Physics 9702 · Paper 4

Energy in SHM

An undamped oscillator is a sealed energy account: kinetic and potential trade places twice per cycle while the total never moves. One formula prices the whole account.

The interchange (a three-marker in words)

The verbal version scores three separate marks (9702/42/M/J/23 Q4(c)): the sum of kinetic and potential energy is constant; at maximum displacement the kinetic energy is zero (all potential); at zero displacement the kinetic energy is maximum (potential minimum).
For a mass between springs the potential energy is elastic; for a pendulum it is gravitational — the interchange story is identical.

Energy against displacement: PE = ½mω²x² (parabola), KE = ½mω²(x₀²−x²) (upside-down parabola). At every x they add to the same flat total ½mω²x₀².

Both energy curves are parabolas in $x$ (PE $\propto x^2$ , KE is the total minus that), and sketching them is itemised: dome shape peaking at $x = 0$ , the peak labelled with the total energy, the ends grounded at $\pm x_0$ — one mark each (9702/42/O/N/24 Q5(c)). The potential-energy version is the same U-shape upside down (9702/41/M/J/25 Q5(c)).
Against time, KE and PE each complete two full cycles per oscillation — the oscillator passes through equilibrium (KE max) twice per period.

The total: E = ½mω²x₀²

E = \tfrac{1}{2}m\omega^2 x_0^2

Symbols

$E$ = total energy of the oscillation (J)
$m$ = oscillating mass (kg)
$\omega$ = angular frequency (rad s⁻¹)
$x_0$ = amplitude (m)

Nothing new — it is the maximum kinetic energy $\tfrac{1}{2}mv_0^2$ with $v_0 = \omega x_0$ substituted. Since energy grows with the squares of both amplitude and frequency, this is the most-tested calculation in the topic — seven appearances in three years.

Worked example

Backwards from energy to period (2023 paper)

A 130 g mass oscillates with total energy 6.4 mJ and amplitude 1.5 cm. Find the period (9702/42/F/M/23 Q3(b)).

$6.4\times10^{-3} = \tfrac{1}{2} \times 0.130 \times \omega^2 \times 0.015^2$ .
$\omega^2 = 438$ , so $\omega = 20.9\ \text{rad s}^{-1}$ .
$T = 2\pi/\omega = 0.30\ \text{s}$ .

Answer

T = 0.30\ \text{s}

One graph, five conclusions

A 2025 question showed only two graphs for a floating block — a straight a–x line and a kinetic-energy dome — and asked for three quantitative conclusions. Everything fell out of this lesson's toolkit: equilibrium position (where the line crosses zero), total energy (the dome's peak),

\omega

(from the gradient), then

v_0 = \omega x_0

and even the mass from

E = \tfrac{1}{2}m\omega^2x_0^2

(9702/41/M/J/25 Q5(b)). Graphs are compressed data — the equations decompress them.

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