Normal modes

Two masses joined by springs can jiggle in a thousand messy-looking ways, yet underneath that mess sit just two clean patterns of motion called normal modes. In a normal mode every part of the system swings at one shared frequency, locked into a fixed relative amplitude — the whole system behaves, for that instant of analysis, like a single simple harmonic oscillator. The payoff is enormous: instead of solving two coupled, tangled equations for $x_1(t)$ and $x_2(t)$ , you solve two independent, boring SHM problems and then add the answers back together. This article takes the two-mass, three-spring system already met in coupled oscillations and asks what its modes actually look like physically, how any motion is built from them, and why nearly-equal mode frequencies produce the beautiful beating exchange of energy between the two masses.

The core insight

A normal mode is a pattern in which every coordinate of a coupled system oscillates at one common frequency with a fixed, time-independent amplitude ratio, and any general motion is a superposition of these modes.

The core idea

Return to the standard system: two equal masses $m$ , each tied to a wall by an end spring $k$ , and tied to each other by a coupling spring $k_c$ . Newton's second law for each mass gives the coupled equations

m\ddot x_1 + (k+k_c)x_1 - k_c x_2 = 0, \qquad m\ddot x_2 + (k+k_c)x_2 - k_c x_1 = 0.

where: $x_1,x_2$ are the displacements of the two masses from equilibrium, $m$ is each mass, $k$ is each end-spring constant, and $k_c$ is the coupling-spring constant. As written, the two equations are tangled — $x_1$ 's equation contains $x_2$ and vice versa — so neither mass on its own performs simple harmonic motion. A normal mode is the special case where a common-frequency trial solution actually works: both masses move as $x_1=a_1\cos(\omega t+\phi)$ , $x_2=a_2\cos(\omega t+\phi)$ , with the ratio $a_1/a_2$ frozen for all time. Substituting this trial form back into the coupled equations (or equivalently changing to the normal coordinates $q_1=x_1+x_2$ , $q_2=x_1-x_2$ , which decouple the equations completely) reveals exactly two such modes for this two-mass system.

Symmetric mode, frequency $\omega_1=\sqrt{k/m}$ : the masses move together, $x_1=x_2$ always. The coupling spring is never stretched relative to its natural length, so it contributes nothing, and each mass simply oscillates against its own end spring — which is why $\omega_1$ has no $k_c$ in it at all.

Antisymmetric mode, frequency $\omega_2=\sqrt{(k+2k_c)/m}$ : the masses move oppositely, $x_1=-x_2$ always. Every excursion of mass 1 stretches the coupling spring by twice as much (both masses move apart), adding extra restoring force on top of the end spring, so $\omega_2$ is always larger than $\omega_1$ whenever $k_c>0$ . The figure below sketches the two shapes side by side: in-phase, equal-amplitude motion for the symmetric mode, and mirror-image, equal-amplitude motion for the antisymmetric mode.

Figure

Two mode shapes for the two mass three spring system symmetric mode with both masses moving right together and antisymmetric mode with masses moving in opposite directions

Any general motion is a superposition of the two normal modes,

x_1(t) = a\cos(\omega_1 t+\phi_1) + b\cos(\omega_2 t+\phi_2), \qquad x_2(t) = a\cos(\omega_1 t+\phi_1) - b\cos(\omega_2 t+\phi_2),

where: $a,b$ are the amplitudes and $\phi_1,\phi_2$ the phases of the symmetric and antisymmetric modes, all four fixed by the initial displacements and velocities of the two masses. When a system is started with only one mass displaced, both modes are excited with equal weight, and their slow drift in and out of phase produces a beat: energy sloshes periodically and almost completely between the two masses at the beat frequency $\lvert\omega_1-\omega_2\rvert$ .

Derivation ANormal mode frequencies by direct substitution

The goal is to show, starting only from the two coupled Newton's-law equations and the assumption that a normal mode has both masses oscillating at one shared frequency with a fixed amplitude ratio, that exactly two such frequencies exist, and to derive their values $\omega_1=\sqrt{k/m}$ and $\omega_2=\sqrt{(k+2k_c)/m}$ purely by algebra, without appealing to the normal-coordinate shortcut.

Step 1. Assume a common-frequency trial solution for both masses, $x_1=a_1\cos(\omega t+\phi)$ and $x_2=a_2\cos(\omega t+\phi)$ , so that $\ddot x_1=-\omega^2 a_1\cos(\omega t+\phi)$ and $\ddot x_2=-\omega^2 a_2\cos(\omega t+\phi)$ . Substituting into the two coupled equations and cancelling the common factor $\cos(\omega t+\phi)$ leaves two linear algebraic equations for the amplitudes: $-m\omega^2 a_1+(k+k_c)a_1-k_c a_2=0$ and $-m\omega^2 a_2+(k+k_c)a_2-k_c a_1=0$ .

Step 2. Rewrite the first equation as $(k+k_c-m\omega^2)a_1 = k_c a_2$ , so the amplitude ratio is $a_1/a_2 = k_c/(k+k_c-m\omega^2)$ . For a genuine normal mode this ratio must be a fixed, real, time-independent number, and by symmetry the second equation gives $a_2/a_1 = k_c/(k+k_c-m\omega^2)$ as well — the same expression. A consistent solution therefore needs $(a_1/a_2)(a_2/a_1)=1$ , which forces $a_1=\pm a_2$ : only two amplitude ratios are possible, $a_1=a_2$ or $a_1=-a_2$ .

Step 3. Try $a_1=a_2$ (the symmetric guess) in $(k+k_c-m\omega^2)a_1=k_c a_1$ . Since $a_1\neq0$ , divide through: $k+k_c-m\omega^2=k_c$ , giving $m\omega^2=k$ , so $\omega_1=\sqrt{k/m}$ .

