Coupled Pendulums

The model with springs

Pendulum $i$ has mass $m_i$, length $L_i$ and point of suspension $(x_i^0,0)$. All swing in the same vertical plane, and the bob sits at $\mathbf r_i=\left(x_i^0+L_i\sin\theta_i,\,-L_i\cos\theta_i\right)$. Neighbouring bobs are connected by springs with spring constant $k$. Each spring has natural length $d_i$ when the pendulums hang vertically, so the springs are neither stretched nor compressed at equilibrium. Then

$$\mathcal L=\sum_i\left[\tfrac12 m_iL_i^2\dot\theta_i^2+m_igL_i\cos\theta_i\right]-\sum_i\tfrac12k\left(|\mathbf r_{i+1}-\mathbf r_i|-d_i\right)^2 .$$

The spring force $\mathbf F$ on bob $i$ enters as the generalised force $Q_i=\mathbf F\cdot\partial\mathbf r_i/\partial\theta_i=L_i\left(F_x\cos\theta_i+F_y\sin\theta_i\right)$, and the equation of motion is

$$m_iL_i^2\,\ddot\theta_i=-m_igL_i\sin\theta_i+Q_i-\gamma\,m_iL_i^2\,\dot\theta_i .$$

Small oscillations and normal modes

For small oscillations the equations become linear, $\mathsf M\ddot{\boldsymbol\theta}=-\mathsf K\boldsymbol\theta$, with $\mathsf M=\mathrm{diag}(m_iL_i^2)$ and

$$K_{ii}=m_igL_i+\sum_{\rm neighbours}k_{\rm eff}L_i^2,\qquad K_{i,i\pm1}=-k_{\rm eff}L_iL_{i\pm1},\qquad k_{\rm eff}=k\cos^2\beta .$$

Here $\beta$ is the inclination of the spring at rest. It is slanted when neighbouring pendulums have different lengths, and then only the horizontal part of the stretch counts. A normal mode is a motion in which all the pendulums swing at the same frequency: $\boldsymbol\theta=\mathbf u\cos\omega t$ gives the eigenvalue problem

$$\mathsf K\mathbf u=\omega^2\,\mathsf M\mathbf u .$$

The page solves it by diagonalising the symmetric matrix $\mathsf M^{-1/2}\mathsf K\mathsf M^{-1/2}$ with Jacobi's method. The bars in the table are the vector $\mathbf u$. With $N$ pendulums there are $N$ normal modes, and any small-amplitude motion is a sum of them. That is the whole secret: what looks like a complicated interplay is simply $N$ independent oscillations superposed.

Two identical pendulums: the energy wanders

With $m_1=m_2=m$, $L_1=L_2=L$, $\omega_0^2=g/L$ and $\kappa=k/m$,

$$\ddot\theta_1=-\omega_0^2\theta_1-\kappa(\theta_1-\theta_2),\qquad \ddot\theta_2=-\omega_0^2\theta_2-\kappa(\theta_2-\theta_1).$$

The sum and the difference separate the equations:

$$\omega_1=\omega_0\quad(\theta_1=\theta_2),\qquad \omega_2=\sqrt{\omega_0^2+2\kappa}\quad(\theta_1=-\theta_2).$$

In the first mode the spring keeps its natural length the whole time, so the pendulums swing as if it were not there. In the second the spring pulls on both, and the oscillation is faster. If only the first pendulum is pushed, $\theta_1(0)=\theta_0$, the two modes are equally strong, and their sum is

$$\theta_1=\theta_0\cos\frac{\Delta\omega\,t}{2}\,\cos\bar\omega t,\qquad \theta_2=\theta_0\sin\frac{\Delta\omega\,t}{2}\,\sin\bar\omega t,$$

with $\bar\omega=(\omega_1+\omega_2)/2$ and $\Delta\omega=\omega_2-\omega_1$. Both pendulums swing at the mean frequency, while the amplitude slowly moves back and forth. The energy has been entirely over in pendulum 2 and is back in pendulum 1 after

$$T_{\rm exch}=\frac{2\pi}{\omega_2-\omega_1}\approx\frac{2\pi\,\omega_0}{\kappa}\qquad(\kappa\ll\omega_0^2).$$

These are beats, just as between two nearly identical tuning forks. Weak coupling gives slow exchange, but the exchange is still complete. With the page's starting values (0.5 kg, 1 m and $k=0.8$ N/m), $T_{\rm exch}\approx 13$ s.

