies of the normal modes are unequal, this means the motion will evolve over time, which leads to the kinetic energy being passed around as stated above. This exchange of energy will happen at a beat frequency $\omega_{2}-\omega_{1}$ . For weakly coupled oscillators (in this case small $k$ ), they are almost independent and so all the normal modes will be very similar in frequency. This leads to the beat frequency being very small. Note that no energy is passed between the normal modes themselves. The designation “normal” means they are independent from each other.

Example 6.7.

Consider the example of two blocks, both of mass $m$ , coupled to three springs, all with spring constant $k$ , from above. Solve for the general motion of each block.

We will solve this with an elementary approach, by looking at the forces on each block. If we set $x_{1}=0$ , and $x_{2}=0$ at the equilibrium positions of each block respectively, then the force from each spring is

$\displaystyle F_{1}$	$\displaystyle=-kx_{1}$	(6.75)
$\displaystyle F_{2}$	$\displaystyle=-k(x_{1}-x_{2})$	(6.76)
$\displaystyle F_{3}$	$\displaystyle=kx_{2}.$	(6.77)

so by Newton II, the equations of motion for each block are

	$\displaystyle m\frac{\operatorname{\mathrm{d}}\!^{2}x_{1}}{\operatorname{% \mathrm{d}}\!t^{2}}+2kx_{1}-kx_{2}=0$		(6.78)
	$\displaystyle m\frac{\operatorname{\mathrm{d}}\!^{2}x_{2}}{\operatorname{% \mathrm{d}}\!t^{2}}-kx_{1}+2kx_{2}=0.$		(6.79)

These are two coupled second-order differential equations, which would be quite hard to solve. Fortunately, by writing the positions of the blocks $x_{1}$ and $x_{2}$ in terms of the normal modes, we can uncouple them and solve them easily! First, we will rearrange slightly and substitute in the resonant frequency of a mass on a spring $\omega_{0}=\sqrt{k/m}$ ,

	$\displaystyle\frac{\operatorname{\mathrm{d}}\!^{2}x_{1}}{\operatorname{\mathrm% {d}}\!t^{2}}+\omega_{0}^{2}x_{1}+\omega_{0}^{2}(x_{1}-x_{2})=0$		(6.80)
	$\displaystyle\frac{\operatorname{\mathrm{d}}\!^{2}x_{2}}{\operatorname{\mathrm% {d}}\!t^{2}}+\omega_{0}^{2}x_{2}-\omega_{0}^{2}(x_{1}-x_{2})=0,$		(6.81)

and then we let $X_{1}=x_{1}+x_{2}$ (the normal mode where the masses are swinging together) and $X_{2}=x_{1}-x_{2}$ (the normal mode where the masses are swinging apart). Then we can get differential equations for $X_{1}$ and $X_{2}$ by adding and subtracting the two equations we have for $x_{1}$ and $x_{2}$ .

	$\displaystyle\frac{\operatorname{\mathrm{d}}\!^{2}X_{1}}{\operatorname{\mathrm% {d}}\!t^{2}}=\frac{\operatorname{\mathrm{d}}\!^{2}x_{1}}{\operatorname{\mathrm% {d}}\!t^{2}}+\frac{\operatorname{\mathrm{d}}\!^{2}x_{2}}{\operatorname{\mathrm% {d}}\!t^{2}}=-\omega_{0}^{2}x_{1}-\omega_{0}^{2}x_{1}=-\omega_{0}^{2}X_{1}$		(6.82)
	$\displaystyle\frac{\operatorname{\mathrm{d}}\!^{2}X_{2}}{\operatorname{\mathrm% {d}}\!t^{2}}=\frac{\operatorname{\mathrm{d}}\!^{2}x_{1}}{\operatorname{\mathrm% {d}}\!t^{2}}-\frac{\operatorname{\mathrm{d}}\!^{2}x_{2}}{\operatorname{\mathrm% {d}}\!t^{2}}=-3\omega_{0}^{2}(x_{1}-x_{2})=-3\omega_{0}^{2}X_{2}.$		(6.83)

These are just the equations of motion for SHM, so we can solve for the amplitude of the normal modes.

	$\displaystyle X_{1}(t)$	$\displaystyle=A_{1}\cos(\omega_{0}t+\phi_{1})$		(6.84)
	$\displaystyle X_{2}(t)$	$\displaystyle=A_{2}\cos(\sqrt{3}\omega_{0}t+\phi_{2}).$		(6.85)

So for the first mode $X_{1}=x_{1}+x_{2}$ we have $x_{1}=x_{2}$ , and for the second mode $X_{2}=x_{1}-x_{2}$ we have $x_{1}=-x_{2}$ . The general solution is a superposition of motion due to normal modes, giving

	$\displaystyle x_{1}(t)$	$\displaystyle=A_{1}^{\prime}\cos(\omega_{1}t+\phi_{1})+A_{2}^{\prime}\cos(% \omega_{2}t+\phi_{2})$		(6.86)
	$\displaystyle x_{2}(t)$	$\displaystyle=A_{1}^{\prime}\cos(\omega_{1}t+\phi_{1})+A_{2}^{\prime}\cos(% \omega_{2}t+\phi_{2}),$		(6.87)

where $\omega_{1}=\omega_{0}=\sqrt{k/m}$ and $\omega_{2}=\sqrt{3}\omega_{0}=\sqrt{3k/m}$ . This system requires four initial conditions to specificy a particular solution i.e. $A_{1}^{\prime}$ , $A_{2}^{\prime}$ , $\phi_{1}$ , and $\phi_{2}$ . These could be $x_{1}(0)$ , $x_{2}(0)$ , $v_{1}(0)$ , and $v_{2}(0)$ .

6.6.1 Degrees of Freedom

In general, for a system of $N$ coupled simple harmonic oscillators, there will be $N$ normal modes ( $N$ ways they can move collectively). Therefore, the general solution will be a superposition of $N$ normal modes and $2N$ initial conditions will be required to specify a particular solution. In the previous example, we defined the variables $X_{1}=x_{1}+x_{2}$ and $X_{2}=x_{1}+x_{2}$ to represent the normal modes. These quantities along with their derivatives $\dot{X}_{1}$ and $\dot{X}_{2}$ are known as the normal coordinates of the system. Each way a system can store energy is known as a degree of freedom, and each degree of freedom has a normal coordinate. In simple harmonic oscillators, the degrees of freedom are called quadratic because the energies are proportional to the squares of the normal coordinates $\alpha_{1}X_{1}^{2}$ , $\alpha_{2}X_{2}^{2}$ , $\beta_{1}\dot{X}_{1}^{2}$ , and $\beta_{2}\dot{X}_{2}^{2}$ for some constants $\alpha_{1}$ , $\alpha_{2}$ , $\beta_{1}$ , and $\beta_{2}$ . Recall that the total energy in each normal mode is constant (if there is no damping) as they do not exchange energy with each other. For a given normal mode, all of the oscillators will pass through their equilibrium points at the same time and they will have a fixed phase and amplitude relationship with each other.