Learn9702Paper 4 · A Level Structured QuestionsOscillationsVelocity and energy in SHM

Oscillations · Subtopic 2 of 3

Velocity and Energy in SHM

In SHM the speed is greatest at equilibrium, $v_0=\omega x_0$ , and zero at the extremes. At displacement $x$ , $v=\pm\omega\sqrt{x_0^2-x^2}$ . Energy continually swaps between kinetic and potential, but the total stays constant (with no damping): $E=\tfrac12m\omega^2x_0^2$ . Kinetic energy is $\tfrac12m\omega^2(x_0^2-x^2)$ and potential energy is $\tfrac12m\omega^2x^2$ .

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

Section 1 of 14

Learning objectives

Use $v=\pm\omega\sqrt{x_0^2-x^2}$ and $v_0=\omega x_0$ .
Interpret displacement, velocity and acceleration graphs.
Describe the interchange of kinetic and potential energy.
Use $E=\tfrac12m\omega^2x_0^2$ .

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

$E_{\text{tot}}=\frac12m\omega^2x_0^2$ ; $E_k=\frac12m\omega^2(x_0^2-x^2)$ ; $E_p=\frac12m\omega^2x^2$ .

1. Energy curves

Energy against displacement

time = 0.0

Parabolas: at

x=\pm x_0

all potential; at

x=0

all kinetic.

Energies in SHM

$E_k=\tfrac12mv^2=\tfrac12m\omega^2(x_0^2-x^2)$
Using $v^2=\omega^2(x_0^2-x^2)$ .
$E_{\text{tot}}=E_k(x=0)=\tfrac12m\omega^2x_0^2$
All kinetic at the centre.
$E_p=E_{\text{tot}}-E_k=\tfrac12m\omega^2x^2$
Subtract.

Worked exampleThe 0.50 kg oscillator

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

Goal: Total; split at 2 cm; where $E_k=E_p$ .
Strategy: Formulas.

$\tfrac12(0.5)(100)(0.0016)=0.040\text{ J}$
Total.
$E_p=0.010,\ E_k=0.030\text{ J}$
At half amplitude: PE is a quarter.
$x^2=\tfrac12x_0^2\Rightarrow x=\frac{4.0}{\sqrt2}=2.8\text{ cm}$
Equal shares.

Answer

0.040

0.010

0.030

2.8

cm.

✓ Check

0.010+0.030=0.040

✓.

Velocity against displacement: an ellipse

time = 0.0

v^2=\omega^2(x_0^2-x^2)

traces an ellipse (a circle here with

\omega=1

). The point goes round once per oscillation.

Common misconceptionEqual energies at half amplitude At

x=\frac{x_0}2

E_p

is only

\frac14

of the total.

Pause and think

Amplitude doubled: total energy?

Summary

Total $\propto x_0^2$ .
KE max at centre; PE max at extremes.

Amplitude: Maximum displacement.

Energy in simple harmonic motion

Kinetic energy

\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

1676

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

10^{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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