Playground Swing

Squatting in time with the swing does exactly nothing

Write l(t) for the distance from the pivot to the rider's centre of mass. Substituting u = l θ into the exact equation l θ″ + 2 l′ θ′ + g sin θ = 0 removes the first-derivative term without any approximation and leaves

u″ + ( g/l − l″/l ) u = 0.

Now squat sinusoidally at frequency W, so l = l₀(1 + ε cos W t). The bracket becomes (ω₀² + ε W² cos) / (1 + ε cos), and at W = ω₀ that is ω₀² — a constant. The gravity term and the inertial term cancel each other exactly. Not to first order in the squat depth: exactly.

The monodromy trace over one pump period is against the theoretical 2, at a squat depth of 0.35 m, and the nonlinear engine run for periods changes the peak amplitude by . The same rider squatting twice per swing multiplies the amplitude by in twelve periods.

That exact cancellation needs a point-mass rider. Give the rider a moment of inertia about their own centre of mass and the trace drops below 2 — still no growth, just no longer exactly marginal. I expected the null to break; it does not. What does leak past it is that a rider who watches the swing instead of a clock cannot produce a pure sinusoid: the autopilot's own motion carries a second harmonic the size of the first, and that is what creeps upward.

The stability chart

Linearised, the squatting swing is Mathieu's equation y″ + (a − 2q cos 2τ) y = 0 with a = 4ω₀²/W² and q = 2ε(ω₀² − W²)/W². The shaded wedges are where solutions grow. Squatting twice per swing puts you at a = 1, dead centre of the widest one, with q = −3ε/2. Squatting once puts you at a = 4 — and, worse, at q = 0, because the numerator vanishes there.

Curves: the transition series of DLMF 28.6. Dots: the same boundaries located by bisecting on the monodromy trace, sharing no code with the series. They agree to on the first tongue and on the second, across values of q.

Growth rate against pumping frequency

Sweep the squat frequency and measure the Floquet exponent of the real engine at each one. There is one peak, it is at times the swing frequency, and only of the sampled band grows at all.

Why children start seated and finish standing

A lean is a torque, so it adds amplitude at a constant rate: the swing climbs in a straight line and it works from a dead stop. A squat is parametric, so it multiplies whatever amplitude is already there: it climbs exponentially and it cannot start at all — θ = 0 is an exact equilibrium of the squatting equation. Give both riders the same tiny nudge and the straight line leads for seconds before the exponential goes past it, and then never looks back. The standing rider runs out of road at the 90-degree slack limit after seconds; at that moment the seated rider is still at degrees.

In this rig the two channels cross at degrees of amplitude: below that the lean is worth more, above it the squat is. Post and colleagues (2007) motion-captured eighteen people pumping from standstill and found the same shape — driven oscillation dominant, parametric subordinate, and the parametric share rising as the swing got higher.

The rider's own moment of inertia is the whole seated story

Work the lean through the Lagrangian and the first-order driving torque comes out as

D = m g I_c L α₀ / M₀,

exactly proportional to I_c, the rider's moment of inertia about their own centre of mass. For a point mass it is identically zero: leaning is then nothing but a change of coordinate, and the swing does not notice. The engine, which was never given this formula, grows at rad/s against a predicted — apart. Set I_c = 0 and the same run gives , smaller by a factor of .

Leaning once per swing also moves the centre of mass off the rope line, and that offset goes as the square of the lean — so it modulates the pivot distance twice per swing and drives the parametric tongue too, at second order. That channel survives at I_c = 0, which is why the point-mass rider is not quite dead.

The exponent is a cube, and I could not find it in print

Take the limit where the rider changes height instantaneously at the bottom. The impulsive force is along the rope, so it exerts no torque about the pivot and ρ² θ′ is conserved across it. Carrying that through gives, per half swing and at any amplitude,

1 − cos θ_next = (l_long / l_short)³ (1 − cos θ).

A cube. The engine reproduces it to relative over four half swings. And because that test feeds the impulse assumption in, there is a second one that does not: ramp the length change smoothly over a time τ and shrink τ. At 10 ms the ordinary integrator lands within of the cube, while the cube and a square differ by — so the test can tell them apart, and it picks the cube.

The only discrete-jump relation I could reach in print is Tea and Falk (1968), equation 4, and as transcribed from the publisher's rasterised first-page preview it is a square. The full text is paywalled. I am not claiming the paper is wrong — I am reporting that three independent routes here give a cube, that I could not read the paper, and that the discrepancy is open.

Ninety degrees, and no further

Released from rest at amplitude A, the chain tension is T = m g (3 cos θ − 2 cos A). It is times body weight at the bottom of a 90-degree swing, and it reaches zero at the turning point exactly when A = 90°. Past that the chains go slack and the swing stops being a pendulum — which is why the game ends the round there. Pump hard enough and one single stroke takes you from degrees straight past the limit.

Counter-intuitively, the angular return per pump grows with amplitude rather than shrinking: per half swing near zero, at 1.2 radians, because 1 − cos θ saturates at 2 while the energy gain does not. I had the sign of that backwards until the harness said so.

What was checked, and against what

Every number on this page is written by the harness into js/figures.js and re-checked from the engine by the page harness. None of them is typed into the HTML.