Interpreting motion equations and simple harmonic motion | Year 12

01 / LEARN

Understand the idea.

Motion is described by derivatives

For motion along a line, $v = x^{'}$ and $a = x^{′′}$ . Velocity is signed, whereas speed is $∣ v ∣$ . A turning point can have zero velocity and nonzero acceleration: the acceleration determines how the velocity is changing at that instant.

When acceleration is given as a function of time, integrate twice and use separate velocity and position conditions. When acceleration depends on displacement, first identify the differential equation; treating displacement as an independent time expression would bypass the unknown motion.

Simple harmonic motion

Simple harmonic motion about zero satisfies $x^{′′} = - ω^{2} x$ , with $ω > 0$ . The acceleration points toward equilibrium. Its general form is $x = A cos (ω t) + B sin (ω t)$ , with period $T = \frac{2 π}{ω}$ .

The initial values give $x (0) = A$ and $v (0) = ω B$ . The amplitude is $R = \sqrt{A^{2} + B^{2}}$ and the maximum speed is $ω R$ . If equilibrium is at $x = c$ , apply these ideas to displacement $x - c$ .

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

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

Worked solution

1
Identify the frequency and solution family.
$ω^{2} = 4, ω = 2, x = A cos (2 t) + B sin (2 t)$
The positive square root defines the angular frequency in the SHM model.
2
Apply the initial displacement.
$x (0) = A = 3$
Cosine equals 1 and sine equals zero at time zero.
3
Differentiate and use the initial velocity.
$v = - 2 A sin (2 t) + 2 B cos (2 t), v (0) = 2 B = 0 \Rightarrow B = 0$
The velocity condition fixes the second independent constant.
4
State the motion and its period.
$x (t) = 3 cos (2 t), T = \frac{2 π}{2} = π$
A full cosine cycle requires its argument to increase by 2 pi.
5
Find the maximum speed.
$v = - 6 sin (2 t), ∣ v ∣_{max} = 6$
The sine factor has maximum absolute value 1; equivalently, maximum speed is angular frequency times amplitude.

Answer

$x = 3 cos 2 t$ ; period $π$ ; maximum speed $6$ .

EXAMPLE 02

Worked example

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

Worked solution

1
Differentiate for velocity.
$x = 2 sin (3 t), v = 6 cos (3 t)$
The chain rule multiplies by the angular frequency 3.
2
Differentiate for acceleration.
$a = - 18 sin (3 t)$
Differentiating cosine introduces both a negative sign and another factor of 3.
3
Express acceleration using displacement.
$a = - 9 (2 sin (3 t)) = - 9 x$
Acceleration is a negative constant multiple of displacement, verifying the simple harmonic motion equation.

Answer

$v = 6 cos 3 t$ , $a = - 18 sin 3 t = - 9 x$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$\frac{2 π}{5}$

Comparing $x^{′′} = - 25 x$ with $x^{′′} = - ω^{2} x$ gives $ω = 5$ . Hence the period is $T = \frac{2 π}{5}$ in the model's time units.

Keep practising

6 worksheet questions · 12 marks

Open worksheet 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 $2 t = \frac{π}{2}$ .
$\frac{π}{4}$
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.
$x = \frac{t^{3}}{3} + t + 4$
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 $\frac{π}{2}$ .

Checking saved progress…

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 ↗

In this lesson

Understand the idea.

Motion is described by derivatives

Simple harmonic motion

See the method in action.

Worked example

Identify the frequency and solution family.

Apply the initial displacement.

Differentiate and use the initial velocity.

State the motion and its period.

Find the maximum speed.

Worked example

Differentiate for velocity.

Differentiate for acceleration.

Express acceleration using displacement.

Try it yourself.

Check your understanding

Keep practising