Why is "Applying the string/open-pipe harmonic series to a pipe closed at one end" a common mistake in Standing waves and resonance?

Why it happens: Students memorise fₙ = nv/2L and use it for every pipe, forgetting that a closed pipe has a different boundary condition. How to avoid it: A closed pipe (node at the closed end, antinode at the open end) fits L = (2n−1)λ/4 and produces ONLY odd harmonics (1st, 3rd, 5th…). Its fundamental is v/4L, and there is no 2nd, 4th… harmonic. Check the ends before choosing a formula. Source: IB Physics Theme C subject reports

Why is "Confusing where nodes and antinodes sit at the ends of a pipe" a common mistake in Standing waves and resonance?

Why it happens: Nodes and antinodes look symmetric on a diagram, so students place them at the wrong ends. How to avoid it: Learn the rule: an OPEN end is a displacement ANTINODE (air free to move); a CLOSED end (or a fixed string end) is a NODE (air/string cannot move). Draw the antinode as a bulge at every open end before doing any calculation. Source: IB Physics Theme C subject reports

Why is "Using λ = L instead of the correct fraction (L = nλ/2 or L = (2n−1)λ/4)" a common mistake in Standing waves and resonance?

Why it happens: Students assume 'one wave fits the length', ignoring that the fundamental is a half-wavelength (string/open) or a quarter-wavelength (closed). How to avoid it: Always sketch the standing wave first. Fundamental of a string/open pipe = ½λ (so λ = 2L); fundamental of a closed pipe = ¼λ (so λ = 4L). Never assume a whole wavelength fits. Source: IB Physics Theme C subject reports

Why is "Taking the distance between adjacent nodes to be a whole wavelength" a common mistake in Standing waves and resonance?

Why it happens: Students see one 'loop' between nodes and assume it is a full wave. How to avoid it: The distance between two adjacent nodes (or two adjacent antinodes) is HALF a wavelength (λ/2); a node to the nearest antinode is a QUARTER (λ/4). Multiply the node spacing by 2 to get λ. Source: IB Physics Theme C subject reports

Why is "Thinking a standing wave transfers energy along the medium" a common mistake in Standing waves and resonance?

Why it happens: It is made of travelling waves, so students assume it must carry energy like them. How to avoid it: A standing wave transfers NO NET energy — the two travelling waves carry equal energy in opposite directions, so it cancels. Energy is stored locally, oscillating between kinetic and potential within each loop. Source: IB Physics Theme C subject reports

IB DP Physics (100452)

Standing waves and resonance

Q: Why is "Confusing the resonance (natural) frequency with the driving frequency in general" a common mistake in Standing waves and resonance?

Why it happens: At resonance the two are equal, so students wrongly treat them as always the same. How to avoid it: The driving frequency is set by the external oscillator and can be anything; the natural frequency is a fixed property of the system. Resonance is the special case where the driving frequency EQUALS a natural frequency — elsewhere they differ and the amplitude is small. Source: IB Physics Theme C subject reports

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Detailed Study Notes

Detailed notes on Wave behaviour for IB DP Physics, covering key concepts, explanations, examples, and exam-focused revision points.

C.4 Standing Waves and Resonance — IB Physics SL Study Notes (Theme C: Wave Behaviour)

How two identical waves travelling in opposite directions superpose to make a standing wave with fixed nodes and antinodes; the harmonics of strings and of open and closed pipes; and resonance — a system driven at its natural frequency oscillating with maximum amplitude — together with the effect of damping.

At a glance

A standing (stationary) wave forms when two waves of the same frequency and amplitude travelling in opposite directions (usually a wave and its reflection) superpose.
Nodes are points of zero/minimum amplitude; antinodes are points of maximum amplitude. Adjacent nodes are λ/2 apart; a node to the nearest antinode is λ/4.
A standing wave transfers no net energy; all points between two adjacent nodes oscillate in phase, but the amplitude varies with position.
String fixed at both ends (and pipe open at both ends): $L = n\lambda/2$ , so $f_n = nv/2L$ — all integer harmonics (f, 2f, 3f…).
Pipe closed at one end: node at the closed end, antinode at the open end, $L = (2n-1)\lambda/4$ , so only odd harmonics (f, 3f, 5f…); fundamental $\lambda = 4L$ .
Resonance: a system driven at its natural frequency oscillates with maximum amplitude.
Damping reduces and broadens the resonance peak — light damping gives a tall, sharp peak; heavy damping gives a low, broad one.
Common traps: giving a closed pipe even harmonics, swapping node/antinode ends, using λ = L, and thinking a standing wave carries energy.

What you’ll learn

Mapped to the 100452 subject guide (2025-onwards).

Explain how a standing wave is formed by the superposition of two identical waves travelling in opposite directions, and locate its nodes and antinodes.
Distinguish a standing wave from a travelling wave in terms of energy transfer, amplitude and phase.
Derive and apply the harmonic frequencies of a string fixed at both ends, fₙ = nv/2L, from L = nλ/2 and v = fλ.
Determine the harmonics of pipes open at both ends (all harmonics, L = nλ/2) and closed at one end (odd harmonics only, L = (2n−1)λ/4).
Describe resonance as maximum-amplitude oscillation when a system is driven at its natural frequency, and explain qualitatively the effect of damping on the resonance peak.

How a standing wave forms: superposition of two waves

Two identical waves going opposite ways add up to give fixed nodes and antinodes.

A standing (stationary) wave is produced when two waves of the same frequency and amplitude travel in opposite directions along the same line and superpose. In almost every real case the two waves are a wave and its reflection — for example a wave sent along a string reflecting off a fixed end, or a sound wave reflecting off the end of a pipe.

The mechanism, step by step:

A travelling wave moves in one direction (say to the right).
It reflects at a boundary and travels back (to the left) with the same frequency, speed and amplitude.
By the principle of superposition, at every point the two displacements add.
At some fixed points the two waves are always in antiphase, so they permanently cancel — these are nodes (zero/minimum amplitude).
Halfway between the nodes the two waves are always in phase, so they reinforce — these are antinodes (maximum amplitude).

The result is a pattern that appears to stand still: the nodes and antinodes stay in fixed positions while the string (or air) between them oscillates.

