An undamped swing trades energy between the moving mass and the stretched spring and never loses any. The total, ½m·v² + ½k·x², is the same at every instant and equals its starting value E₀. Plotted as a path in the plane of position against velocity, that conservation is an ellipse with semi-axes A and A·ω₀, traced once per period.
Core ideas
Three ideas: the two kinds of energy, the proof that their sum is fixed, and the picture the proof draws.
The moving mass carries kinetic energy ½m·v² where v = x′; the stretched spring stores potential energy ½k·x². At the ends of the swing the mass is momentarily still, v = 0, and the energy is all in the spring. Passing through the rest position, x = 0, the spring is relaxed and the energy is all in the motion. In between it is shared.
Differentiate E = ½m·v² + ½k·x² with the chain rule, using x′ = v and v′ = x″. The rate of change is E′ = m·v·x″ + k·x·v = v·(m·x″ + k·x), and the bracket is zero by the equation. So E has the value E₀ it had at the start, for ever.
Plot each instant as the point (x, v). Because E is fixed, the points satisfy ½m·v² + ½k·x² = E₀, which is an ellipse centred on the origin, and the mass travels round it once in each period. The plane of x against v is the phase plane; “From Second Order to a System” in the previous course first drew it, and the next course returns to it in earnest.
The material
The energy, the proof that it is fixed, the ellipse it gives, and two starts that show where the energy sits at the ends and the middle of a swing.
For the mass on a spring with position x and velocity v = x′, the energy is E = ½m·v² + ½k·x²: kinetic energy plus the energy stored in the spring.
Its starting value is E₀ = ½m·v₀² + ½k·x₀², an exact fraction when the mass, the spring constant and the start are fractions.
The factor ½ appears in both terms and it is the first thing to check in any answer. Without it the two terms are still in the right ratio, but the number is twice the true energy.
If m·x″ + k·x = 0, then E = ½m·v² + ½k·x² is constant along every solution: E(t) = E₀ for all t.
With v = x′, the derivative of ½m·v² is m·v·v′ = m·v·x″ by the chain rule, and the derivative of ½k·x² is k·x·x′ = k·x·v.
So E′ = v·(m·x″ + k·x). By the equation the bracket is 0, so E′ = 0 and E does not change. With a damper the bracket is −c·x′ and E′ = −c·v², which is never positive: the energy only falls, and it is the subject of the next module.
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Divide ½m·v² + ½k·x² = E₀ by E₀. The result is x²/(2E₀/k) + v²/(2E₀/m) = 1. The first denominator is exactly A², because A² = 2E₀/k, and the second is A²·k/m = (A·ω₀)². So the path is
The mass goes round it clockwise: where v > 0, at the top, x is increasing, so the point moves to the right. The shape of the ellipse is set by ω₀ alone, and its size by the start.
With m = 1, k = 4, start (3, 8): E₀ = ½·64 + ½·4·9 = 32 + 18 = 50. The semi-axes are A = 5 and A·ω₀ = 10.
Check the ends: at x = 5 and v = 0 the energy is ½·4·25 = 50, and at x = 0 and v = 10 it is ½·100 = 50. Both are E₀, so both points are on the ellipse.
With m = 2, k = 5, start (1, 0): E₀ = ½·2·0 + ½·5·1 = 5/2. The amplitude squared is A² = 2E₀/k = 1, which is x₀² as it should be for a release from rest. The ellipse is tall and thin or short and wide according to ω₀.
The mass does not appear in E₀ for a start at rest. It appears only in how fast the energy is exchanged.
Everything the previous lesson computed about a swing can now be read off the ellipse. The amplitude is where it crosses the position axis, the peak speed is where it crosses the velocity axis, and the area it encloses, π·A·(A·ω₀) = 2π·E₀/√(mk), is fixed by the energy, the mass and the spring.
Interactive
Free, damped and forced motion, with every exact quantity
Method
Four moves. Check the factor of one half at the first of them.
E₀ = ½m·v₀² + ½k·x₀². Square first, then multiply, then halve. For m = 1, k = 4, start (3, 8) it is 32 + 18 = 50.
A² = 2E₀/k. It must agree with x₀² + v₀²/ω₀² from the previous lesson; for E₀ = 50 and k = 4 it is 25.
A along the position axis and A·ω₀ along the velocity axis. Here 5 and 10, so the ellipse is twice as tall as it is wide.
At the position axis (v = 0, x = A) and the velocity axis (x = 0, v = A·ω₀) recompute the energy. Both must equal E₀.
The first preset: m = 1, k = 4, x(0) = 3, x′(0) = 8.
Both ends check. The starting point (3, 8) is on the ellipse too: 9/25 + 64/100 = 36/100 + 64/100 = 1. A point that failed this would mean an arithmetic slip in E₀, A or ω₀.
You should be able to compute E₀ = ½m·v₀² + ½k·x₀² exactly, show that E′ = v·(m·x″ + k·x) is zero, write the ellipse x²/A² + v²/(A·ω₀)² = 1 with its semi-axes, and say where the energy sits at the ends and the middle of a swing.
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