Different pendulums

Different lengths: tuning decides everything

If the lengths differ, the natural frequencies $\omega_{01}=\sqrt{g/L_1}$ and $\omega_{02}=\sqrt{g/L_2}$ differ. For equal masses and weak coupling, at most the fraction

$$\eta\approx\frac{4\kappa^2}{\left(\omega_{01}^2-\omega_{02}^2\right)^2+4\kappa^2}$$

of the energy reaches the other pendulum, with $\kappa=k_{\rm eff}/m$ (the slanting spring counts only with $k\cos^2\beta$). If the difference in $\omega^2$ is large compared with the coupling, the other pendulum stays almost still. The formula holds for weak coupling but shows the trend for stronger coupling too. Make the second pendulum 30 cm longer with $k=0.8$ N/m: only about half the energy gets across, and the exchange is quicker. Turn $k$ down to 0.2 N/m, and the second pendulum gets less than a tenth. This is resonance in its purest form: energy moves efficiently only between oscillators that are tuned to each other.

Different masses: like an elastic collision

If the lengths are equal but the masses differ, the tuning is perfect, because both pendulums have $\omega_0=\sqrt{g/L}$. Even so, not all the energy gets across. The slow mode is still $\theta_1=\theta_2$, but in the fast one the common centre of mass stays still: $m_1\theta_1=-m_2\theta_2$. If pendulum 1 is pushed, the largest fraction of the energy that reaches pendulum 2 is

$$\eta=\frac{4m_1m_2}{(m_1+m_2)^2}.$$

This is exactly the fraction that an elastic head-on collision transfers from one ball to another at rest. A heavy and a light ball can never exchange all the energy, neither in a collision nor through a spring. Try 0.5 kg and 2 kg: $\eta=0.64$, and the simulation gives about 62%.

Pendulums on a taut string

The classic home experiment has no springs. The pendulums hang from their own points on a string stretched between two chairs, and they swing perpendicular to the string. The coupling goes through the string itself: when a pendulum swings out, it drags its point of suspension a little with it, and the string passes the pull on to its neighbours.

The equations

Let $y_j$ be the sideways displacements of the suspension points and $T$ the horizontal tension in the string. A piece of string of length $a$ acts as a sideways spring of stiffness $T/a$. The suspension points are massless, so the forces on them must cancel. The pendulum string pulls with the weight $m_jg$ obliquely down towards the bob, and for small swings the horizontal part is $m_jg\,\theta_j$:

$$\mathsf S\,\mathbf y=g\,\mathsf D\,\boldsymbol\theta,\qquad S_{jj}=\frac{T}{a_{j-1}}+\frac{T}{a_j},\quad S_{j,j\pm1}=-\frac{T}{a},\qquad \mathsf D=\mathrm{diag}(m_j).$$

The bob sits a distance $y_j+L_j\theta_j$ out to the side, and the only horizontal force on it is the pull of the string, $-m_jg\,\theta_j$. So $m_j(\ddot y_j+L_j\ddot\theta_j)=-m_jg\theta_j$. Substituting $\mathbf y$ gives

$$\mathsf A\,\ddot{\boldsymbol\theta}=-g\,\boldsymbol\theta,\qquad \mathsf A=\mathrm{diag}(L_j)+g\,\mathsf S^{-1}\mathsf D .$$

The string thus makes the pendulums effectively longer, and by how much depends on how the other pendulums are swinging. That is the coupling. The normal modes are found from the symmetric problem $\mathsf{DA}\,\ddot{\boldsymbol\theta}=-g\,\mathsf D\boldsymbol\theta$.