Key spacings (learn these — they turn a measured pattern into a wavelength):

Distance between two adjacent nodes (or two adjacent antinodes) $= \dfrac{\lambda}{2}$ .
Distance from a node to the nearest antinode $= \dfrac{\lambda}{4}$ .

A string fixed at both ends must have a node at each end. The fundamental (n = 1) has one antinode; each higher harmonic adds one more antinode. The solid and faded curves show the two extreme positions the string swings between.

Standing wave = superposition of two identical waves travelling in opposite directions.
Nodes: always-zero amplitude (waves in antiphase). Antinodes: maximum amplitude (waves in phase).
Adjacent nodes are λ/2 apart; node to nearest antinode is λ/4.

See the full worked example for standing waves and resonance →

Standing wave vs travelling wave: the three differences

Standing: no net energy, amplitude varies with position, in phase between nodes. Travelling: carries energy, constant amplitude, phase varies.

Examiners love a "state the differences" question here, and they want contrasts — each point must say something about both wave types.

Feature	Standing (stationary) wave	Travelling (progressive) wave
Energy	Transfers no net energy along the medium	Transfers energy in the direction of travel
Amplitude	Varies with position — zero at nodes, maximum at antinodes	Same for every point
Phase	All points between two adjacent nodes are in phase; adjacent loops are in antiphase	Phase varies continuously with position
Wave profile	Pattern does not move — nodes and antinodes stay put	Profile moves through the medium
Wavelength	Twice the node-to-node distance	Distance between successive points in phase

Why no net energy? A standing wave is built from two travelling waves carrying equal energy in opposite directions. The two energy flows cancel, so there is no net transport — energy is simply stored, sloshing between kinetic and potential form within each loop.

Why "in phase between nodes"? Watch one loop of a vibrating string: every particle in that loop reaches its highest point at the same instant and its lowest point at the same instant. They differ only in amplitude, not in timing (phase). Cross a node into the next loop and everything moves the opposite way — that loop is in antiphase.

Energy: standing wave transfers none; travelling wave transfers energy.
Amplitude: varies with position (standing) vs constant (travelling).
Phase: in phase between adjacent nodes (standing) vs continuously varying (travelling).

Harmonics on a string fixed at both ends

Both ends are nodes, so L = nλ/2 and the string sounds all integer harmonics fₙ = nv/2L.

A string clamped at both ends (guitar, violin, sonometer) must have a node at each fixed end — the string cannot move there. The only standing waves that fit are those whose length is a whole number of half-wavelengths:

$L = \frac{n\lambda}{2}, \qquad n = 1, 2, 3, \dots$

Rearranging gives the allowed wavelengths and, with $v = f\lambda$ , the allowed harmonic frequencies:

$\lambda_n = \frac{2L}{n}, \qquad f_n = \frac{v}{\lambda_n} = \frac{nv}{2L}$

Harmonic	n	Antinodes	Wavelength	Frequency
Fundamental (1st)	1	1	$2L$	$\dfrac{v}{2L} = f_1$
2nd harmonic	2	2	$L$	$2f_1$
3rd harmonic	3	3	$\dfrac{2L}{3}$	$3f_1$
$n$ th harmonic	$n$	$n$	$\dfrac{2L}{n}$	$nf_1$

Key features:

A string fixed at both ends produces all integer harmonics: $f_1, 2f_1, 3f_1, \dots$
The $n$ th harmonic has n antinodes and n + 1 nodes (counting the two ends).
The fundamental is the lowest note; its wavelength is twice the string length.

Worked micro-example: a 0.80 m string carries waves at 320 m s⁻¹. Fundamental $f_1 = \dfrac{320}{2 \times 0.80} = 200\,\text{Hz}$ , so the harmonics are 200, 400, 600 Hz… — a neat 1 : 2 : 3 pattern.

Both ends fixed → nodes at both ends → L = nλ/2.
fₙ = nv/2L; all integer harmonics are present.
nth harmonic has n antinodes and n + 1 nodes.

See the full worked example for standing waves and resonance →

Deriving the harmonic frequencies fₙ = nv/2L (first principles)

Combine the boundary condition L = nλ/2 with the wave equation v = fλ to get fₙ = nv/2L.

You should be able to build the harmonic formula from just two facts you already know — the boundary condition and the wave equation. Doing so means you never have to memorise $f_n = \dfrac{nv}{2L}$ as a bare formula, and it makes the closed-pipe case (below) easy too.

Step 1 — the boundary condition. A string fixed at both ends has a node at each end. Between the ends you can fit 1, 2, 3… complete "loops", and each loop is half a wavelength long. So the length must be a whole number of half-wavelengths: $L = n \times \frac{\lambda_n}{2} \quad\Rightarrow\quad \lambda_n = \frac{2L}{n}, \qquad n = 1, 2, 3, \dots$

Step 2 — bring in the wave equation. Every wave obeys $v = f\lambda$ , so $f = \dfrac{v}{\lambda}$ . Substitute the allowed wavelengths from Step 1: $f_n = \frac{v}{\lambda_n} = \frac{v}{\left(\dfrac{2L}{n}\right)} = \boxed{\frac{nv}{2L}}$

Step 3 — read off the physics. Putting $n = 1$ gives the fundamental $f_1 = \dfrac{v}{2L}$ , and every higher harmonic is an integer multiple of it: $f_n = n f_1$ . The whole harmonic series drops straight out of "a whole number of half-wavelengths fits the length".

The closed pipe by the same method. A pipe closed at one end has a node at the closed end and an antinode at the open end, so the shortest fit is a quarter-wavelength, and each allowed pattern adds another half-wavelength. That gives $L = (2n-1)\dfrac{\lambda}{4}$ , and the same substitution $f = v/\lambda$ yields $f_n = \frac{(2n-1)v}{4L}, \qquad n = 1, 2, 3, \dots$ — i.e. only the odd multiples $1, 3, 5, \dots$ of the fundamental $\dfrac{v}{4L}$ .

Each harmonic fits a whole number of half-wavelength "loops" between the fixed ends. Writing L = nλ/2 and substituting into f = v/λ gives fₙ = nv/2L directly.

Boundary condition (nodes at both ends): L = nλ/2, so λₙ = 2L/n.
Wave equation f = v/λ then gives fₙ = nv/2L.
Closed pipe (node + antinode): L = (2n−1)λ/4 → fₙ = (2n−1)v/4L (odd only).

