Interpreting motion equations and simple harmonic motion | Year 12

01 / LEARN

The idea, step by step

Motion is described by derivatives

For position $x (t)$ on a line, velocity is $v = x^{'}$ and acceleration is $a = x^{''}$ . Velocity carries a sign, whereas speed is $∣ v ∣$ . An instant with zero velocity need not be an instant with zero acceleration.

If acceleration is given as a function of time, integrate twice and apply the velocity and position conditions separately. If the acceleration is described in terms of displacement, a differential equation may be needed instead.

Simple harmonic motion

The equation $x^{''} = - ω^{2} x$ describes simple harmonic motion about $x = 0$ , with $ω > 0$ . Acceleration always points toward equilibrium. A solution is $x = A cos (ω t) + B sin (ω t)$ , and its period is $2 π / ω$ .

At t=0, x (0)=A and v (0)=B\omega. The amplitude is $\sqrt{A^{2} + B^{2}}$ , and maximum speed is $ω \sqrt{A^{2} + B^{2}}$ . If the equilibrium is at $x = c$ , apply the same ideas to displacement $x - c$ .

A common trap

At an endpoint the object is at rest, so its acceleration is zero.

Instead: In SHM velocity is zero at an endpoint, but restoring acceleration has greatest magnitude there.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

A particle satisfies $x^{''} = - 4 x$ , $x (0) = 3$ , $v (0) = 0$ . Find x (t), period and maximum speed.

$ω = 2$ . Write $x = A cos 2 t + B sin 2 t$ .
$A = 3$ and $2 B = 0$ , so $B = 0$ .
The amplitude is 3, giving maximum speed $2 (3) = 6$ .

x = 3 cos 2 t

; period

π

; maximum speed

6

Example 2

Worked example

For $x = 2 sin 3 t$ , find velocity and acceleration, and verify the motion equation.

Differentiate once for velocity and twice for acceleration.
Replace $2 sin 3 t$ by x to obtain $a = - 9 x$ .
The period is $2 π / 3$ and acceleration points toward equilibrium.

v = 6 cos 3 t

a = - 18 sin 3 t = - 9 x

Check your understanding

Find the period if $x^{''} = - 25 x$ .

Show answer

$2 π / 5$

$ω = 5$ ; use $T = 2 π / ω$ .

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

For $x = 4 cos (2 t)$ m, find velocity and acceleration.
2 marks
Solve $x^{''} = - 9 x$ , $x (0) = 0$ , $v (0) = 6$ .
2 marks
Find amplitude and period of $x = 3 cos t + 4 sin t$ .
2 marks
For $x = 5 cos 2 t$ , find the first positive time at equilibrium.
2 marks
A particle has $a (t) = 2 t$ , $v (0) = 1$ , $x (0) = 4$ . Find its position.
2 marks
A model has $x^{''} = - 16 (x - 2)$ . State its equilibrium and period.
2 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

For $x = 4 cos (2 t)$ m, find velocity and acceleration.
2 marks
Apply the chain rule to each derivative.
The acceleration is $- 4 x$ .
$v = - 8 sin 2 t$ m/s; $a = - 16 cos 2 t$ m/s $2$ .
Solve $x^{''} = - 9 x$ , $x (0) = 0$ , $v (0) = 6$ .
2 marks
Write $x = A cos 3 t + B sin 3 t$ .
$A = 0$ and $3 B = 6$ , so $B = 2$ .
$x = 2 sin 3 t$
Find amplitude and period of $x = 3 cos t + 4 sin t$ .
2 marks
Amplitude is $\sqrt{3^{2} + 4^{2}} = 5$ .
The angular frequency is 1.
Amplitude $5$ ; period $2 π$ .
For $x = 5 cos 2 t$ , find the first positive time at equilibrium.
2 marks
Equilibrium requires $cos 2 t = 0$ .
The first positive solution is $2 t = π / 2$ .
$π / 4$
A particle has $a (t) = 2 t$ , $v (0) = 1$ , $x (0) = 4$ . Find its position.
2 marks
Integrate to $v = t^{2} + 1$ .
Integrate again and apply the position condition.
$x = t^{3} / 3 + t + 4$
A model has $x^{''} = - 16 (x - 2)$ . State its equilibrium and period.
2 marks
Displacement from equilibrium is $u = x - 2$ .
Then $u^{''} = - 16 u$ , so $ω = 4$ .
Equilibrium $x = 2$ ; period $π / 2$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics topic 6.1–6.2, with the approved school-level rates, differential-equation and motion applications.

Official curriculum source ↗

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