Standing waves and resonance (QCE Physics Unit 2)

Syllabus dot point

“Explain the formation of standing waves in strings (fixed at both ends) and in air columns (open and closed pipes), and solve problems involving the resonant frequencies of mechanical systems”

9 min readUpdated 22 May 2026

Quick answer

Standing waves form when two travelling waves of the same frequency superpose in opposite directions, producing fixed nodes and antinodes; a string fixed at both ends and an open pipe both have all-integer resonances $f_n = n v / 2L$ , while a closed pipe has only odd resonances $f_n = n v / 4L$ .

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What this dot point is asking
How a standing wave forms
String fixed at both ends
Open pipe (both ends open)
Closed pipe (one end closed)
Resonance
How this appears in IA1 and EA
Examples in context
Try this
Exam-style questions

What this dot point is asking

QCAA wants you to explain how a standing wave forms (superposition of two waves travelling in opposite directions with the same frequency), to identify nodes and antinodes, and to derive the resonant-frequency series for three standard systems: a string fixed at both ends, an open pipe, and a closed pipe.

How a standing wave forms

When a wave reflects off a boundary, the reflected wave travels back through the incident wave. The two waves superpose. At points where they are always in antiphase, the displacements cancel (a node). At points where they are always in phase, the displacements add to maximum (an antinode).

The result is a wave pattern with fixed nodes and antinodes that does not propagate. Energy is trapped in the standing-wave region. Only frequencies that fit the boundary conditions survive; others decay through destructive interference.

String fixed at both ends

Both ends are nodes (the string cannot move). The string supports a series of resonant modes:

L = n \frac{\lambda_n}{2}, \quad n = 1, 2, 3, \ldots

\lambda_n = \frac{2L}{n}, \quad f_n = \frac{n v}{2L}

The fundamental ( $n = 1$ ) has $\lambda_1 = 2L$ and $f_1 = v / 2L$ . Higher modes are integer multiples: $f_2 = 2 f_1$ , $f_3 = 3 f_1$ , and so on. The series contains all integers (a full harmonic series).

Open pipe (both ends open)

Both ends are antinodes (air can move freely). Same length-wavelength relation as the string:

L = n \frac{\lambda_n}{2}, \quad f_n = \frac{n v}{2L}, \quad n = 1, 2, 3, \ldots

All harmonics present. A flute is approximately an open pipe.

Closed pipe (one end closed)

The closed end is a node; the open end is an antinode. Quarter-wavelengths fit the length:

L = n \frac{\lambda_n}{4}, \quad n = 1, 3, 5, \ldots

\lambda_n = \frac{4L}{n}, \quad f_n = \frac{n v}{4L}

Only odd harmonics are present ( $f_1, 3f_1, 5f_1, \ldots$ ). A clarinet is approximately a closed pipe at low register, which is why a clarinet sounds different from a flute of the same length playing the same fundamental.

Resonance

Resonance occurs when an applied periodic force has a frequency equal to one of the natural frequencies of the system. Energy is transferred efficiently into the standing wave and amplitude builds up over many cycles.

Examples: a tuning fork held above a tube of adjustable length will resonate when the air column matches a closed-pipe harmonic; pushing a child on a swing at the swing's natural frequency builds amplitude.

Resonant frequencies of a closed organ pipe

An organ pipe is $1.20$ m long and closed at one end. Take the speed of sound as $343$ m s $^{-1}$ . Find the fundamental frequency and the first three audible resonances.

Step 1: Recall the boundary conditions for a closed pipe

A closed pipe has a node (no air movement) at the sealed end and an antinode (maximum air movement) at the open end. This fixed-node-to-antinode geometry means only odd harmonics fit: $n = 1, 3, 5, \ldots$ . Even harmonics are excluded because they would require an antinode at the closed end.

Step 2: Write the resonant-frequency formula for a closed pipe

The fundamental wavelength satisfies $L = \lambda_1/4$ , so the general formula is:

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

Step 3: Calculate the fundamental frequency

Substitute $n = 1$ , $v = 343$ m s $^{-1}$ and $L = 1.20$ m:

f_1 = \frac{(1)(343)}{4 \times 1.20} = \frac{343}{4.80} = 71.5 \text{ Hz}

Step 4: Find the next two odd harmonics

Each higher resonance is an odd-integer multiple of $f_1$ :

f_3 = 3f_1 = 3 \times 71.5 = 214 \text{ Hz}

f_5 = 5f_1 = 5 \times 71.5 = 357 \text{ Hz}

The even harmonics ( $n = 2, 4$ ) are absent from a closed pipe.

