Physics 9702 · Paper 4

What Makes Motion Simple Harmonic

Not every back-and-forth is . One condition on the acceleration decides — and stating it exactly is a banked two-marker.

The definition and a = −ω²x

Motion is simple harmonic when the acceleration is directly proportional to the displacement from equilibrium and always in the opposite direction to the displacement (equivalently: always directed towards a fixed point). One mark per half, recycled word-for-word across sessions (9702/41/O/N/23 Q4(a)) (9702/41/O/N/24 Q4(a)) (9702/42/M/J/25 Q5(a)).

a = -\omega^2 x

Symbols

$a$ = acceleration of the oscillator (m s⁻²)
$x$ = displacement from equilibrium (m)
$\omega^2$ = the (positive) constant of proportionality (rad² s⁻²)

The minus sign is the physics: displaced right, the restoring push points left, always back towards equilibrium. Without the minus the object would run away and never return. The constant being called $\omega^2$ is no accident — a stiffer restoring force means larger $\omega$ , so faster oscillations.

For s.h.m., an a-x graph is a straight line through the origin with negative gradient −ω². If it is curved, does not pass through the origin, or has positive gradient, the motion is not simple harmonic.

Shown a graph instead of asked for words? The two marks map onto two features: a straight line through the origin shows $a \propto x$ , and the negative gradient shows they are always opposite (9702/42/M/J/24 Q4(a)) (9702/42/O/N/25 Q5(a)(i)). The gradient itself is $-\omega^2$ , which turns any such graph into a period: $T = 2\pi\sqrt{x_0/a_0}$ gave 1.9 s from a line ending at (1.2 cm, −13 cm s−²) (9702/42/O/N/25 Q5(a)(ii)).
Sketching the line yourself? Stop it at $\pm x_0$ — the oscillator never goes further, and extending the line loses a mark (9702/42/M/J/23 Q4(b)).

Where the ω comes from: the shadow of a circle

SHM and circular motion share $\omega$ for a reason a 2025 question made students derive (9702/41/O/N/25 Q1(c)): shine parallel light past a ball moving in a circle of radius $R$ at angular speed $\omega$ , and watch its shadow on a screen.

The shadow's displacement from the centre is $x = R\sin\theta$ .
Steady rotation means $\theta = \omega t$ .
Substitute: $x = R\sin(\omega t)$ — exactly the form $x = x_0\sin(\omega t)$ , so the shadow moves with simple harmonic motion.

One circle, viewed edge-on, is an oscillation. The question finished with numbers: a 0.46 m diameter circle at 1.9 rad s−¹ gives a shadow with $x_0 = 0.23$ m, $T = 3.3$ s and $a_0 = \omega^2 x_0 = 0.83$ m s−² (9702/41/O/N/25 Q1(d)).

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What Makes Motion Simple Harmonic

Not every back-and-forth is . One condition on the acceleration decides — and stating it exactly is a banked two-marker.

The definition and a = −ω²x

Motion is simple harmonic when the acceleration is directly proportional to the displacement from equilibrium and always in the opposite direction to the displacement (equivalently: always directed towards a fixed point). One mark per half, recycled word-for-word across sessions (9702/41/O/N/23 Q4(a)) (9702/41/O/N/24 Q4(a)) (9702/42/M/J/25 Q5(a)).

a = -\omega^2 x

Symbols

$a$ = acceleration of the oscillator (m s⁻²)
$x$ = displacement from equilibrium (m)
$\omega^2$ = the (positive) constant of proportionality (rad² s⁻²)

The minus sign is the physics: displaced right, the restoring push points left, always back towards equilibrium. Without the minus the object would run away and never return. The constant being called $\omega^2$ is no accident — a stiffer restoring force means larger $\omega$ , so faster oscillations.

Shown a graph instead of asked for words? The two marks map onto two features: a straight line through the origin shows $a \propto x$ , and the negative gradient shows they are always opposite (9702/42/M/J/24 Q4(a)) (9702/42/O/N/25 Q5(a)(i)). The gradient itself is $-\omega^2$ , which turns any such graph into a period: $T = 2\pi\sqrt{x_0/a_0}$ gave 1.9 s from a line ending at (1.2 cm, −13 cm s−²) (9702/42/O/N/25 Q5(a)(ii)).
Sketching the line yourself? Stop it at $\pm x_0$ — the oscillator never goes further, and extending the line loses a mark (9702/42/M/J/23 Q4(b)).

Where the ω comes from: the shadow of a circle

SHM and circular motion share $\omega$ for a reason a 2025 question made students derive (9702/41/O/N/25 Q1(c)): shine parallel light past a ball moving in a circle of radius $R$ at angular speed $\omega$ , and watch its shadow on a screen.

The shadow's displacement from the centre is $x = R\sin\theta$ .
Steady rotation means $\theta = \omega t$ .
Substitute: $x = R\sin(\omega t)$ — exactly the form $x = x_0\sin(\omega t)$ , so the shadow moves with simple harmonic motion.

One circle, viewed edge-on, is an oscillation. The question finished with numbers: a 0.46 m diameter circle at 1.9 rad s−¹ gives a shadow with $x_0 = 0.23$ m, $T = 3.3$ s and $a_0 = \omega^2 x_0 = 0.83$ m s−² (9702/41/O/N/25 Q1(d)).