Coupled Oscillators & Normal Modes

Two carts that will not leave each other alone

Put one cart on an air track, tie it to a wall with a spring, pull it aside and let go. You get a sine wave at one frequency, forever. Nothing surprising.

Now put a second cart next to it, tie that one to the opposite wall, and join the two carts with a third, much weaker spring. Pull only the first cart aside and release. For a few seconds it swings and the second cart barely stirs. Then the second one starts to build. A minute later the first cart is hanging dead still and the second is doing all the swinging. Then it comes back. Nobody pushed anything; the energy walked from one cart to the other through a spring that is barely there.

The Lagrangian gave you a machine for writing equations of motion for systems with several coordinates. This lesson is about what to do with those equations once you have them, when the coordinates refuse to separate. The answer is always the same: find the special motions in which the whole system rings at a single frequency. Those are the normal modes, and once you have them, every other motion in the system is just a sum of them.

Why the equations refuse to separate

Take two equal masses $m$ on a line. Each is tied to its nearby wall by a spring of stiffness $k$, and a coupling spring of stiffness $k_c$ joins them. Let $x_1$ and $x_2$ be displacements from equilibrium, both measured to the right.

The wall spring on cart 1 is stretched by $x_1$ and pulls back with $-kx_1$. The middle spring's length has changed by $x_2 - x_1$: if cart 2 has moved right more than cart 1, the middle spring is stretched and pulls cart 1 to the right. So it contributes $+k_c(x_2-x_1)$ to cart 1 and the opposite to cart 2. That gives

Neither line can be solved by itself: each one contains the other cart's coordinate. That is what "coupled" means, and it is the entire difficulty.

There is a cheap trick here that shows you the way out. Add the two equations. The $k_c$ terms cancel exactly, and with $s = x_1+x_2$ you get $m\ddot s = -ks$. Subtract them instead. With $d = x_1-x_2$ the coupling terms add rather than cancel, and you get $m\ddot d = -(k+2k_c)d$.

Two ordinary, uncoupled oscillator equations. Each combination has its own single frequency:

$s$ and $d$ are called normal coordinates. The physical motions they describe are the normal modes. In the $s$ mode ($d=0$, so $x_1=x_2$) the carts move together, the middle spring never changes length, and it might as well not exist: hence $\omega_+$ has no $k_c$ in it at all. In the $d$ mode ($s=0$, so $x_2=-x_1$) the carts move oppositely, the middle spring is stretched by $2x_1$, and each cart feels an extra restoring force $2k_cx_1$.

The antiphase mode is always the faster one, because it is the one that deforms the coupling spring. Turn the coupling up and the two frequencies spread apart, with the in-phase one pinned in place.

Worked example 1: the numbers for a real pair

Two carts of mass $m=0.50$ kg, wall springs $k=200$ N/m, coupling spring $k_c = 50$ N/m.

In-phase:

Antiphase:

That is the whole calculation for the easy version of this problem. Write down the two shapes, ask what each does to the coupling spring, read off the effective stiffness.

Any motion at all is a sum of the modes

The equations are linear, so if two motions solve them, so does their sum. The general solution is

Four constants $A,B,\phi_+,\phi_-$ for four initial conditions (two positions, two velocities). The sign flip on $B$ is the only difference between the lines, and it is just the statement that the second mode is antiphase.

Now the opening mystery is not mysterious. Hold cart 2 at rest, pull cart 1 aside by $a$, release both. You need $x_1(0)=a$, $x_2(0)=0$, which forces $A=B=a/2$, and zero initial velocities force $\phi_\pm=0$. Then

using the identity $\cos P+\cos Q = 2\cos\frac{P-Q}{2}\cos\frac{P+Q}{2}$. The second factor is a fast wobble at the average frequency. The first is a slowly changing envelope. Cart 2 gets the same expression with a sine in place of the first cosine, so when one envelope is at its largest the other is at zero. That is a beat, and it is the energy sloshing back and forth.

Two timescales matter, and people mix them up constantly. The envelope of $x_1$ vanishes when $\frac{\omega_--\omega_+}{2}t=\pi/2$, so the energy is fully handed over after

and the whole pattern repeats (cart 1 loud again) after twice that,

If the modes are quoted in hertz rather than rad/s, those become $t_{\text{transfer}} = 1/(2\Delta f)$ and $T_{\text{beat}}=1/\Delta f$.

