PhysicsLab

Oscillations · Energy in SHM

Energy in SHM Lab

E = ½mω²x₀² = 0.395 J
x = 0.200 mKE 0.000 JPE 0.395 JKE+PE 0.395 J

Energy vs displacement

Cross-over sits on x₀/√2 at ½E; at x₀/2 the cyan dot sits on ¾E.

ⓘ more

Both parabolas come straight from v = ±ω√(x₀²−x²): squaring it gives KE = ½mω²(x₀²−x²), and PE = ½kx² = ½mω²x² is what is left over. Their sum is the flat emerald line, so the two dots always trade height and never change the total.

Energy vs time — and x vs t below it

Count the numbered bands: 4 energy cycles above, 2 motion cycles below.

f_motion

1.00 Hz

f_energy

2.00 Hz

f_energy / f_motion

2.00

ⓘ more

With x = x₀cos ωt, PE = ½mω²x₀²cos²ωt = ½E(1 + cos 2ωt) and KE = ½E(1 − cos 2ωt). The cos 2ωt is the whole story: angular frequency 2ω, period T/2, frequency 2f. Count the numbered bands — four energy cycles sit above two motion cycles, and both panels share one time axis so the count is not a claim, it is visible.

Try first: snap to x = x₀/2 — KE reads ¾E.

Standard fractions — snap and read off

x/x₀ = 1.000

KE/E = 1 − (x/x₀)² = 1 − (1.000)² = 0.000

PE/E = (x/x₀)² = 1.000

x = 0 → all KEKE = ¾E, PE = ¼EKE = PE = ½Ex = ±x₀ → all PE
x0.200 m
v = ±ω√(x₀²−x²)0.000 m s⁻¹
KE = ½mv²0.000 J
PE = ½kx²0.395 J
KE + PE0.395 J

Controls

ω = 2πf6.28 rad s⁻¹
k = mω²19.7 N m⁻¹
T = 1/f1.00 s
v₀ = x₀ω1.26 m s⁻¹

E ∝ x₀² and E ∝ f²

E now
0.395 J
baseline
0.395 J
(x₀/x₀ᵣ)²1.000
(f/fᵣ)²1.000
m/mᵣ1.000
product1.000
E/Eᵣ measured1.000

Pin a baseline, then drag anything: the product row and the measured row stay equal.

What this confirms
  • ›KE = ½mω²(x₀² − x²) — LO (f) — the syllabus gives v = ±ω√(x₀²−x²); square it into ½mv² and you get the cyan inverted parabola. The KE readout and ½mv² from the v readout agree at every snap.
  • ›PE = ½mω²x² = ½kx² — the violet upright parabola, zero at the centre and E at the extremes, with k = mω² shown live in Controls.
  • ›E = ½mω²x₀² and KE + PE is constant — LO (h) — the emerald line is flat on both graphs, the 3D column never changes height, and the KE+PE chip holds still while the two parts trade.
  • ›Energy graphs run at 2f while x runs at f — LO (g), (h) — four numbered energy bands sit directly above two numbered motion bands on one shared time axis, and f_energy/f_motion reads 2.00 for every f you choose.
  • ›x = x₀/2 → KE = ¾E, PE = ¼E — the arithmetic 1 − (0.500)² = 0.750 is printed, and the cyan dot lands on the ¾E gridline. Standard exam fraction, watched rather than memorised.
  • ›KE = PE at x = x₀/√2 — the two parabolas cross exactly on the 0.707x₀ marker at ½E; the KE = PE badge lights only there.
  • ›E ∝ x₀² and E ∝ f² — the ×2 buttons report E ×4.00 with both energies shown, and the pinned-baseline ratio panel keeps (x₀/x₀ᵣ)²(f/fᵣ)²(m/mᵣ) equal to the measured E/Eᵣ for any drag.
  • ›Vertical spring: ½kx² about equilibrium absorbs gravity — GPE and EPE split differently at the lowest, equilibrium and highest points, yet the Total column is identical, because the kex and mgx terms cancel when x is measured from equilibrium.