The v–x graph is an ellipse
● VERBATIM — examined word-for-wordFrom v = ±ω√(x₀² − x²). Cuts the x-axis at ±x₀ (v = 0 at extremes), the v-axis at ±ωx₀ (v_max at equilibrium); traversed clockwise (x right, v up). Label questions: release point sits on the x-axis at the release displacement; first return to equilibrium is the v-axis intercept. With damping the curve spirals inward — same centre, shrinking amplitude.
(x/x₀)² + (v/ωx₀)² = 1What this actually means
Plot v against x across one cycle and you trace an ellipse: it cuts the x-axis at ±x₀ (speed zero at the extremes) and the v-axis at ±ωx₀ (top speed through equilibrium), traversed clockwise with x rightward and v upward.
It is the equation v = ±ω√(x₀² − x²) drawn whole, or equivalently the circular-motion phasor with its axes scaled.
Labelling questions: the release point sits ON the x-axis at the release displacement (v = 0 there); the first pass through equilibrium is the v-axis intercept. With damping, the ellipse becomes an inward spiral about the same centre.
Prove it — watch it be true
- Watch the phasor tip trace its circle
- Its horizontal projection is x, its vertical rate is v: together they draw the ellipse
- At the sides v = 0 (extremes); at top speed x = 0 (equilibrium)