Worked example 2: unequal masses, no symmetry to lean on

Symmetry guessing dies the moment the two sides stop matching. A cart of $m_1=1.0$ kg and a cart of $m_2=2.0$ kg each have a wall spring of $k=400$ N/m, joined by $k_c=200$ N/m. Find the mode frequencies and the shape of the slower one.

The equations of motion are built exactly as before, only with different masses:

A normal mode means both carts moving at one frequency, so try $x_1 = A_1\cos\omega t$ and $x_2=A_2\cos\omega t$. Each $\ddot x$ brings down $-\omega^2$, the cosines divide out, and you are left with two algebraic equations:

with $k+k_c=600$ N/m. Multiply them together and cancel $A_1A_2$:

Expand: $360000-1200\omega^2-600\omega^2+2\omega^4=40000$, so $2\omega^4-1800\omega^2+320000=0$, and dividing by 2,

This is a quadratic in $\omega^2$. The discriminant is $900^2-4(160000)=810000-640000=170000$, whose square root is $412.3$. So

giving $\omega_- = \sqrt{656.2}=25.6$ rad/s and $\omega_+=\sqrt{243.8}=15.6$ rad/s. Two masses, two modes, as promised.

For the shape, put $\omega^2=243.8$ back into the first equation, which you never used on its own:

The heavy cart swings $1.78$ times as far as the light one in the slow mode. That should feel right: this is the mode where they move in the same direction, and the mode adjusts until the momenta balance out sensibly.

The recipe generalises with no changes. Assume every coordinate goes as $\cos\omega t$, get a set of linear equations in the amplitudes, and demand that the determinant of the coefficient matrix vanish (otherwise the only solution is all amplitudes zero). For three masses that determinant is a cubic in $\omega^2$; for the chain of $1.0$, $2.0$ and $3.0$ kg between four springs of $500$ N/m it gives modes at $11.8$, $24.5$ and $34.6$ rad/s.

Chains, rings, and free systems

Three shortcuts cover most of what appears in competition.

A uniform chain between fixed walls. $N$ equal masses $m$ joined by $N+1$ identical springs $k$ to each other and to two walls. The mode shapes are sampled sine waves that vanish at the walls, and the frequencies are

You can check it against the case you already did: $N=1$ gives $2\sqrt{k/m}\sin(\pi/4)=\sqrt{2k/m}$, which is right for one mass between two springs. Use this whenever the chain is uniform; setting up a five-by-five determinant by hand when the formula exists is a way to lose twenty minutes.

A free system. If nothing is tied to a wall, one mode has $\omega=0$: the whole thing translates. Two free masses $m_1,m_2$ joined by a spring $k$ have that zero mode plus one real one at $\omega=\sqrt{k/\mu}$, where $\mu = m_1m_2/(m_1+m_2)$ is the reduced mass. For $1.0$ and $2.0$ kg with $k=300$ N/m, $\mu = 2/3$ kg and $\omega=\sqrt{450}=21.2$ rad/s.

A ring. Three equal masses joined head to tail in a loop by three identical springs has the zero translation mode plus a doubly degenerate mode at $\omega=\sqrt{3k/m}$. For $m=0.50$ kg and $k=200$ N/m that is $\sqrt{1200}=34.6$ rad/s. The counting works: three masses, three modes, one of them zero.

Worked example 3: two pendulums of different lengths

This one combines everything: a coupling that is not symmetric because the two oscillators are not identical.

Two bobs of mass $m=0.40$ kg hang on strings of length $L_1=1.00$ m and $L_2=1.20$ m, joined by a spring $k=3.0$ N/m that is unstretched when both hang straight down. Take $g = 9.80$ m/s². Use the horizontal bob displacements $x_1,x_2$ as coordinates.

A pendulum of length $L$ displaced by $x$ feels a restoring force $-mgx/L$, so on its own it has $\omega^2 = g/L$. The spring adds a coupling exactly like the carts:

Divide by $m$ and write $\omega_1^2 = g/L_1 + k/m$ and $\omega_2^2 = g/L_2 + k/m$ for the diagonal entries, with $k/m$ appearing on the off-diagonal too. Numbers: $k/m = 3.0/0.40 = 7.5$ s⁻², $g/L_1 = 9.80$ s⁻², $g/L_2 = 9.80/1.20 = 8.167$ s⁻². So $\omega_1^2 = 17.30$ and $\omega_2^2=15.667$ s⁻².