Final answer: $f_1 = 71.5$ Hz; $f_3 = 214$ Hz; $f_5 = 357$ Hz. Only odd harmonics are present.

Common traps

Confusing the closed-pipe series with the open-pipe series: Closed pipes only have odd harmonics. Drawing the antinode at the wrong end is the most common error.
Treating the string fundamental as $\lambda_1 = L$: A string fixed at both ends has nodes at the ends, so the fundamental has only one antinode and $\lambda_1 = 2L$ , not $L$ .
Forgetting end correction in air columns: Real open ends have antinodes slightly outside the physical pipe. QCAA problems use ideal pipes unless they specify a correction.
Treating resonance as one specific frequency: Resonance happens at every natural frequency, not just the fundamental.

How this appears in IA1 and EA

IA1: Often a stimulus showing the standing-wave pattern on a string or in a pipe and asking for either the harmonic number or the length given a frequency and speed.
EA Paper 1: Multiple choice on which harmonics are present in closed pipes versus open pipes.
EA Paper 2: A two-part question on a musical-instrument system, typically asking for wave speed and then a higher harmonic.

Examples in context

Example 1. A Queensland Symphony Orchestra cellist at QPAC tunes the C string ( $f_1 = 65.4 \text{ Hz}$ ) to a string length $0.69 \text{ m}$ fixed at both ends. Wave speed is $v = 2L f_1 = 2 \times 0.69 \times 65.4 = 90.3 \text{ m s}^{-1}$ . The first overtone (2nd harmonic) sits at $130.8 \text{ Hz}$ . The cellist adjusts tension to alter $v$ and so $f_1$ , a direct application of the QCAA Unit 2 harmonics relation $f_n = nv/(2L)$ .

Example 2. A Bundaberg sugar mill operator notices a vibration peak at $58 \text{ Hz}$ in a steam pipe open at both ends, length $2.9 \text{ m}$ . For an open pipe, $f_n = nv/(2L)$ ; with $v = 340 \text{ m s}^{-1}$ , $f_1 = 58.6 \text{ Hz}$ , confirming the resonance is the fundamental. Engineers add a damping clamp at the midpoint (a pressure node) to suppress it. QCAA EA Unit 2 thematic items pair industrial stimulus with this kind of harmonic identification.

Try this

Q1. State the relationship between length and wavelength for the fundamental of a string fixed at both ends. [2 marks]

Cue. $L = \lambda/2$ , so $\lambda = 2L$ .

Q2. A pipe closed at one end has length $0.85 \text{ m}$ . Using $v = 340 \text{ m s}^{-1}$ , calculate the first three resonant frequencies. [3 marks]

Cue. Closed-pipe odd harmonics only; $f_1 = v/(4L) = 100 \text{ Hz}$ ; $f_3 = 300 \text{ Hz}$ ; $f_5 = 500 \text{ Hz}$ .

Q3. A guitar string of length $0.65 \text{ m}$ has wave speed $260 \text{ m s}^{-1}$ . (a) Calculate $f_1$ and $f_3$ . (b) Explain why pressing the string at the $12$ th fret (midpoint) doubles the fundamental frequency. (c) Discuss one factor that limits the precision of a QCAA IA2 resonance experiment. [3+3+2 marks; ISMG: Knowledge and conceptual understanding, Evaluation]

Cue. (a) $f_1 = 200 \text{ Hz}$ , $f_3 = 600 \text{ Hz}$ ; (b) halves $L$ so doubles $f_1$ ; (c) endpoint position uncertainty.

Exam-style questions

Questions in the style of QCAA 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.

Year 11 SAC-style5 marks

A guitar string of length

0.65

m is fixed at both ends and produces a fundamental frequency of

196

Hz (the G string). (a) Find the wave speed on the string. (b) Find the frequency of the third harmonic.

Show worked answer →

(a) Wave speed. For a string fixed at both ends, $\lambda_1 = 2L = 2 \times 0.65 = 1.30$ m.

$v = f_1 \lambda_1 = (196)(1.30) = 255$ m s $^{-1}$ .

(b) Third harmonic. $f_n = n f_1$ .

$f_3 = 3 \times 196 = 588$ Hz.

Markers reward the explicit relationship $\lambda_1 = 2L$ , substitution into $v = f \lambda$ , and units throughout.

Practise this

Sources & how we know this

Physics General senior syllabus 2025 — QCAA