Wave motion and superposition: WACE Year 12 Physics Unit 4

Syllabus dot point

“Apply the wave model to the wave equation, superposition, standing waves, resonance and beats”

8 min readUpdated 2 Jun 2026

Quick answer

Every periodic wave obeys $v=f\lambda$ . When two waves overlap their displacements add (superposition), and when a wave reflects and interferes with itself in a bounded medium you get standing waves at the natural frequencies. Resonance occurs when a driving frequency matches a natural frequency, and beats arise when two close frequencies superpose.

What this dot point is asking

WACE wants you to use the wave model quantitatively, combine overlapping waves by superposition, and explain standing waves on strings and in air columns, plus the related ideas of resonance and beats. The unifying tool is that displacements add point by point, while the wave speed is fixed by the medium.

The wave equation

For any periodic wave the speed, frequency and wavelength are linked by

v=f\lambda,

where $f$ is in hertz and $\lambda$ in metres. The frequency is set by the source; the speed is set by the medium. So if a wave passes into a new medium and slows down, its frequency stays the same and its wavelength shrinks. The period is $T=\frac{1}{f}$ .

Superposition

When two or more waves meet, the resultant displacement at each point is the vector sum of the individual displacements. This is the principle of superposition. Where crests align with crests the waves reinforce (constructive interference); where a crest meets a trough they cancel (destructive interference). After overlapping, the waves pass through unchanged.

Standing waves

A standing wave forms when a wave reflects and superposes with itself in a bounded medium, producing fixed nodes (zero displacement) and antinodes (maximum displacement). Only certain wavelengths fit the boundaries, giving a discrete set of harmonics.

For a string fixed at both ends, length $L$ must hold a whole number of half-wavelengths,

L=\frac{n\lambda}{2}\quad\Rightarrow\quad f_n=\frac{nv}{2L},\quad n=1,2,3,\dots

The same applies to a pipe open at both ends. A pipe closed at one end must have a node at the closed end and an antinode at the open end, so it holds odd quarter-wavelengths and only odd harmonics are present,

f_n=\frac{nv}{4L},\quad n=1,3,5,\dots

Resonance

Every system has natural frequencies (its harmonics). When an external driving force oscillates at one of these natural frequencies, energy transfers efficiently and the amplitude grows large: this is resonance. It explains why a guitar body amplifies a plucked string and why air in a tube sings at particular pitches.

Beats

When two waves of slightly different frequencies $f_1$ and $f_2$ superpose, the amplitude rises and falls periodically. The number of these loudness pulses per second is the beat frequency,

f_{beat}=|f_1-f_2|.

Musicians use beats to tune instruments: the beats slow and vanish as two notes are brought into unison.

Harmonics of a string and a beat frequency

Fundamental, third harmonic and tuning beats

A string of length $0.80\ \text{m}$ carries waves at $v=240\ \text{m s}^{-1}$ and is fixed at both ends.

Fundamental frequency ( $n=1$ ):

f_1=\frac{v}{2L}=\frac{240}{2(0.80)}=150\ \text{Hz}.

Third harmonic ( $n=3$ ):

f_3=3f_1=450\ \text{Hz}.

If this $150\ \text{Hz}$ string is sounded with a $154\ \text{Hz}$ tuning fork, the beat frequency is

f_{beat}=|154-150|=4\ \text{Hz},

so you hear four pulses each second until the string is tuned.

Constructive and destructive interference in detail

The principle of superposition is the engine behind every interference effect in the unit, so it pays to state it precisely. When two waves overlap, the resultant displacement at each point and instant is the algebraic sum of the individual displacements. If the waves are in phase (path difference a whole number of wavelengths), crest aligns with crest and the amplitudes add, giving constructive interference and maximum intensity. If they are exactly out of phase (path difference a half-integer number of wavelengths), crest aligns with trough and they cancel, giving destructive interference and, for equal amplitudes, zero intensity. A crucial point is that the waves emerge unchanged after overlapping; superposition is temporary. This single rule explains two-slit fringes, the nodes and antinodes of standing waves, and the loud-quiet cycle of beats, which is why it is the most important idea to get right in the wave strand.

Choosing the right boundary model

Sketch the standing wave before calculating. Decide whether the medium has two fixed (or two open) ends, which support all harmonics, or one closed end, which supports only odd harmonics with a different formula. The diagram fixes how many half- or quarter-wavelengths fit the length.

Using the both-ends formula for a closed pipe

A common slip is applying $f_n=\frac{nv}{2L}$ to a pipe closed at one end. A closed pipe needs a node at the closed end and an antinode at the open end, so its fundamental fits only a quarter-wavelength and it produces only odd harmonics with $f_n=\frac{nv}{4L}$ . Using the both-ends formula doubles the predicted fundamental frequency. Always check the boundary conditions and draw the nodes and antinodes first.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2022-style6 marks

A sound wave of frequency

512\ \text{Hz}

travels through air at

343\ \text{m s}^{-1}

. (a) Calculate its wavelength. (b) The wave then passes into water, where it travels at

1480\ \text{m s}^{-1}

. Calculate its new wavelength and explain why the frequency does not change.

Show worked answer →

A 6 mark calculation rewards both wavelengths and a source-frequency explanation.

(a) Wavelength in air: $\lambda=\dfrac{v}{f}=\dfrac{343}{512}=0.67\ \text{m}$ .
(b) Wavelength in water: The frequency is set by the source and does not change on entering a new medium, so $\lambda=\dfrac{v}{f}=\dfrac{1480}{512}=2.9\ \text{m}$ .
Why frequency is unchanged: The number of wave cycles per second arriving at the boundary must equal the number leaving (cycles cannot pile up or disappear), so the frequency is preserved. The speed changes with the medium, so the wavelength changes to keep $v=f\lambda$ satisfied.

Markers reward $\lambda=v/f$ in air ( $0.67\ \text{m}$ ), the unchanged frequency, and the new wavelength of $2.9\ \text{m}$ in water.

WACE 2020-style5 marks

Two tuning forks of frequencies

440\ \text{Hz}

and

443\ \text{Hz}

are sounded together. (a) Calculate the beat frequency. (b) Explain, in terms of superposition, why beats are heard.

Show worked answer →

A 5 mark answer rewards the beat-frequency calculation and a superposition explanation.

(a) Beat frequency. $f_{\text{beat}}=|f_1-f_2|=|443-440|=3\ \text{Hz}$ , so three beats are heard per second.

(b) Why beats occur. The two waves of slightly different frequency superpose. They drift in and out of phase: when they are in phase the displacements add (constructive interference) giving a loud sound, and when they are out of phase they partially cancel (destructive interference) giving a quiet sound. This regular rise and fall in amplitude is heard as beats, repeating at the difference frequency.

Markers reward $f_{\text{beat}}=|f_1-f_2|=3\ \text{Hz}$ and the in-phase/out-of-phase superposition causing the loud-quiet variation.

Practise this

Exam-style questions 2 on this page
Flashcards Physics deck

Sources & how we know this

Physics ATAR course: syllabus and support materials — SCSA
Physics: past ATAR course examinations — SCSA