Two identical pendulums: the sag is the key

If two identical pendulums hang a distance $a$ apart and a distance $a_0$ from the chairs, the two modes are

$$\omega_1^2=\frac{g}{L+h}\quad(\text{in phase}),\qquad \omega_2^2=\frac{g}{L+h\,\dfrac{1/a_0}{1/a_0+2/a}}\quad(\text{in antiphase}),\qquad h=\frac{mg\,a_0}{T}.$$

Here $h$ is precisely the vertical sag that the pendulums' weight gives the string at the suspension points, since the vertical forces are governed by the same matrix $\mathsf S$. When the pendulums swing in phase, they thus become one sag longer. When they swing in antiphase, the middle section holds them in place, and the addition is smaller: a third of $h$ when $a=a_0$. The difference gives the beats.

It makes a practical rule of thumb. With 40 cm pendulums and a string that sags 5 cm, $T_{\rm exch}\approx 34$ s. With a 3 cm sag it takes just under a minute. A tight string gives slow exchange, and a slack string a fast one.

Different masses spoil the tuning

Here there is a difference from the spring coupling. A heavy pendulum makes the string sag more and so becomes effectively longer than a light pendulum of the same length. Two pendulums of equal length but different mass are therefore no longer tuned, and the exchange becomes both incomplete and quicker. Try it with one nut and with three.

Many pendulums: a chain and a waveguide

For $N$ identical pendulums with free ends the normal modes are standing waves:

$$u_j^{(n)}=\cos\frac{n\pi\left(j-\tfrac12\right)}{N},\qquad \omega_n^2=\frac gL+\frac{4k}{m}\sin^2\frac{n\pi}{2N},\qquad n=0,1,\dots,N-1 .$$

The table numbers them from 1 to $N$. The first has all the pendulums in phase; the last has neighbours in antiphase. All the frequencies lie in a band between $\sqrt{g/L}$ and $\sqrt{g/L+4k/m}$. With spacing $a$ between the pendulums and wavenumber $q=n\pi/(Na)$, the dispersion relation is

$$\omega^2=\omega_0^2+\frac{4k}{m}\sin^2\frac{qa}{2}.$$

Gravity gives a lower cut-off frequency $\omega_0$. Below it no wave can travel along the chain, and an oscillation driven at one end dies away exponentially. It is the same mathematics as in a waveguide with a cut-off frequency. In the continuum limit the equation becomes the Klein–Gordon equation $\ddot\theta=c^2\,\partial_x^2\theta-\omega_0^2\theta$ with $c^2=ka^2/m$.

Try it: add pendulums until there are eight, press Push the first, and watch in the graph how the energy travels down the chain as a wave packet and is thrown back from the end.

Where coupled oscillations turn up

Huygens' clocks. In 1665 Christiaan Huygens lay ill and noticed that two pendulum clocks hanging on the same beam always ended up swinging in antiphase after half an hour or so. He disturbed one of them, and they found each other again. The coupling went through the tiny movements of the beam. The experiment was repeated and explained in 2002.

Molecules. CO₂ is three masses coupled by two springs. The symmetric stretch, in which the carbon atom stays still, is infrared-inactive. The asymmetric stretch at 2349 cm⁻¹ and the bend at 667 cm⁻¹ absorb infrared light, and it is they that make CO₂ a greenhouse gas.

The Wilberforce pendulum. A weight on a helical spring can both bob up and down and twist about itself. The shape of the spring couples the two motions, and if they are tuned the weight slowly alternates between bobbing and twisting, exactly like the two pendulums.

Quantum mechanics. The formula for $\eta$ with different lengths is the same as Rabi's formula for a two-state system, and the same mathematics governs the tunnelling of the ammonia molecule and the oscillations of neutrinos between flavours.

A pendulum the size of the Earth. In inertial navigation the position error oscillates like a pendulum with the Earth's radius as its length and a period of 84.4 minutes. See the Schuler pendulum on the gyroscope page.