Pipes: open at both ends vs closed at one end

Open pipe = antinodes at both ends, all harmonics. Closed pipe = node at closed end, odd harmonics only.

Air columns in pipes form longitudinal standing waves, but we draw them as transverse "displacement" patterns for clarity. The rule for the ends is:

an open end is a displacement antinode (the air is free to move), and
a closed end is a displacement node (the air cannot move there).

Pipe open at both ends — behaves exactly like a string. Antinodes at both ends means $L = \dfrac{n\lambda}{2}$ , so $f_n = \frac{nv}{2L}, \qquad n = 1, 2, 3, \dots \quad(\text{all harmonics}).$ The fundamental has $\lambda = 2L$ .

A pipe open at both ends has a displacement antinode at each open end. Like a string, it supports every integer harmonic: f₁, 2f₁, 3f₁, …

Pipe closed at one end — only odd harmonics. A node at the closed end and an antinode at the open end means the shortest fit is a quarter-wavelength ( $\lambda = 4L$ for the fundamental). Only odd numbers of quarter-wavelengths fit: $L = (2n-1)\frac{\lambda}{4} \quad\Rightarrow\quad f_n = \frac{(2n-1)v}{4L}, \qquad n = 1, 2, 3, \dots$ This gives frequencies $f_1, 3f_1, 5f_1, \dots$ — the even harmonics are missing. A closed pipe of a given length therefore sounds an octave lower (half the frequency) than the same pipe with both ends open.

A pipe closed at one end has a node (red dot) at the closed end and an antinode at the open end. Only odd harmonics fit — there is no 2nd or 4th harmonic, so the sequence is f₁, 3f₁, 5f₁, …

Open end = displacement antinode; closed end = displacement node.
Open pipe: L = nλ/2, fₙ = nv/2L, all harmonics (like a string).
Closed pipe: L = (2n−1)λ/4, fₙ = (2n−1)v/4L, odd harmonics only; fundamental λ = 4L.

See the full worked example for standing waves and resonance →

Resonance and damping

Driving a system at its natural frequency gives maximum amplitude; damping lowers and broadens the peak.

Every oscillating system — a string, an air column, a mass on a spring, a swing — has one or more natural frequencies at which it likes to vibrate once disturbed. The harmonics we have been calculating are the natural frequencies of a string or pipe.

Resonance is what happens when you drive (force) the system with a periodic push whose frequency equals a natural frequency. Energy is then transferred to the system most efficiently on every cycle, and its amplitude builds up to a maximum. This is why:

pushing a swing in time with its natural swing makes it go higher and higher,
a singer can shatter a glass by sounding its natural frequency,
a wind instrument only sounds loudly at the pipe's harmonic frequencies.

The resonance curve. If you plot the steady oscillation amplitude against the driving frequency, you get a peak centred on the natural frequency $f_0$ . Away from $f_0$ the amplitude is small; at $f_0$ it is largest.

Damping is any process (friction, air resistance, energy radiated as sound) that removes energy from the oscillation. Its effect on the resonance curve is:

Light damping → a tall, sharp peak (large maximum amplitude, resonance over a narrow band of frequencies).
Heavier damping → a lower, broader peak, shifted very slightly to a lower frequency.
Critical / heavy (over-)damping → the system barely resonates at all; the peak is flattened right out.

All three systems resonate at the same natural frequency f₀, but as the damping increases the resonance peak becomes lower and broader — heavier damping means a smaller maximum amplitude spread over a wider band of frequencies.

One phrase to keep straight: the driving frequency is whatever you choose to force the system at; the natural frequency is a fixed property of the system. They are equal only at resonance — do not treat "resonance frequency" and "driving frequency" as the same thing in general.

Resonance: maximum amplitude when driven at a natural frequency.
Resonance curve = amplitude vs driving frequency, peaking at f₀.
More damping → lower, broader peak (and slightly lower peak frequency).

See the full worked example for standing waves and resonance →

Quick recap

A standing wave forms when two waves of the same frequency and amplitude travel in opposite directions and superpose, giving fixed nodes (zero amplitude) and antinodes (maximum amplitude).
Standing wave vs travelling wave: no net energy transfer, amplitude varies with position, in phase between adjacent nodes — versus energy transferred, constant amplitude, phase varying with position.
String fixed at both ends (and pipe open at both ends): L = nλ/2, so fₙ = nv/2L — all integer harmonics.
Pipe closed at one end: node at closed end, antinode at open end, L = (2n−1)λ/4, so fₙ = (2n−1)v/4L — odd harmonics only; fundamental λ = 4L.
Adjacent nodes (or antinodes) are λ/2 apart; a node to its nearest antinode is λ/4 apart.
Resonance is maximum-amplitude oscillation when a system is driven at its natural frequency; damping lowers and broadens the resonance peak.

Memorise this

Verbatim phrases, formulae and definitions IB DP mark schemes credit (key for AO1 knowledge marks on Paper 1).

Standing wave = two identical waves in opposite directions superposing; no net energy transfer.
Node = zero/min amplitude; antinode = max amplitude. Node–node = λ/2; node–antinode = λ/4.
String / open pipe: L = nλ/2 → fₙ = nv/2L (ALL harmonics). Fundamental λ = 2L.
Closed pipe: L = (2n−1)λ/4 → fₙ = (2n−1)v/4L (ODD harmonics only). Fundamental λ = 4L.
Open end = antinode; closed/fixed end = node.
Resonance: driven at the natural frequency → maximum amplitude.
More damping → lower, broader resonance peak.
A standing wave transfers no net energy; points between adjacent nodes are in phase.

How it’s examined

Standing waves and resonance appear across all three papers. Paper 1A (MCQ): identifying nodes/antinodes on a diagram, working out which harmonic a pattern shows, comparing open and closed pipes, and recognising that node spacing is λ/2 — numbers are kept simple for the no-calculator paper. Paper 1B (data-based): a resonance-tube experiment (finding the speed of sound from the difference between successive resonance lengths, so the end correction cancels) or reading a resonance curve. Paper 2: structured calculations of harmonic frequencies for strings and both pipe types (3–6 marks), 'state the differences between standing and travelling waves', and 'explain resonance/the effect of damping' descriptive questions; occasionally an extended-response combining a boundary-condition sketch, a frequency calculation, and a resonance/damping explanation. Command terms: state, identify, determine, calculate, sketch, describe, explain. Examiner reports repeatedly flag: giving a closed pipe even harmonics, mislabelling which end is a node vs an antinode, using λ = L, treating node spacing as a whole wavelength, claiming a standing wave transfers energy, and confusing the driving frequency with the natural frequency. Always sketch the standing-wave pattern first and mark the boundary conditions — most marks follow directly from a correct sketch.

