Energy in simple harmonic motion

The equation of motion m dv/dt = −kx is written in terms of time. Multiply both sides by v = dx/dt. By the chain rule the left side becomes mv dv/dt = d/dt(½mv²), and the right side becomes −kx dx/dt = −d/dt(½kx²). Collect them and d/dt(½mv² + ½kx²) = 0: the combination is frozen for as long as the motion obeys Hooke's law. Work gives the same ½kx². As the mass moves from x₁ to x₂, the spring does work W = ∫ −kx dx over that interval, which equals ½kx₁² − ½kx₂² and depends only on the endpoints, so it defines a potential energy U(x) = ½kx² with U = 0 at equilibrium. Nothing here used the sinusoidal solution. That is why energy can answer questions about position and speed that never mention t, and why the same multiplication, with any extra force F included, returns dE/dt = Fv: the power that force feeds in or drains out.

Substitute x = A cos(ωt + φ) and v = −ωA sin(ωt + φ). Using mω² = k, U = ½kA² cos²(ωt + φ) and K = ½kA² sin²(ωt + φ), and because cos² + sin² = 1 the sum is ½kA² at every instant. Rewrite cos²θ as ½(1 + cos 2θ) and each part is seen to oscillate at 2ω about E/2: the energy exchange repeats every half period, never dips below zero, and averages E/2 for each part. Differentiate to get the rate of exchange: dU/dt = −ωE sin[2(ωt + φ)]. Its largest size, ωE, comes where x = ±A/√2, not at equilibrium, where the spring exerts no force, nor at a turning point, where the mass is momentarily still. A 0.40 kg glider on a 160 N m⁻¹ spring with a 5.0 cm amplitude has ω = 20 rad s⁻¹ and E = 0.20 J, so the spring passes energy to the glider at up to 4.0 W, and U runs from E to 0 and back in T/2 = 0.157 s.

Set ½mv² + ½kx² equal to its value at a turning point, ½kA², and solve: v² = (k/m)(A² − x²), so v = ±ω√(A² − x²). Energy tells you the speed at a place but not the direction, which is why the sign stays open; the mass passes each interior point twice per cycle, once each way. The relation is lopsided. Well out from the centre, at x = 0.6A, the mass keeps √(1 − 0.36) = 80% of its top speed, and the speed collapses only in the last stretch: at x = 0.9A it is √0.19 = 44% of ωA, and at 0.99A it is 14%. Because the mass lingers where it is slow, it spends most of each cycle near the ends. So the time average of K over a cycle is E/2, less than the 2E/3 you get by averaging K = E(1 − x²/A²) evenly over every position between −A and +A.

Draw U(x) = ½kx² as a parabola and E as a horizontal line. At any x, the height of the parabola is U and the gap from the parabola up to the line is K = E − U. Kinetic energy cannot be negative, so motion is allowed only where the line lies above the curve, and the two crossings, where U(x) = E, are the turning points x = ±A. The slope of the curve gives the force, F = −dU/dx = −kx, always pointing downhill toward equilibrium, and the curvature d²U/dx² = k together with the mass sets ω = √(k/m). Raise the line and the crossings spread, but only as √E, because U climbs as x². A tap on a moving mass raises its line without touching the parabola, so the diagram shows at once how far the next swing will reach, and a slow loss to friction lowers the line cycle by cycle, pulling both turning points in toward the centre.

Plot the state as a point with x across and v up. Dividing ½mv² + ½kx² = E through by E gives x²/A² + v²/(ωA)² = 1, an ellipse with semi-axes A and ωA, so a whole oscillation is one closed curve. The point runs clockwise: at x = +A it has v = 0, then v turns negative as the mass heads back, so the point drops below the axis on its way to x = 0. Different energies give nested ellipses that never cross, because a state (x, v) has one energy and one future. This is the fastest route from a snapshot to the amplitude. A glider with ω = 20 rad s⁻¹ caught at x₀ = 3.0 cm moving at 0.80 m s⁻¹ has v₀/ω = 4.0 cm, so it lies on the ellipse with A = √(3.0² + 4.0²) cm = 5.0 cm. Plot momentum p = mv instead of v and the area enclosed, π × A × mωA, equals E × T, the energy times the period.

Two assumptions sit under E = ½kA², and real oscillators break both. The first is that nothing removes energy. Add a drag force −bv and the multiplication by v gives dE/dt = −bv², which is never positive: E drains fastest where the mass moves fastest, and the phase-space point spirals inward instead of closing its ellipse. Because E ∝ A², losses look larger in energy than in amplitude; an amplitude that shrinks by 5% each cycle means 1 − 0.95² = 9.75% of the energy gone per cycle. The second assumption is that U is exactly ½kx². A real spring stiffens or yields at large stretch, and a pendulum's U = mgL(1 − cos θ) falls below the parabola ½mgLθ², by 2.3% at θ = 30° and 8.8% at 60°. Energy is still conserved, so K = E − U(x) still gives the speed. What fails is the sinusoid, and with it the amplitude-independent period: a 30° swing already takes about 1.7% longer than 2π√(L/g).