When two oscillators are connected so that the motion of one affects the force on the other, their coordinates are coupled. The individual masses no longer oscillate independently, but the system can still have special collective motions called normal modes.

Consider two identical masses $m$ connected to fixed walls by springs of stiffness $k$ and to each other by a coupling spring of stiffness $\kappa$:

$$\text{wall}-k-m_1-\kappa-m_2-k-\text{wall}.$$

Let $x_1$ and $x_2$ be small displacements from equilibrium. Newton's second law gives

$$m\ddot x_1=-kx_1-\kappa(x_1-x_2),$$

$$m\ddot x_2=-kx_2-\kappa(x_2-x_1).$$

Each equation contains both coordinates, so the motions are coupled.

In-phase normal mode

Suppose the two masses move together with equal displacement:

$$x_1=x_2.$$

Then the coupling spring is never stretched, because

$$x_1-x_2=0.$$

Each mass therefore obeys

$$m\ddot x=-kx,$$

so the mode frequency is

$$\boxed{\omega_+=\sqrt{\frac{k}{m}}}.$$

The mode shape is proportional to

$$\boxed{(x_1,x_2)=(1,1)}.$$

Both masses move in phase.

Out-of-phase normal mode

Now suppose the masses move with equal magnitude in opposite directions:

$$x_1=-x_2.$$

The coupling spring is stretched twice as much as either displacement. Substituting $x_2=-x_1$ into the first equation gives

$$m\ddot x_1=-(k+2\kappa)x_1.$$

Thus

$$\boxed{\omega_- =\sqrt{\frac{k+2\kappa}{m}}}$$

with mode shape

$$\boxed{(x_1,x_2)=(1,-1)}.$$

Because the coupling spring now deforms strongly, this mode is stiffer and therefore has the higher natural frequency.

Why these motions are special

In a normal mode, every coordinate oscillates sinusoidally at the same frequency, and the ratios between their amplitudes remain fixed. The whole system therefore preserves one characteristic shape while its overall amplitude changes in time.

The two combinations

$$q_+=x_1+x_2,$$

$$q_-=x_1-x_2$$

make this explicit. Adding and subtracting the coupled equations gives

$$\ddot q_++\frac{k}{m}q_+=0,$$

$$\ddot q_-+\frac{k+2\kappa}{m}q_-=0.$$

The original coupled problem has become two independent harmonic oscillators in the normal coordinates $q_+$ and $q_-$.

General motion as a superposition of modes

Because the equations are linear, a general free motion can contain both modes:

$$x_1(t)=A_+\cos(\omega_+t+\phi_+)+A_-\cos(\omega_-t+\phi_-),$$

$$x_2(t)=A_+\cos(\omega_+t+\phi_+)-A_-\cos(\omega_-t+\phi_-).$$

Initial conditions determine how much of each mode is present. Unless the system is started in one pure normal mode, the individual masses generally do not execute a single-frequency sinusoid.

Worked example

Let

$$m=1.0,\mathrm{kg},\qquad k=9.0,\mathrm{N/m},\qquad \kappa=8.0,\mathrm{N/m}.$$

The in-phase frequency is

$$\omega_+=\sqrt{\frac{9}{1}}=3.0,\mathrm{rad/s}.$$

The out-of-phase frequency is

$$\omega_- =\sqrt{\frac{9+2(8)}{1}}=\sqrt{25}=5.0,\mathrm{rad/s}.$$

If both masses are displaced equally and released, only the $3.0,\mathrm{rad/s}$ mode is excited. If they are displaced equally in opposite directions, only the $5.0,\mathrm{rad/s}$ mode is excited. A generic initial displacement excites both frequencies.

From two oscillators to many

A system with $N$ independent coupled coordinates generally has $N$ normal modes in the linear approximation. Each mode has its own natural frequency and collective motion pattern. This is why complex structures, molecules, strings, and other distributed systems can possess many resonant frequencies.

Normal modes are therefore the motions that reveal the hidden independent oscillators inside a coupled linear system.