Sources: IB Diploma Programme Physics Guide (first assessment 2025) — Theme C: Wave behaviour (C.4 Standing waves and resonance); IB Physics Data Booklet (2025); IB Physics subject reports and specimen papers (2023–2025). Last reviewed 2026-07-21.

Take this whole topic with you

Step-by-step worked examples — Standing waves and resonance

Step-by-step solutions to past-paper-style questions on standing waves and resonance, written exactly the way a tutor would explain them at the board.

Question type:

1Node spacing gives the wavelength
Getting startedDirect calculation• nodes & antinodes, AO2
Question
A standing wave is set up on a stretched string. Adjacent nodes are found to be 0.15 m apart, and the string oscillates at 500 Hz. Calculate (a) the wavelength of the wave and (b) the speed of the two travelling waves that form it. (3 marks)
Step-by-step solution
1. Step 1
  The distance between two adjacent nodes is half a wavelength, so $\lambda = 2 \times (\text{node spacing})$ .
  $\lambda = 2 \times 0.15 = 0.30\,\text{m}$
2. Step 2
  The travelling waves have the same frequency as the standing wave. Use $v = f\lambda$ .
  $v = f\lambda = 500 \times 0.30 = 150\,\text{m s}^{-1}$
Answer
(a) λ = 0.30 m; (b) v = 150 m s⁻¹.
Examiner tip
Mark scheme: (1) node spacing = λ/2 so λ = 0.30 m; (2) correct use of v = fλ; (3) v = 150 m s⁻¹ with unit. The single most common slip is taking the node spacing to be a whole wavelength — it is a half.
2Fundamental frequency of a string
Getting startedDirect calculation• harmonics, string, AO2
Question
A guitar string of length 0.80 m is fixed at both ends. Waves travel along it at 320 m s⁻¹. Calculate (a) the fundamental frequency (first harmonic) and (b) the frequency of the third harmonic. (3 marks)
Step-by-step solution
1. Step 1
  A string fixed at both ends has $L = \dfrac{n\lambda}{2}$ , so the harmonic frequencies are $f_n = \dfrac{nv}{2L}$ . The fundamental is $n = 1$ .
  $f_1 = \frac{v}{2L} = \frac{320}{2 \times 0.80} = 200\,\text{Hz}$
2. Step 2
  The harmonics of a string are integer multiples of the fundamental, so the third harmonic is $f_3 = 3f_1$ .
  $f_3 = 3 \times 200 = 600\,\text{Hz}$
Answer
(a) f₁ = 200 Hz; (b) f₃ = 600 Hz.
Examiner tip
Mark scheme: (1) f₁ = v/2L; (2) f₁ = 200 Hz; (3) f₃ = 3f₁ = 600 Hz. A string fixed at both ends supports ALL integer harmonics, so f₃ is simply three times the fundamental.
3Fundamental of a closed pipe
Getting startedDirect calculation• harmonics, closed pipe, AO2
Question
A pipe of length 0.165 m is closed at one end and open at the other. The speed of sound in the air inside is 340 m s⁻¹. Calculate the fundamental frequency of the pipe. (3 marks)
Step-by-step solution
1. Step 1
  A pipe closed at one end has a node at the closed end and an antinode at the open end. For the fundamental this is a quarter of a wavelength, so $L = \dfrac{\lambda}{4}$ , giving $\lambda = 4L$ .
  $\lambda = 4L = 4 \times 0.165 = 0.66\,\text{m}$
2. Step 2
  Use $v = f\lambda$ to find the frequency.
  $f = \frac{v}{\lambda} = \frac{340}{0.66} = 515\,\text{Hz}$
Answer
f ≈ 515 Hz.
Examiner tip
Mark scheme: (1) fundamental of a closed pipe is λ/4 so λ = 4L; (2) λ = 0.66 m; (3) f ≈ 515 Hz with unit. Using λ = 2L (the string/open-pipe result) here is the classic error — a closed pipe's fundamental is a quarter, not a half, of a wavelength.
4First three harmonics of an open pipe
Building confidenceMulti-step problem• harmonics, open pipe, AO2
Question
An organ pipe of length 0.68 m is open at both ends. The speed of sound is 340 m s⁻¹. Calculate the frequencies of the first three harmonics. (4 marks)
Step-by-step solution
1. Step 1
  A pipe open at both ends has an antinode at each open end, exactly like a string it satisfies $L = \dfrac{n\lambda}{2}$ , so $f_n = \dfrac{nv}{2L}$ with $n = 1,2,3,\dots$ (all harmonics).
  $f_1 = \frac{v}{2L} = \frac{340}{2 \times 0.68} = 250\,\text{Hz}$
2. Step 2
  Second harmonic ( $n = 2$ ):
  $f_2 = 2f_1 = 2 \times 250 = 500\,\text{Hz}$
3. Step 3
  Third harmonic ( $n = 3$ ):
  $f_3 = 3f_1 = 3 \times 250 = 750\,\text{Hz}$
Answer
f₁ = 250 Hz, f₂ = 500 Hz, f₃ = 750 Hz.
Examiner tip
Mark scheme: (1) recognise open pipe behaves like a string, f₁ = v/2L = 250 Hz; (2) f₂ = 500 Hz; (3) f₃ = 750 Hz; (4) all with units. Open pipes give ALL integer harmonics — the sequence is 1f, 2f, 3f, exactly like a string.
5Same tube, open vs closed
Building confidenceMulti-step problem• open pipe, closed pipe, AO2, AO3
Question
A tube is 0.50 m long. The speed of sound is 340 m s⁻¹. Calculate the fundamental frequency when the tube is (a) open at both ends and (b) closed at one end. (c) State how the two fundamentals compare. (4 marks)
Step-by-step solution
1. Step 1
  Open at both ends: fundamental has $L = \dfrac{\lambda}{2}$ , so $f = \dfrac{v}{2L}$ .
  $f_{\text{open}} = \frac{340}{2 \times 0.50} = 340\,\text{Hz}$
2. Step 2
  Closed at one end: fundamental has $L = \dfrac{\lambda}{4}$ , so $f = \dfrac{v}{4L}$ .