What you need

Two chairs with backs, 2–3 metres of thin, smooth string (kitchen string, builder's line or fishing line), 6–12 identical nuts, scissors, a tape measure and a stopwatch, e.g. on your phone. M10–M16 nuts are good, because they are heavy enough to pull the string straight and small enough not to be caught by the air. An M12 nut weighs about 16 g, but do weigh them on kitchen scales.

What to do

1. Stretch out the string. Stand the chairs back to back, about 1 m apart. Tie the string round both chair backs at the same height so that it is slightly taut. You adjust the tension later by pushing the chairs a little further apart or closer together.
2. Make two pendulums. Cut two pieces of string about 60 cm long. Thread each piece through three nuts and tie a double knot, so that the nuts sit like a small weight at the end.
3. Hang them up. Tie the pendulums to the taut string about 30 cm apart and about 30 cm from each chair. A half hitch that can be slid along makes it easy to adjust the length afterwards.
4. Make them the same length. Measure from the taut string to the middle of the nuts, and adjust until both pendulums are exactly the same length, e.g. 40 cm. This is the most important step: a difference of half a centimetre is visible.
5. Check the sag. The string should sag a little at the pendulums – a couple of centimetres. If it is completely taut, the exchange happens very slowly. If it hangs a lot, it happens quickly, but the pendulums easily start to wobble.
6. Start the experiment. Hold one pendulum still, and pull the other out 5–10 cm, perpendicular to the taut string. Release it gently, and then let go of the first one too, without pushing it.

After a few swings the still pendulum begins to move. After half a minute to a minute the first pendulum is almost still and the second is swinging fully. Then the motion goes back again.

Experiments you can do

The period of a pendulum. Time 10 full swings of one pendulum while the other is held still, and divide by 10. Compare with $2\pi\sqrt{L/g}$. For 40 cm it is 1.27 s.

The exchange period. Time from the moment the first pendulum is still until it is still the next time. That is $T_{\rm exch}$. Repeat a few times and take the average.

The two normal modes. Pull both pendulums out equally far to the same side and release them together. Now there is no exchange, because they are swinging in the first normal mode. Then pull them out to opposite sides: the second normal mode, which is slightly faster. Time 20 swings of each, $T_1$ and $T_2$, and calculate

$$T_{\rm exch}=\frac{T_1T_2}{T_1-T_2}.$$

Compare with the time you measured directly. The difference in period is small, so time many swings.

The tension. Push the chairs a little further apart, so that the string becomes tighter and the sag smaller. The exchange becomes slower. Slacken it, and it becomes faster.

The tuning. Make one pendulum 3–5 cm shorter. Now only part of the motion gets across, and the first pendulum never comes completely to rest.

The mass. Make the pendulums the same length again, but give one of them one nut and the other three. The exchange is again incomplete. On the string this is because the heavy pendulum makes the string sag more and becomes effectively longer.

The chain. Hang up three, four or more identical pendulums at equal spacing, and push the end one. Watch the motion travel down the row and back.

Measurement sheet

Tips

If the nuts start to rotate or swing in ellipses, start with a smaller displacement and release more gently. If you want the pendulums to stay in one plane, each weight can hang from two strings joined in a V from two points on the taut string. Use a smooth string: a string that twists up will turn the weight. And keep the doors shut, because even a slight draught disturbs it.

References

• A. P. French, Vibrations and Waves, Norton (1971), ch. 5.
• H. Goldstein, C. Poole & J. Safko, Classical Mechanics, 3rd ed., Addison-Wesley (2002), ch. 6.
• C. Huygens, Horologium Oscillatorium, Paris (1673).
• M. Bennett, M. F. Schatz, H. Rockwood & K. Wiesenfeld, ‘Huygens's clocks’, Proc. R. Soc. Lond. A 458, 563 (2002).
• R. E. Berg & T. S. Marshall, ‘Wilberforce pendulum oscillations and normal modes’, Am. J. Phys. 59, 32 (1991).