Step 4. Try $a_1=-a_2$ (the antisymmetric guess) in $(k+k_c-m\omega^2)a_1=k_c a_2=-k_c a_1$ . Divide through by $a_1\neq0$ : $k+k_c-m\omega^2=-k_c$ , giving $m\omega^2=k+2k_c$ , so $\omega_2=\sqrt{(k+2k_c)/m}$ . Because $k_c>0$ , $\omega_2>\omega_1$ always, matching the physical picture that the antisymmetric pattern feels extra restoring force from the coupling spring.

Worked example

Using the standard numbers $m=0.5\ \text{kg}$ , $k=20\ \text{N/m}$ , $k_c=5\ \text{N/m}$ , so $\omega_1=\sqrt{40}\approx6.32\ \text{rad/s}$ and $\omega_2=\sqrt{60}\approx7.75\ \text{rad/s}$ : the system starts from rest with $x_1(0)=0.02\ \text{m}$ , $x_2(0)=0$ . Estimate the time at which the energy has first fully transferred from mass 1 to mass 2, and check the antisymmetric mode by substitution.

Step 1. Starting from rest with only $x_1$ displaced excites both modes equally, so $x_1(t)=A_0\cos\!\left(\tfrac{\omega_2-\omega_1}{2}t\right)\cos\!\left(\tfrac{\omega_1+\omega_2}{2}t\right)$ with $A_0=0.02\ \text{m}$ — a fast oscillation inside a slowly varying envelope $A_0\left|\cos\!\left(\tfrac{\omega_2-\omega_1}{2}t\right)\right|$ .

Step 2. Mass 1's envelope first vanishes (all the energy is now in mass 2) when $\tfrac{\omega_2-\omega_1}{2}t=\tfrac{\pi}{2}$ , i.e. $t=\pi/(\omega_2-\omega_1)$ . Numerically $\omega_2-\omega_1=7.75-6.32=1.43\ \text{rad/s}$ , so $t=\pi/1.43\approx2.20\ \text{s}$ . The full beat period, when the energy has returned to mass 1, is twice this, about $4.4\ \text{s}$ .

Step 3. Sanity-check the antisymmetric mode itself: with $x_1=0.01\cos(7.75\,t)$ , $x_2=-0.01\cos(7.75\,t)$ , substitute into $m\ddot x_1+(k+k_c)x_1-k_c x_2$ . Since $\ddot x_1=-\omega_2^2 x_1$ , the left side becomes $x_1\big[-m\omega_2^2+(k+k_c)-k_c(-1)\big]=x_1\big[-(k+2k_c)+k+2k_c\big]=0$ , confirming $\omega_2=\sqrt{(k+2k_c)/m}$ is exactly the frequency that makes $x_1=-x_2$ self-consistent.

Answer: the energy first fully transfers to mass 2 at $t\approx2.20\ \text{s}$ , with the full beat period about $4.4\ \text{s}$ , and the antisymmetric-mode substitution checks out exactly.

Common traps

Treating any solution of the coupled equations as a "normal mode" — a true mode needs one shared frequency and a fixed, time-independent amplitude ratio between the coordinates, not just some periodic motion.
Forgetting $\omega_1=\sqrt{k/m}$ has no $k_c$ dependence — the coupling spring does nothing in the symmetric mode, so it cannot appear in that frequency.
Confusing "superposition of modes" with "the modes interact" — once excited, normal modes evolve completely independently; beats arise from adding two separate steady oscillations, not from energy leaking between the modes themselves.
Applying the beat picture when $\omega_1$ and $\omega_2$ are very different — beats are a useful visual only when the two frequencies are close; otherwise the "envelope" period is comparable to the oscillation period itself and the slow-envelope idea breaks down.

Ask the tutor: if you want to see how the mode-shape argument changes for unequal masses or unequal end springs, ask the AI tutor.

How is a normal mode different from just any oscillation of the coupled system?

Almost any starting condition produces motion where $x_1(t)$ and $x_2(t)$ look messy and their amplitude ratio drifts with time — that is not a normal mode. A normal mode is the special case where the trial solution $x_1=a_1\cos(\omega t+\phi)$ , $x_2=a_2\cos(\omega t+\phi)$ actually satisfies both coupled equations, which forces the amplitude ratio $a_1/a_2$ to stay fixed forever, not just at one instant. There are exactly two such special patterns for this two-mass system, and every other motion is built by adding them together.

If the two masses in the antisymmetric mode always have equal and opposite displacement, does the centre of mass move?

No. Since $x_1=-x_2$ at every instant in the antisymmetric mode, and the two masses are equal, the centre of mass $\tfrac{m x_1+m x_2}{2m}=\tfrac{x_1+x_2}{2}$ stays exactly at zero throughout the motion. In the symmetric mode, by contrast, $x_1=x_2$ always, so the centre of mass itself oscillates at $\omega_1$ — the whole system shifts back and forth together. This distinction is a fast way to sanity-check which mode you are looking at from a description of the motion.

Do beats mean the two masses are exchanging energy back and forth forever with no loss?

In this idealised, undamped model, yes — the total mechanical energy is conserved and simply relocates periodically between the two masses, fully to mass 2 at $t\approx2.20\ \text{s}$ in the worked example and back to mass 1 by about $4.4\ \text{s}$ , repeating indefinitely. Real coupled oscillators always have some damping, so in practice the beat envelope itself slowly shrinks and the whole system eventually settles down, but that decay is a separate topic from the ideal normal-mode picture developed here.

Normal modes

The core idea

Worked example

Common traps

Before this

Where this leads