  $f_{\text{closed}} = \frac{340}{4 \times 0.50} = 170\,\text{Hz}$
3. Step 3
  Compare: the closed-pipe fundamental is exactly half the open-pipe fundamental (one octave lower), because a closed pipe fits only a quarter-wavelength where an open pipe fits a half.
Answer
(a) 340 Hz; (b) 170 Hz; (c) the closed pipe's fundamental is half that of the open pipe (an octave lower).
Examiner tip
Mark scheme: (1) f_open = v/2L = 340 Hz; (2) f_closed = v/4L = 170 Hz; (3) closed = ½ open; (4) units throughout. Closing one end of a pipe of the same length halves the fundamental — a favourite comparison question.
6Identifying a harmonic from the pattern
Building confidenceGraph or diagram• harmonics, string, AO2, AO3
Question
A string of length 1.2 m fixed at both ends vibrates at 375 Hz. A photograph shows the standing wave has three antinodes. Calculate (a) the wavelength, (b) the wave speed on the string, and (c) the fundamental frequency. (4 marks)
Step-by-step solution
1. Step 1
  Three antinodes on a string fixed at both ends means the third harmonic, $n = 3$ , so $L = \dfrac{3\lambda}{2}$ , giving $\lambda = \dfrac{2L}{3}$ .
  $\lambda = \frac{2 \times 1.2}{3} = 0.80\,\text{m}$
2. Step 2
  Wave speed from $v = f\lambda$ using the 375 Hz vibration.
  $v = f\lambda = 375 \times 0.80 = 300\,\text{m s}^{-1}$
3. Step 3
  The fundamental is $f_1 = \dfrac{v}{2L}$ (or simply $f_3 / 3$ ).
  $f_1 = \frac{300}{2 \times 1.2} = 125\,\text{Hz}$
Answer
(a) λ = 0.80 m; (b) v = 300 m s⁻¹; (c) f₁ = 125 Hz.
Examiner tip
Mark scheme: (1) 3 antinodes → n = 3, λ = 2L/3 = 0.80 m; (2) v = fλ = 300 m s⁻¹; (3) f₁ = 125 Hz; (4) units. Check: f₃ = 3 × 125 = 375 Hz ✓ — counting the antinodes to get n is the assessed skill.
7Two consecutive closed-pipe resonances
StretchMulti-step problem• closed pipe, harmonics, AO2, AO3
Question
A pipe closed at one end resonates strongly at 550 Hz and again at the next resonance of 770 Hz, with no resonance in between. The speed of sound is 340 m s⁻¹. Determine (a) the fundamental frequency of the pipe and (b) its length. (5 marks)
Step-by-step solution
1. Step 1
  A closed pipe resonates only at odd multiples of its fundamental: $f_1, 3f_1, 5f_1, 7f_1,\dots$ . Two consecutive resonances therefore differ by $2f_1$ .
  $2f_1 = 770 - 550 = 220\,\text{Hz} \Rightarrow f_1 = 110\,\text{Hz}$
2. Step 2
  Check which harmonics these are: $550 / 110 = 5$ (the 5th harmonic) and $770 / 110 = 7$ (the 7th harmonic) — both odd, consecutive, and consistent.
  $550 = 5f_1,\quad 770 = 7f_1$
3. Step 3
  The fundamental of a closed pipe is $f_1 = \dfrac{v}{4L}$ , so rearrange for the length $L$ .
  $L = \frac{v}{4f_1} = \frac{340}{4 \times 110} = 0.77\,\text{m}$
Answer
(a) f₁ = 110 Hz; (b) L ≈ 0.77 m.
Examiner tip
Mark scheme: (1) consecutive closed-pipe resonances differ by 2f₁; (2) f₁ = 110 Hz; (3) identify 5th and 7th harmonics; (4) L = v/4f₁; (5) L ≈ 0.77 m with unit. The key insight is that a closed pipe skips even harmonics, so the gap between resonances is 2f₁, not f₁.
8Measuring the speed of sound with a resonance tube
StretchMulti-step problem• resonance, closed pipe, AO2, AO3, practical
Question
In a resonance-tube experiment a tuning fork of frequency 480 Hz is held over a tube closed at the water surface. The air column resonates first when its length is 0.170 m and again (the next resonance) at 0.525 m. Determine the speed of sound in the air. Explain why using the difference between the two lengths avoids the end-correction error. (4 marks)
Step-by-step solution
1. Step 1
  Consecutive resonances of a closed air column are half a wavelength apart, so $\dfrac{\lambda}{2}$ equals the difference in the two lengths.
  $\frac{\lambda}{2} = 0.525 - 0.170 = 0.355\,\text{m} \Rightarrow \lambda = 0.710\,\text{m}$
2. Step 2
  Use $v = f\lambda$ with the tuning-fork frequency.
  $v = f\lambda = 480 \times 0.710 = 341\,\text{m s}^{-1}$
3. Step 3
  The antinode sits slightly above the open end (the 'end correction'), so the first length is not exactly $\lambda/4$ . But the difference between two successive resonance lengths is always exactly $\lambda/2$ , because the same fixed end correction is present in both and cancels out.
Answer
v ≈ 341 m s⁻¹; taking the difference of the two lengths (= λ/2) cancels the constant end correction.
Examiner tip
Mark scheme: (1) λ/2 = length difference = 0.355 m; (2) λ = 0.710 m; (3) v = fλ = 341 m s⁻¹; (4) explanation that the end correction cancels in the difference. Using only the first length (0.170 m ≠ λ/4 exactly) would give a systematically low speed.

Model Answers — Standing waves and resonance

High-scoring sample answers for standing waves and resonance on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). For a standing wave, define (a) a node and (b) an antinode.
Model answer
(a) A node is a point on a standing wave where the amplitude of oscillation is always zero (a minimum) — the two superposing waves are permanently in antiphase there and cancel.

