Physics 9702 · Paper 4

The Equation Family

Five equations describe every simple harmonic motion. They are one family: pick the right member, and every P4 calculation is two lines.

Displacement against time: sin or cos?

x = x_0\sin(\omega t) \quad\text{or}\quad x = x_0\cos(\omega t)

Symbols

$x_0$ = amplitude (m)
$\omega$ = angular frequency (rad s⁻¹)
$t$ = time (s)

Same motion, different starting points. sin starts at equilibrium ( $x=0$ at $t=0$ , moving fast); cos starts at the extreme ( $x=x_0$ , momentarily at rest). Exam sentences tell you which: “released from its maximum displacement” → cos; “passes through equilibrium at $t=0$ ” → sin (9702/42/M/J/24 Q4(c)).

One motion, three graphs. v leads x by a quarter cycle (v is biggest where x crosses zero), and a is the exact mirror of x — that mirror is the a = −ω²x definition drawn out in time.

Velocity is the gradient of $x(t)$ : $v = v_0\cos(\omega t)$ for the sin start — velocity runs a quarter-cycle ahead of displacement. Acceleration mirrors displacement exactly (that is $a = -\omega^2 x$ again).

Common mistake

\omega t

is in radians. A calculator left in degrees turns

0.10\sin(4.71)

= −0.10 m into a nonsense

-8.2\times10^{-3}

m. Switch to radian mode before every SHM paper — this exact slip is printed as a warning in the coursebook.

Speeds and accelerations at a glance

v = \pm\omega\sqrt{x_0^2 - x^2}, \qquad v_0 = \omega x_0, \qquad a_0 = \omega^2 x_0

Symbols

$v$ = velocity at displacement x (± = two travel directions) (m s⁻¹)
$v_0$ = maximum speed (at x = 0) (m s⁻¹)
$a_0$ = maximum acceleration (at x = ±x₀) (m s⁻²)

The first equation gives the speed anywhere; set $x=0$ and it collapses to $v_0 = \omega x_0$ . Fastest through the middle, momentarily still at the ends; acceleration does the opposite — zero at the middle, biggest at the ends.

v = ±ω√(x₀² − x²) drawn as a graph: a closed loop around the origin. It meets v = 0 exactly at x = ±x₀ (the ends of the swing) and x = 0 at v = ±v₀ (full speed through the middle).

Worked example

Unpacking a given equation (2025 paper)

An oscillator has $v = 0.56\cos(16t)$ (SI units). Find the period, the amplitude, and the equation for $v$ in terms of $x$ (9702/42/M/J/25 Q5(b)).

Match the pattern $v = v_0\cos(\omega t)$ : $v_0 = 0.56$ m s−¹, $\omega = 16$ rad s−¹, so $T = 2\pi/16 = 0.39\ \text{s}$ .
$x_0 = v_0/\omega = 0.56/16 = 0.035\ \text{m}$ .
$v = \pm 16\sqrt{0.035^2 - x^2}$ .

Answer

T = 0.39\ \text{s},\ x_0 = 3.5\ \text{cm},\ v = \pm16\sqrt{0.035^2 - x^2}

Your turn— tap to reveal the worked answer (9702/42/F/M/24 Q3(b))

An object rests on a platform oscillating vertically with $\omega = 39$ rad s−¹. What is the largest amplitude for which the object never leaves the platform, and where would contact first be lost?

Contact is lost when the platform accelerates downward faster than free fall. The largest downward acceleration is $a_0 = \omega^2 x_0$ at the top of the motion, so the limit is $x_0 = g/\omega^2 = 9.81/39^2 = 6.4 \times 10^{-3}\ \text{m}$ — and contact is first lost at the top.

