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Standard fractions
At x = x₀/2 → KE = ¾E_total. KE = PE at x = x₀/√2 ≈ 0.707x₀. One line each: KE/E = 1 − (x/x₀)².
What this actually means
The one-line tool: KE/E = 1 − (x/x₀)². No mass, no ω, no timing needed.
At half amplitude the kinetic energy is ¾ of the total, not half, because energy is quadratic in x. And KE equals PE at x = x₀/√2 ≈ 0.707x₀, not at x₀/2.
Both results drop straight out of the one-liner; quote it, substitute, done.
The trap
Assuming KE = PE at half amplitude — the crossover is at x₀/√2.
Prove it — watch it be true
- Pause the mass at half amplitude
- Read the bars: KE is three-quarters of the total
- Nudge outward until the bars match: that happens near 0.707x₀