(b) An antinode is a point on a standing wave where the amplitude of oscillation is a maximum — the two superposing waves meet in phase, so their displacements add constructively.
Why this scores
Why this scores 2/2. (1) node = point of zero/minimum amplitude; (2) antinode = point of maximum amplitude. The word to include is amplitude — describing them as points of 'no wave' or 'biggest wave' is too vague for the mark.
Question 2
3 marks
Q (3 marks). A wire of length 0.75 m is fixed at both ends. Transverse waves travel along it at 270 m s⁻¹. Calculate the fundamental frequency of the wire.
Model answer
For a string fixed at both ends the fundamental (first harmonic) fits half a wavelength into the length: $L = \dfrac{\lambda}{2}$ , so $\lambda = 2L = 1.5\,\text{m}$ .

$f_1 = \frac{v}{2L} = \frac{270}{2 \times 0.75}$

$f_1 = \frac{270}{1.5} = 180\,\text{Hz}$

The fundamental frequency is 180 Hz.
Why this scores
Why this scores 3/3. (1) recognise fundamental is λ/2, λ = 1.5 m (or use f = v/2L directly); (2) correct substitution; (3) f₁ = 180 Hz with unit. Both ends are fixed so both are nodes — that is what forces the half-wavelength fit.
Question 3
3 marks
Q (3 marks). State three differences between a standing (stationary) wave and a travelling (progressive) wave.
Model answer
Energy transfer: a travelling wave transfers energy along its direction of travel; a standing wave transfers no net energy (energy is stored, oscillating between kinetic and potential locally).

Amplitude: in a travelling wave every point oscillates with the same amplitude; in a standing wave the amplitude varies with position, from zero at nodes to a maximum at antinodes.

Phase: in a travelling wave the phase changes continuously with position (adjacent points are slightly out of phase); in a standing wave all points between two adjacent nodes oscillate in phase (and points in adjacent loops are in antiphase).
Why this scores
Why this scores 3/3. One mark for each valid contrasting pair: (1) energy transfer vs none; (2) constant amplitude vs position-dependent amplitude; (3) continuously varying phase vs in-phase between nodes. Any three of these earn full marks; each must contrast both wave types, not just describe one.
Question 4
4 marks
Q (4 marks). A pipe of length 0.40 m is closed at one end. The speed of sound is 340 m s⁻¹. (a) Calculate the fundamental frequency. (b) State the frequency of the next possible harmonic, and explain why the pipe cannot produce a harmonic at twice the fundamental.
Model answer
(a) A closed pipe has a node at the closed end and an antinode at the open end, so the fundamental fits a quarter-wavelength: $L = \dfrac{\lambda}{4}$ , giving $\lambda = 4L = 1.6\,\text{m}$ .

$f_1 = \frac{v}{4L} = \frac{340}{4 \times 0.40} = 212.5\,\text{Hz} \approx 213\,\text{Hz}$

(b) The next harmonic is the third harmonic:

$f_3 = 3f_1 = 3 \times 212.5 = 637.5\,\text{Hz} \approx 638\,\text{Hz}$

A closed pipe can only support standing waves that have a node at the closed end and an antinode at the open end. This boundary condition is satisfied only by odd numbers of quarter-wavelengths, so only odd harmonics (1st, 3rd, 5th…) exist. A frequency of $2f_1$ would require an even harmonic, which cannot meet the node-at-one-end/antinode-at-the-other condition, so it is not produced.
Why this scores
Why this scores 4/4. (1) f₁ = v/4L = 213 Hz; (2) next harmonic is the 3rd, f₃ = 638 Hz; (3) states the boundary condition (node at closed end, antinode at open end); (4) explains that only odd harmonics satisfy it. Answering 'f₂ = 425 Hz' is the trap — even harmonics do not exist in a closed pipe.
Question 5
5 marks
Q (5 marks). An organ pipe open at both ends is 0.60 m long. The speed of sound is 340 m s⁻¹. (a) State the boundary condition at each open end. (b) Calculate the frequencies of the first three harmonics. (c) State how many antinodes the third harmonic has.
Model answer
(a) At each open end there is a displacement antinode (the air is free to move, so the amplitude is a maximum there).

(b) With an antinode at each end the pipe satisfies $L = \dfrac{n\lambda}{2}$ , so $f_n = \dfrac{nv}{2L}$ :

$f_1 = \frac{v}{2L} = \frac{340}{2 \times 0.60} = 283\,\text{Hz}$ $f_2 = 2f_1 = 567\,\text{Hz}, \qquad f_3 = 3f_1 = 850\,\text{Hz}$

(c) The third harmonic ( $n = 3$ ) has three antinodes along the pipe (with two nodes between them, plus the antinodes at the two open ends counting within the pattern).
Why this scores
Why this scores 5/5. (1) antinode at each open end; (2) f₁ = v/2L = 283 Hz; (3) f₂ = 567 Hz; (4) f₃ = 850 Hz; (5) third harmonic has 3 antinodes. An open pipe supports all integer harmonics, so the sequence is 1f : 2f : 3f — unlike a closed pipe.
Question 6
4 marks
Q (4 marks). (a) Explain what is meant by resonance. (b) A mass–spring system is driven by an oscillator whose frequency is slowly increased through the system's natural frequency. Describe how the amplitude of the mass changes, and explain the effect of increasing the amount of damping on the resonance peak.
Model answer
(a) Resonance occurs when a system is driven (forced to oscillate) at a frequency equal to its natural frequency. At this driving frequency the system absorbs energy most efficiently and oscillates with maximum amplitude.

(b) As the driving frequency is increased towards the natural frequency, the amplitude of the mass rises to a maximum at resonance, then falls again as the driving frequency moves beyond the natural frequency — giving a peak in the amplitude-versus-frequency graph.

Increasing the damping:

lowers the maximum amplitude at resonance (the peak is shorter), and

makes the peak broader and shifts it to a slightly lower frequency.

So heavy damping gives a low, broad resonance peak, while light damping gives a tall, sharp peak.
Why this scores
Why this scores 4/4. (1) resonance = driven at the natural frequency; (2) amplitude is maximum at resonance; (3) amplitude rises to a peak then falls as frequency passes through resonance; (4) more damping → lower and broader peak. The discriminator is linking heavier damping to both a lower and a broader peak.
Question 7
6 marks
Q (6 marks). A string of length 0.90 m is fixed at both ends and driven by a vibrator. As the driving frequency is slowly raised from zero, the first standing wave (one antinode) appears at 140 Hz. (a) Calculate the wave speed on the string. (b) Calculate the next two frequencies at which a standing wave forms. (c) Explain, in terms of resonance, why standing waves appear only at these particular frequencies.
Model answer
(a) One antinode means the fundamental (first harmonic), $n = 1$ , with $L = \dfrac{\lambda}{2}$ , so $\lambda = 2L = 1.8\,\text{m}$ .

