← all cardsTopic 8 · 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
- Pull the mass down a distance A from equilibrium and release
- Read the total energy bar the moment of release: all potential, ½kA²
- Watch it convert fully to KE at the centre: ½mv_max² equals the same number