The determinant condition for a two-by-two system of this shape has a tidy closed form worth memorising:

The mean of the two uncoupled values, plus or minus a term that mixes the detuning with the coupling. Here the mean is $(17.30+15.667)/2 = 16.483$, the half-difference is $(17.30-15.667)/2=0.8167$, and

So $\omega^2 = 16.483 - 7.544 = 8.939$, giving the lower mode $\omega=\sqrt{8.939}=2.99$ rad/s, and $\omega^2 = 24.03$ for the upper, $\omega = 4.90$ rad/s.

That formula also tells you what detuning does. If the two oscillators are identical the square root is just $k/m$ and the splitting is symmetric about the common frequency. Make them different and the splitting grows, but the modes stop being fifty-fifty mixtures: each mode leans towards the oscillator it started closest to. Which is why detuned oscillators never fully hand their energy over. Writing $\kappa$ for the coupling rate and $\delta$ for half the frequency difference, the largest fraction of energy that ever reaches the second oscillator is

With $\kappa = 0.40$ rad/s and $\delta = 0.60$ rad/s, that is $0.16/(0.16+0.36)=0.16/0.52 = 0.308$. Not even a third of the energy gets across, no matter how long you wait. Detuning kills transfer; that is the whole reason a tuning fork rings a matched fork across the room and ignores a mismatched one.

Driving a coupled system

Push cart 1 with a force $F\cos\omega t$ and let the damping be tiny. In steady state both carts oscillate at the drive frequency $\omega$, so put $x_1 = X_1\cos\omega t$, $x_2 = X_2\cos\omega t$ into the equations. Cart 2 has no drive of its own, so its equation is

That single expression answers a family of problems. For $m=0.50$ kg, $k=200$ N/m, $k_c=50$ N/m, drive at the in-phase mode $\omega=20.0$ rad/s: $m\omega^2 = 0.50\times400=200$, the denominator is $250-200=50$, and the ratio is $50/50 = 1.0$. The carts move together, which is what driving at a mode frequency should do: the drive builds up that mode and nothing else. Drive at $22.0$ rad/s instead, between the modes: $m\omega^2=0.50\times484=242$, the denominator is $250-242 = 8$, and the ratio is $50/8 = 6.25$. The undriven cart swings six times as far as the driven one.

Push that further and you get the vibration absorber. The denominator of $X_1$ (not $X_2$) carries a factor $k_a - m_a\omega^2$ when you bolt a small mass $m_a$ on a spring $k_a$ to a shaking machine. Tune it so $\omega^2 = k_a/m_a$ at the drive frequency and the machine stops dead: the absorber moves exactly enough to cancel the drive. To quiet a machine driven at $14.14$ rad/s with an absorber of $m_a = 0.40$ kg you need

Recognising the technique

Reach for normal modes the moment a problem has more than one thing that can oscillate and a spring, string, or torque linking them. The specific giveaways:

• Two or more masses, and the equation for each contains the other's coordinate. That is the definition of coupled, and guessing single-frequency motion is the only general way out.
• The words in phase, antiphase, symmetric, antisymmetric, or a question asking for "the higher mode" or "the lowest mode". Count degrees of freedom first, then hunt for exactly that many frequencies.
• Beats, energy transfer, or sloshing. Any question about how long until the other oscillator has the energy is really asking for two mode frequencies and their difference. Find $\omega_-$ and $\omega_+$, subtract, then use $\pi/\Delta\omega$ for transfer and $2\pi/\Delta\omega$ for the full beat.
• A uniform line of $N$ identical masses and identical springs. Use the sine formula, not a determinant.
• Nothing attached to a wall. Expect a zero-frequency mode and use the reduced mass for the remaining one.
• A driven system where the question asks for an amplitude ratio or how to make something stop shaking. Write the undriven body's equation alone.

The general strategy never changes: choose coordinates measured from equilibrium, write one equation of motion per coordinate, assume everything goes as $\cos\omega t$, and demand that the resulting homogeneous system have a nonzero solution. Symmetry, when the system has it, just lets you skip to the answer.

The same idea (find the special shapes in which a whole extended system oscillates coherently) is what makes fields tractable too. Next you use it on space rather than on a handful of masses, in boundary-value problems in E&M.