$v = f\lambda = 140 \times 1.8 = 252\,\text{m s}^{-1}$

(b) A string fixed at both ends supports all integer harmonics, $f_n = nf_1$ :

$f_2 = 2 \times 140 = 280\,\text{Hz}, \qquad f_3 = 3 \times 140 = 420\,\text{Hz}$

(c) The string has a set of natural frequencies — the harmonics $f_1, f_2, f_3,\dots$ — determined by its length, tension and mass per unit length. A large-amplitude standing wave forms only when the driving frequency equals one of these natural frequencies, i.e. at resonance. At other frequencies the reflected waves do not reinforce consistently, so no stable standing wave (only small, irregular vibration) is seen.
Why this scores
Why this scores 6/6. (a) 2 marks: λ = 2L = 1.8 m, v = 252 m s⁻¹. (b) 2 marks: f₂ = 280 Hz, f₃ = 420 Hz. (c) 2 marks: standing waves are resonances that occur only when the driving frequency matches a natural frequency (harmonic) of the string. Linking the discrete frequencies to resonance at the natural frequencies is the assessed reasoning.
Question 8
6 marks
Q (6 marks). A wire fixed at both ends has a fundamental frequency of 150 Hz. It is observed to resonate strongly at 750 Hz. (a) Determine which harmonic 750 Hz corresponds to. (b) The wire is 0.60 m long. Calculate the wavelength of the standing wave at 750 Hz and the wave speed on the wire. (c) State the number of nodes (including the two ends) present in this standing wave.
Model answer
(a) For a wire fixed at both ends the harmonics are integer multiples of the fundamental, $f_n = nf_1$ :

$n = \frac{f_n}{f_1} = \frac{750}{150} = 5$

So 750 Hz is the fifth harmonic.

(b) For the fifth harmonic, $L = \dfrac{5\lambda}{2}$ , so:

$\lambda = \frac{2L}{5} = \frac{2 \times 0.60}{5} = 0.24\,\text{m}$

$v = f\lambda = 750 \times 0.24 = 180\,\text{m s}^{-1}$

(Check: $f_1 = v/2L = 180 / 1.2 = 150\,\text{Hz}$ ✓.)

(c) The $n$ th harmonic of a string has $n + 1$ nodes counting both fixed ends. For $n = 5$ that is 6 nodes (with 5 antinodes between them).
Why this scores
Why this scores 6/6. (a) 2 marks: n = f_n/f₁ = 5, the fifth harmonic. (b) 3 marks: λ = 2L/5 = 0.24 m and v = 180 m s⁻¹. (c) 1 mark: 6 nodes (n + 1). The self-check f₁ = v/2L = 150 Hz confirms the answer — examiners reward a consistency check.
Question 9
10 marks
Q (10 marks — extended response). A hollow tube of length 0.50 m contains air in which the speed of sound is 340 m s⁻¹. (a) Explain how a standing wave is formed inside a tube when a loudspeaker plays a suitable note at one end. (b) The tube is closed at one end. Sketch (describe) the displacement pattern of the fundamental and calculate its frequency. (c) The same tube is now open at both ends. Calculate the fundamental frequency and explain why it is higher than in part (b). (d) The closed tube is driven by a loudspeaker whose frequency is slowly increased from zero. Explain, in terms of resonance and natural frequencies, what is heard, and describe how heavier damping of the air column would change the loudness of each resonance.
Model answer
(a) Formation of the standing wave. The loudspeaker sends a sound (longitudinal travelling) wave down the tube. This wave reflects at the far end and travels back. The reflected wave and the incoming wave have the same frequency, speed and amplitude but travel in opposite directions, so by the principle of superposition they combine. Where they are always in antiphase the displacements cancel (a node); where they are always in phase they reinforce (an antinode). A stable pattern of fixed nodes and antinodes — a standing wave — is set up when the tube length matches the boundary conditions.

(b) Closed tube — fundamental. With one end closed there is a displacement node at the closed end and a displacement antinode at the open end. The fundamental fits a quarter-wavelength in the tube (node → antinode), so $L = \dfrac{\lambda}{4}$ and $\lambda = 4L = 2.0\,\text{m}$ . The pattern is a single node at the closed end swelling to a single antinode at the open end (no interior node).

$f_{\text{closed}} = \frac{v}{4L} = \frac{340}{4 \times 0.50} = 170\,\text{Hz}$

(c) Open tube — fundamental. With both ends open there is a displacement antinode at each end and a node in the middle, so the fundamental fits a half-wavelength: $L = \dfrac{\lambda}{2}$ , $\lambda = 2L = 1.0\,\text{m}$ .

$f_{\text{open}} = \frac{v}{2L} = \frac{340}{2 \times 0.50} = 340\,\text{Hz}$

This is higher (in fact exactly twice, an octave up) than the closed-tube fundamental because the open tube fits a half-wavelength into the same length whereas the closed tube fits only a quarter-wavelength. A shorter wavelength means, for the same wave speed, a higher frequency ( $f = v/\lambda$ ).

(d) Resonance and damping. As the driving frequency rises from zero, most frequencies produce only a faint, unstable vibration. But whenever the driving frequency equals one of the tube's natural frequencies — for the closed tube these are the odd harmonics $f_1 = 170\,\text{Hz}$ , $f_3 = 510\,\text{Hz}$ , $f_5 = 850\,\text{Hz}\dots$ — the air column resonates: the standing wave builds to a large amplitude and a loud note is heard. So a series of loud resonances is heard at these specific frequencies, separated by quiet regions.

