§-Syllabus dot point

WAPhysicsSyllabus dot point

How do reflected waves superpose to form standing waves and resonance?

Explain standing waves on strings and in pipes and relate harmonics to resonance

A focused answer to the WACE Year 12 Physics Unit 4 content point on standing waves. How superposition of waves travelling in opposite directions forms nodes and antinodes, the harmonic series on strings and in pipes, and the condition for resonance.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

What this dot point is asking

WACE wants you to explain standing waves as the superposition of two travelling waves, locate nodes and antinodes, and find the allowed frequencies for strings and pipes. This is where superposition produces something that does not appear to move along at all.

How a standing wave forms

When a wave reflects off a boundary, the reflected wave travels back through the incoming wave. Where the two are always in phase, constructive interference gives an antinode of maximum oscillation; where they are always out of phase, destructive interference gives a node of zero displacement. The pattern of nodes and antinodes stays fixed in space, so the wave appears to stand still while points between nodes oscillate.

Strings fixed at both ends

A string fixed at both ends must have a node at each end. The longest wave that fits has a single antinode in the middle, so its wavelength is twice the string length: $\lambda_1=2L$ . Successive harmonics fit more half-wavelengths, giving

\lambda_n=\frac{2L}{n},\qquad f_n=\frac{nv}{2L},\qquad n=1,2,3,\dots

The fundamental (first harmonic) is $f_1=v/2L$ , and higher harmonics are whole-number multiples of it.

Pipes

An open pipe (open at both ends) has an antinode at each end and behaves like the string case, with $f_n=nv/2L$ for all integers $n$ . A pipe closed at one end has a node at the closed end and an antinode at the open end, so only odd harmonics exist:

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

This is why a closed pipe of a given length sounds an octave lower than an open pipe of the same length.

Resonance

Every system has natural frequencies equal to its harmonic series. When an external driver oscillates at one of these frequencies, energy is fed in efficiently each cycle and the standing wave grows to a large amplitude. This is resonance, responsible for the loud note of a tuned instrument and, destructively, for structures shaking apart when driven at a natural frequency.

Standing waves versus travelling waves

A frequent short-answer question asks how a standing wave differs from a travelling wave, and the contrast is worth memorising. In a travelling wave, energy and the wave profile move along the medium, every point has the same amplitude (just reached at different times), and there are no permanent nodes. In a standing wave, the profile does not progress, energy is stored rather than transported along the medium, and there are fixed nodes (zero amplitude) and antinodes (maximum amplitude). All points between adjacent nodes oscillate in phase but with different amplitudes, and points on opposite sides of a node oscillate exactly out of phase. Stating these differences precisely, especially the fixed nodes and the lack of net energy transport, is what distinguishes a full-mark answer.

Resonance in practice and its dangers

Resonance has both useful and destructive consequences that examiners like to see linked to real situations. Musical instruments rely on it: a wind player or string player excites the natural frequencies of an air column or string, and the resonant body amplifies those harmonics to produce a loud, sustained note. A radio tuner resonates electrically at one chosen frequency to select a station. Destructively, machinery driven near a structural natural frequency can shake itself apart, and bridges or buildings driven by wind or marching feet at a resonant frequency can suffer dangerously large oscillations, which is why soldiers break step on bridges and why structures are designed so their natural frequencies avoid common driving frequencies. In every case the underlying physics is the same: energy is fed in efficiently each cycle when the driving frequency matches a natural frequency, so the amplitude builds up.

Counting half-wavelengths

To find a harmonic, sketch the pattern, count nodes and antinodes against the boundary conditions, and read off how many half-wavelengths fit the length. Mislabelling an open pipe as closed (or vice versa) changes which harmonics exist, so identify the ends first.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20237 marksA pipe closed at one end has a length of

0.40\ \text{m}

. The speed of sound in air is

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

. (a) Calculate the fundamental frequency of the pipe. (b) Calculate the frequency of the next harmonic it can produce. (c) An open pipe of the same length is also sounded; calculate its fundamental frequency and compare it with that of the closed pipe.

Show worked answer →

A 7 mark calculation rewards the closed-pipe fundamental, its next (odd) harmonic and the open-pipe comparison.

(a) Closed-pipe fundamental: For a pipe closed at one end, $f_1=\dfrac{v}{4L}=\dfrac{343}{4(0.40)}=\dfrac{343}{1.6}=214\ \text{Hz}$ .
(b) Next harmonic: A closed pipe supports only odd harmonics, so the next is the third harmonic: $f_3=3f_1=3\times214=643\ \text{Hz}$ .
(c) Open pipe: For an open pipe, $f_1=\dfrac{v}{2L}=\dfrac{343}{2(0.40)}=\dfrac{343}{0.8}=429\ \text{Hz}$ . The open pipe's fundamental is twice that of the closed pipe (one octave higher), because the closed pipe fits only a quarter-wavelength while the open pipe fits a half-wavelength.

Markers reward $f_1=v/4L=214\ \text{Hz}$ for the closed pipe, the odd third harmonic $643\ \text{Hz}$ , and the open-pipe $429\ \text{Hz}$ with the factor-of-two (octave) comparison.

WACE 20205 marksExplain how a standing wave is formed on a string fixed at both ends, and explain why only certain frequencies produce standing waves.

Show worked answer →

A 5 mark explanation needs the superposition mechanism and the boundary condition.

Formation. A wave travelling along the string reflects at the fixed ends and travels back, superposing with the incoming wave. Where the two waves are always in phase, constructive interference gives antinodes of maximum oscillation; where they are always out of phase, destructive interference gives nodes of zero displacement. The pattern is fixed in space, forming a standing wave.

Why only certain frequencies. Each fixed end must be a node (it cannot move). Only wavelengths that fit a whole number of half-wavelengths between the ends satisfy this, giving $\lambda_n=2L/n$ and frequencies $f_n=nv/2L$ . Other frequencies do not produce a stable node-at-each-end pattern, so the reflected waves do not reinforce and no standing wave builds.

Markers reward reflection and superposition forming nodes and antinodes, the node-at-each-fixed-end condition and the resulting allowed wavelengths or frequencies.