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Double Pendulum Explorer

explore deterministic chaos and the butterfly effect in coupled motion

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why double pendulums become unpredictable

This is deterministic chaos, not randomness. Start two runs just 0.1° apart: after a while they no longer look related because separation grows roughly exponentially (a positive Lyapunov exponent). Energy keeps shuttling between the two arms through nonlinear coupling, so tiny phase differences get amplified into completely different future states. The equations are exact; what collapses is the prediction horizon.
This pendulum does not just go left-right; it keeps changing its mind about which arm leads the dance.

just two rods and gravity

There is no motor, no noise generator, no trick — only two arms, two pivots, and gravity. The wild motion comes entirely from the second joint, which lets each arm push the other around. Try starting with a small nudge near the bottom: it swings almost like an ordinary pendulum. Chaos only takes over once you give it enough energy for the arms to swing over the top.

the tip can fall faster than gravity

Release the pendulum from rest with the arms held out horizontally and watch the outer bob: for an instant it can accelerate downward faster than a freely dropped ball — more than g. The rigid arms act as a lever and whip, so the inner arm flings the outer one down harder than gravity alone ever could. It is the same effect that lets a hinged stick beat a falling coin to the floor.
Gravity starts the fall, but geometry decides who lands first.

Problem solved in full

  1. The period of two clean oscillations from release at rest 7 steps

    Released from rest at θ₁ = 20° and θ₂ = 10°, with m₁ = m₂ = 1, l₁ = l₂ = 1 and g = 9.81, and damping c = 0. Angles this small are not chaotic — the motion is a sum of two clean oscillations. Find the period of each, then decide which of them the cycle marks on the panel are counting.

    Open this problem in the tool

    1. Cut the equations of motion above down to size. Small angles mean sin θ becomes θ and cos 2Δ becomes 1, and the squared-velocity terms are second order in a small amplitude, so they go. Both denominators collapse to 2m₁ + m₂ − m₂ = 2, and what survives is linear and still coupled: neither bob moves without pushing the other.

      θ¨1=−2kθ1+kθ2,θ¨2=2kθ1−2kθ2,k=gℓ\ddot\theta_1=-2k\theta_1+k\theta_2,\qquad \ddot\theta_2=2k\theta_1-2k\theta_2,\qquad k=\frac{g}{\ell}

    2. Set a yardstick before solving anything. A single 1 m pendulum swings at ω₀ = √(g/l), giving the 2.01 s the panel prints as its period hint and stamps on every cycle mark. Hold on to it, because nothing below is going to equal it.

      k=9.811=9.81 s−2,ω0=k=3.132 rad/s,T0=2πω0=2.01 sk=\frac{9.81}{1}=9.81\ \mathrm{s^{-2}},\quad \omega_0=\sqrt{k}=3.132\ \mathrm{rad/s},\quad T_0=\frac{2\pi}{\omega_0}=2.01\ \mathrm{s}

    3. A normal mode is a motion in which both angles oscillate at one shared frequency and keep a fixed ratio. Substitute θᵢ = aᵢ cos ωt and each of the two equations hands you that ratio a₂/a₁, arrived at from opposite ends.

      θi=aicos⁡ωt ⟹ (2k−ω2)a1=k a2,(2k−ω2)a2=2k a1\theta_i=a_i\cos\omega t\ \Longrightarrow\ (2k-\omega^{2})a_1=k\,a_2,\quad (2k-\omega^{2})a_2=2k\,a_1

    4. The two expressions for one ratio must agree, and setting them equal eliminates both amplitudes. What is left is a quadratic in ω², and a quadratic has 2 roots. That is where the second period comes from.

      (2k−ω2)2=2k2⟹ω2=(2∓2) k=5.747, 33.49 s−2(2k-\omega^{2})^{2}=2k^{2}\quad\Longrightarrow\quad \omega^{2}=(2\mp\sqrt2)\,k=5.747,\ 33.49\ \mathrm{s^{-2}}

    5. So this pendulum has 2 periods, 2.62 s and 1.09 s, and the 2.01 s on the cycle marks is neither. It lies between them and belongs to neither one.

      ω−=2.397,  ω+=5.787 rad/s⟹T−=2.62 s,  T+=1.09 s\omega_-=2.397,\ \ \omega_+=5.787\ \mathrm{rad/s}\quad\Longrightarrow\quad T_-=2.62\ \mathrm{s},\ \ T_+=1.09\ \mathrm{s}

    6. Each root carries its own shape back. The slow root gives a₂/a₁ = +1.414: both bobs swing the same way, the lower one 1.414 times wider. The fast root gives −1.414: they swing in opposition and the joint between them scissors.

      a2a1=2k−ω2k=±2=±1.414\frac{a_2}{a_1}=\frac{2k-\omega^{2}}{k}=\pm\sqrt2=\pm1.414

    7. A start of 20° and 10° is neither shape, so both modes are running at once. Split it into a slow amplitude A and a fast amplitude B — the top bob needs A + B = 20°, the bottom needs √2(A − B) = 10° — and the slow mode takes 13.54° of the start while the fast one takes 6.46°.

      A+B=20∘,  2 (A−B)=10∘⟹A=13.54∘,  B=6.46∘A+B=20^\circ,\ \ \sqrt2\,(A-B)=10^\circ\quad\Longrightarrow\quad A=13.54^\circ,\ \ B=6.46^\circ

    Answer

    2.62 s and 1.09 s, and their ratio is exactly 1 + √2. An irrational ratio means the 2 modes never come back into step, so the figure the lower bob traces never quite repeats — and this is the tame small-angle case, with no chaos in it anywhere. It also settles what the 2.01 s cycle marks are: a single-pendulum ruler laid against a 2-frequency motion, which is why the angles logged at each mark keep drifting instead of returning to 20.0° and 10.0°. Push the release to 145° and even the 2 frequencies go, because sin 145° = 0.5736 while the angle itself is 2.531 rad — wrong by a factor of 4.4, and the linearisation this whole derivation stands on has nothing left to say.

    T−T+=2+22−2=1+2=2.4142…\frac{T_-}{T_+}=\sqrt{\frac{2+\sqrt2}{2-\sqrt2}}=1+\sqrt2=2.4142\ldots
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