Answers · The physics underneath

Why does pushing a swing at just the right moment make it go much higher?

In shortA swing goes higher when each push arrives in step with the motion already there, so many small pushes add up instead of cancelling.

The swing goes higher because each push arrives when the swing is already moving away, so the push does positive work and adds a little energy every cycle; the large amplitude is the accumulation of many small additions, not the size of one shove. A gentle push in step can build a large swing, and a hard shove at the wrong moment can remove energy rather than add it.

The motion is an oscillator with a push and a leak:

m \ddot x + b \dot x + k x = F \cos(\omega t)

Here \(m\) is mass, \(k\) the restoring strength, \(b\) the leak, and \(F\cos(\omega t)\) the push. The system's own rate is \(\omega_0=\sqrt{k/m}\), and the pushing rate as a multiple of it is \(r=\omega/\omega_0\). The leak parameter is \(d=b/(m\omega_0)\). At a steady swing the amplitude is \(A=(F/k)/\sqrt{(1-r^2)^2+(dr)^2}\).

Why timing multiplies the push

At \(r=1\) the push is in step with the velocity already present, so \(F\cos(\omega t)\) and \(\dot x\) have the same sign over each cycle and the work \(F\dot x\) per cycle is positive. Off that rate the two drift out of phase: for part of the cycle the push helps, for part it hinders, and the net work per cycle is small. That is why the response curve is a peak rather than a slope. The push size sets the scale of the whole curve; the timing decides whether the pushes add or cancel before that scale is applied.

FIG 1

pushing rate, multiple of own rate	denominator	steady amplitude
0.5	0.7649	1.31
0.8	0.4327	2.31
1.0	0.3000	3.33
1.2	0.5685	1.76

The natural guess is that a harder push always wins. It does raise the whole response curve in proportion to \(F\), but it does not move the peak. With \(d=0.3\), one unit of force at the own rate gives \(3.33\). At half the own rate it gives \(1.31\), so a push twice as hard there gives \(2.61\), still below the gentle on-time push. At \(r=0.8\), a push twice as hard gives \(4.62\), which beats \(3.33\), so a large enough wrong-timed push can win in absolute size; what it cannot do is win per unit of effort, because most of its work is undone on the next part of the cycle. A push that arrives half a cycle late actively subtracts energy.

What sets the height

The peak is not reached on the first push. If each cycle adds \(P\) and the leak removes fraction \(r\) of the current swing, then the next amplitude is given by

a_{k+1}=a_k+P-r a_k

With \(P=1\), the steady value is \(1/r\). At \(r=0.2\) it is \(5\); the sequence from rest is \(1,1.8,2.44,2.95,3.36\), closing on \(5\). At \(r=0.05\) it is \(20\), and at \(r=0.5\) it is \(2\). A small leak gives a tall swing but takes longer to get there, because the same push is being stored for more cycles before the loss catches up with it.

FIG 2

cycle	r = 0.05	r = 0.2	r = 0.5
1	1.000	1.000	1.000
2	1.950	1.800	1.500
5	4.524	3.362	1.938
10	8.025	4.463	1.998

How narrow the timing has to be

The same leak fixes the width of the peak. For this model \(Q=1/d\), and the fractional width is \(\Delta f/f_0=1/Q=d\). A system with \(Q=1000\) responds over a band \(1/1000=0.001\) of its own rate, which is \(0.1\%\); at \(f_0=500\) Hz that band is \(500\times0.001=0.5\) Hz. A system with \(Q=10000\) has width \(0.0001\), one part in ten thousand, so at \(100\,000\,000\) Hz it accepts a band of \(100\,000\,000\times0.0001=10\,000\) Hz. Tall and narrow are the same fact read twice: lowering the leak raises the peak and narrows the band together.

The linear formula stops being accurate when the swing is large enough that the restoring force is no longer proportional to displacement, or when the leak changes with amplitude. Then the peak can shift and a harder push can change the shape of the response, not just its height. If the motion itself changes the push, as a twisting deck changes the airflow around it, the system can supply its own timing and no external rate has to match; that is a different mechanism and it is not capped by the leak in the same way.

Worth keeping

Timing decides whether pushes add or cancel; push size only scales the whole response.
A small leak gives a tall, narrow response and a slow build-up, while a large leak gives a low, broad response that arrives quickly.
At resonance the steady amplitude is the push per cycle divided by the fraction lost per cycle.

Where this goes next

This question is one step of Waves, Modes and Frequency, a course that works through its subject from the beginning. The page that teaches this part of it is Pushing at the Right Moment, and it reads the whole way through whether or not you have an account.

What an account changes is the practice: problems standing under every idea, marked with a reason rather than a tick, in a copy of the course that is yours.

Read that page The whole course