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Oscillations · Subtopic 2 of 3

Velocity and Energy in SHM

In SHM the speed is greatest at equilibrium, v0=ωx0v_0=\omega x_0, and zero at the extremes. At displacement xx, v=±ωx02−x2v=\pm\omega\sqrt{x_0^2-x^2}. Energy continually swaps between kinetic and potential, but the total stays constant (with no damping): E=12mω2x02E=\tfrac12m\omega^2x_0^2. Kinetic energy is 12mω2(x02−x2)\tfrac12m\omega^2(x_0^2-x^2) and potential energy is 12mω2x2\tfrac12m\omega^2x^2.

Cambridge 9702 Paper 4Edexcel YPH11 WPH15AQA 7408 Paper 1≈ 35 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Use v=±ωx02−x2v=\pm\omega\sqrt{x_0^2-x^2} and v0=ωx0v_0=\omega x_0.
  • Interpret displacement, velocity and acceleration graphs.
  • Describe the interchange of kinetic and potential energy.
  • Use E=12mω2x02E=\tfrac12m\omega^2x_0^2.
Pause and think

A mass on a spring is pulled down twice as far before release. How does its total energy change?

Section 2 of 14

Core idea

Start hereA swing's energy trade

At the top of a swing you pause: all potential energy. At the bottom you rush: all kinetic. The total stays the same (until friction wins), swapping back and forth twice per cycle.

The big idea

Etot=12mω2x02E_{\text{tot}}=\frac12m\omega^2x_0^2; Ek=12mω2(x02−x2)E_k=\frac12m\omega^2(x_0^2-x^2); Ep=12mω2x2E_p=\frac12m\omega^2x^2.

1. Energy curves

Energy against displacement
−1−0.50.510.20.40.60.811.20displacement xenergypotential ½mω²x²kinetictotal
Parabolas: at x=±x0x=\pm x_0 all potential; at x=0x=0 all kinetic.
Energies in SHM
  1. Ek=12mv2=12mω2(x02−x2)E_k=\tfrac12mv^2=\tfrac12m\omega^2(x_0^2-x^2)
    Using v2=ω2(x02−x2)v^2=\omega^2(x_0^2-x^2).
  2. Etot=Ek(x=0)=12mω2x02E_{\text{tot}}=E_k(x=0)=\tfrac12m\omega^2x_0^2
    All kinetic at the centre.
  3. Ep=Etot−Ek=12mω2x2E_p=E_{\text{tot}}-E_k=\tfrac12m\omega^2x^2
    Subtract.
Worked exampleThe 0.50 kg oscillator

ω=10\omega=10 rad s−1^{-1}, amplitude 4.0 cm.

Goal
Total; split at 2 cm; where Ek=EpE_k=E_p.
Strategy
Formulas.
  1. 12(0.5)(100)(0.0016)=0.040 J\tfrac12(0.5)(100)(0.0016)=0.040\text{ J}
    Total.
  2. Ep=0.010, Ek=0.030 JE_p=0.010,\ E_k=0.030\text{ J}
    At half amplitude: PE is a quarter.
  3. x2=12x02⇒x=4.02=2.8 cmx^2=\tfrac12x_0^2\Rightarrow x=\frac{4.0}{\sqrt2}=2.8\text{ cm}
    Equal shares.
Answer0.0400.040 J; 0.0100.010/0.0300.030 J; 2.82.8 cm.
✓ Check0.010+0.030=0.0400.010+0.030=0.040 ✓.
Velocity against displacement: an ellipse
−2−112−1−0.50.510xv
v2=ω2(x02−x2)v^2=\omega^2(x_0^2-x^2) traces an ellipse (a circle here with ω=1\omega=1). The point goes round once per oscillation.
Common misconceptionEqual energies at half amplitude At x=x02x=\frac{x_0}2, EpE_p is only 14\frac14 of the total.
Pause and think

Amplitude doubled: total energy?

Summary

  • Total ∝x02\propto x_0^2.
  • KE max at centre; PE max at extremes.
Key words
Amplitude
Maximum displacement.
Energy in simple harmonic motion
−1−0.50.510.20.40.60.811.20displacement x / x₀energy / total energykineticpotentialtotal
Kinetic energy 12mω2(x02−x2)\tfrac12m\omega^2(x_0^2-x^2) is greatest at the centre; potential energy is greatest at the extremes. Without damping their sum stays constant.
HistoryHooke's anagram. In 16761676 Robert Hooke published "ceiiinosssttuv", an anagram, to claim priority without revealing his idea. Two years later he decoded it: "ut tensio, sic vis" (as the extension, so the force). That linear restoring force is what makes a mass on a spring simple harmonic.
Fun factAtoms are springs too. Atoms in a crystal vibrate about their positions like masses on springs, at around 101310^{13} Hz. Their vibrational energy is part of the solid's internal energy. Heating the crystal increases their amplitude, which is why solids expand and eventually melt.
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