Common mistake

Mixed units kill more marks here than physics does. A 2023 paper gave

v = \pm9.7\sqrt{11.6 - x^2}

in cm: this gives the amplitude directly,

\sqrt{11.6} = 3.4

cm, but the maximum acceleration needed metres —

a_0 = 9.7^2 \times 0.034 = 3.2\ \text{m s}^{-2}

(9702/42/M/J/23 Q4(a)). Decide the unit system before substituting, not after.

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The Equation Family

Five equations describe every simple harmonic motion. They are one family: pick the right member, and every P4 calculation is two lines.

Displacement against time: sin or cos?

x = x_0\sin(\omega t) \quad\text{or}\quad x = x_0\cos(\omega t)

Symbols

$x_0$ = amplitude (m)
$\omega$ = angular frequency (rad s⁻¹)
$t$ = time (s)

Same motion, different starting points. sin starts at equilibrium ( $x=0$ at $t=0$ , moving fast); cos starts at the extreme ( $x=x_0$ , momentarily at rest). Exam sentences tell you which: “released from its maximum displacement” → cos; “passes through equilibrium at $t=0$ ” → sin (9702/42/M/J/24 Q4(c)).

One motion, three graphs. v leads x by a quarter cycle (v is biggest where x crosses zero), and a is the exact mirror of x — that mirror is the a = −ω²x definition drawn out in time.

Velocity is the gradient of $x(t)$ : $v = v_0\cos(\omega t)$ for the sin start — velocity runs a quarter-cycle ahead of displacement. Acceleration mirrors displacement exactly (that is $a = -\omega^2 x$ again).

Common mistake

\omega t

is in radians. A calculator left in degrees turns

0.10\sin(4.71)

= −0.10 m into a nonsense

-8.2\times10^{-3}

m. Switch to radian mode before every SHM paper — this exact slip is printed as a warning in the coursebook.

Speeds and accelerations at a glance

v = \pm\omega\sqrt{x_0^2 - x^2}, \qquad v_0 = \omega x_0, \qquad a_0 = \omega^2 x_0

Symbols

$v$ = velocity at displacement x (± = two travel directions) (m s⁻¹)
$v_0$ = maximum speed (at x = 0) (m s⁻¹)
$a_0$ = maximum acceleration (at x = ±x₀) (m s⁻²)

The first equation gives the speed anywhere; set $x=0$ and it collapses to $v_0 = \omega x_0$ . Fastest through the middle, momentarily still at the ends; acceleration does the opposite — zero at the middle, biggest at the ends.

v = ±ω√(x₀² − x²) drawn as a graph: a closed loop around the origin. It meets v = 0 exactly at x = ±x₀ (the ends of the swing) and x = 0 at v = ±v₀ (full speed through the middle).

Worked example

Unpacking a given equation (2025 paper)

An oscillator has $v = 0.56\cos(16t)$ (SI units). Find the period, the amplitude, and the equation for $v$ in terms of $x$ (9702/42/M/J/25 Q5(b)).

Match the pattern $v = v_0\cos(\omega t)$ : $v_0 = 0.56$ m s−¹, $\omega = 16$ rad s−¹, so $T = 2\pi/16 = 0.39\ \text{s}$ .
$x_0 = v_0/\omega = 0.56/16 = 0.035\ \text{m}$ .
$v = \pm 16\sqrt{0.035^2 - x^2}$ .

Answer

T = 0.39\ \text{s},\ x_0 = 3.5\ \text{cm},\ v = \pm16\sqrt{0.035^2 - x^2}

Your turn— tap to reveal the worked answer (9702/42/F/M/24 Q3(b))

Common mistake

Mixed units kill more marks here than physics does. A 2023 paper gave

v = \pm9.7\sqrt{11.6 - x^2}

in cm: this gives the amplitude directly,

\sqrt{11.6} = 3.4

cm, but the maximum acceleration needed metres —

a_0 = 9.7^2 \times 0.034 = 3.2\ \text{m s}^{-2}

(9702/42/M/J/23 Q4(a)). Decide the unit system before substituting, not after.