← all cardsTopic 10 · Oscillations · Energy
Energy of the oscillation = ½kA²
● VERBATIM — examined word-for-wordPull the mass a distance A below equilibrium, release from rest → total oscillation energy is ½kA² (A measured from equilibrium, k the spring constant). Then v_max = ωA, ω = √(k/m). This sidesteps every GPE-vs-EPE headache the question tries to cause.
E_osc = ½kA² = ½m v_max²What this actually means
Pull the mass a distance A below equilibrium and release from rest: the oscillation's total energy is ½kA², with A measured from equilibrium.
From there the chain is short: ½mv_max² = ½kA² gives v_max, and ω = √(k/m) gives everything else.
This is the vertical-spring shortcut in its most examinable form. No GPE-versus-EPE ledger required at any point.
Prove it — watch it be true
- On the energy-vs-displacement plot read the extreme x = ±x₀: Total is entirely potential there, which is ½kA² with A measured from equilibrium
- Read x = 0: the same Total is now entirely kinetic, so ½mv_max² = ½kA² and v_max = ωA
- Press the E ∝ x₀² doubling button: E tracks A² exactly, with no GPE-versus-EPE ledger opened at any point