If the air column were more heavily damped (for example by adding an absorbing material), each resonance would be less loud — the peak amplitude at each natural frequency is reduced — and the resonances would be less sharp (loud over a broader band of frequencies rather than at a single sharp value). Very heavy damping would make the resonances barely noticeable.
Why this scores
Why this scores 10/10. (a) 3 marks: incoming and reflected waves of equal frequency/amplitude travel in opposite directions and superpose to give fixed nodes and antinodes. (b) 2 marks: node at closed end/antinode at open end, λ = 4L, f = 170 Hz. (c) 2 marks: antinodes at both ends, λ = 2L, f = 340 Hz, and reasoning that a shorter fitted wavelength gives a higher frequency. (d) 3 marks: resonance occurs only at the natural (odd-harmonic) frequencies giving loud notes; heavier damping lowers and broadens each resonance. This is a model Paper 2 extended response — the discriminators are (i) getting the closed vs open boundary conditions and factor-of-two relationship right, and (ii) explaining damping in terms of both reduced peak loudness and broader (less sharp) resonance.

Key Formulae — Standing waves and resonance

The formulae you need to memorise for standing waves and resonance on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Harmonics on a string (or open pipe)
$f_n = \frac{nv}{2L}, \qquad n = 1, 2, 3, \dots$
$f_n$
frequency of the nth harmonic (Hz)
$n$
harmonic number (any positive integer)
$v$
wave speed on the string / speed of sound (m s⁻¹)
$L$
length of the string or pipe (m)
When to use
For a string fixed at both ends, or a pipe open at both ends — both fit L = nλ/2 and give ALL integer harmonics.
Example
String, L = 0.80 m, v = 320 m s⁻¹: f₁ = 320/(2×0.80) = 200 Hz; f₃ = 600 Hz.
Wavelengths of a string / open pipe
$\lambda_n = \frac{2L}{n}, \qquad n = 1, 2, 3, \dots$
$\lambda_n$
wavelength of the nth harmonic (m)
$L$
length of the string or open pipe (m)
$n$
harmonic number
When to use
To find the wavelength that fits a string (both ends fixed) or an open pipe. The fundamental (n = 1) is λ = 2L.
Example
Third harmonic on a 1.2 m string: λ₃ = 2(1.2)/3 = 0.80 m.
Harmonics of a closed pipe (odd only)
$f_n = \frac{(2n-1)v}{4L}, \qquad n = 1, 2, 3, \dots$
$f_n$
frequency of the nth allowed harmonic (Hz)
$(2n-1)$
the odd harmonic numbers 1, 3, 5, …
$v$
speed of sound in the pipe (m s⁻¹)
$L$
length of the pipe (m)
When to use
For a pipe closed at one end (node at the closed end, antinode at the open end): L = (2n−1)λ/4, so ONLY odd harmonics exist. Fundamental λ = 4L.
Example
Closed pipe, L = 0.50 m, v = 340 m s⁻¹: f₁ = 340/(4×0.50) = 170 Hz; next is f₃ = 510 Hz.
Spacing of nodes and antinodes
$d_{\text{node-node}} = \frac{\lambda}{2}, \qquad d_{\text{node-antinode}} = \frac{\lambda}{4}$
$\lambda$
wavelength of the standing wave (m)
$d_{\text{node-node}}$
distance between two adjacent nodes (or two adjacent antinodes)
$d_{\text{node-antinode}}$
distance from a node to the nearest antinode
When to use
To find a wavelength from a measured pattern: adjacent nodes are half a wavelength apart; a node and its nearest antinode are a quarter apart.
Example
Adjacent nodes 0.15 m apart → λ = 2 × 0.15 = 0.30 m.
Wave equation
$v = f\lambda$
$v$
wave speed (m s⁻¹)
$f$
frequency (Hz)
$\lambda$
wavelength (m)
When to use
Links the speed, frequency and wavelength of the travelling waves that make up a standing wave — used to convert between a measured wavelength and a frequency.
Example
λ = 0.30 m at 500 Hz → v = 500 × 0.30 = 150 m s⁻¹.

Key Definitions and Keywords — Standing waves and resonance

Definitions to memorise and the exact keywords mark schemes credit for standing waves and resonance answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Standing (stationary) wave
Examiner keyword
A wave pattern formed when two waves of the same frequency and amplitude travel in opposite directions and superpose, producing fixed nodes and antinodes. No net energy is transferred along it.
Travelling (progressive) wave
Examiner keyword
A wave that transfers energy through a medium in the direction of travel; every point oscillates with the same amplitude and the phase varies continuously with position.
Principle of superposition
Examiner keyword
When two or more waves overlap, the resultant displacement at each point is the vector sum of the individual displacements. This is what produces the nodes and antinodes of a standing wave.
Node
Examiner keyword
A point on a standing wave where the amplitude of oscillation is always zero (a minimum), because the two superposing waves are permanently in antiphase there.
Antinode
Examiner keyword
A point on a standing wave where the amplitude of oscillation is a maximum, because the two superposing waves meet in phase and reinforce.
Harmonic
Examiner keyword
One of the discrete frequencies at which a standing wave can be set up on a system. A string/open pipe allows all integer harmonics (f, 2f, 3f…); a closed pipe allows only odd harmonics (f, 3f, 5f…).
Fundamental frequency (first harmonic)
Examiner keyword
The lowest frequency at which a standing wave forms on a system — the first harmonic. For a string/open pipe f₁ = v/2L; for a closed pipe f₁ = v/4L.
Overtone
Any harmonic above the fundamental. The first overtone is the next allowed harmonic above f₁ (the 2nd harmonic for a string/open pipe, the 3rd harmonic for a closed pipe).
Resonance
Examiner keyword
The large-amplitude response of a system when it is driven (forced to oscillate) at a frequency equal to one of its natural frequencies, at which it absorbs energy most efficiently.
Natural frequency
Examiner keyword
A frequency at which a system oscillates freely once disturbed. A system has a set of natural frequencies (its harmonics); driving it at one of them produces resonance.
Driving (forcing) frequency
Examiner keyword
The frequency of the external periodic force applied to a system. Resonance occurs only when the driving frequency equals a natural frequency — the two are equal at resonance but are otherwise distinct.
Damping
Examiner keyword
The dissipation of an oscillating system's energy (e.g. by friction or drag). Heavier damping lowers and broadens the resonance peak; light damping gives a tall, sharp peak.
Boundary condition
Examiner keyword
The requirement fixing what must occur at each end of the system: a fixed string end or a closed pipe end is a node; a free string end or an open pipe end is an antinode. These conditions determine the allowed harmonics.
In phase (within a loop)
All the points between two adjacent nodes of a standing wave oscillate in phase — they reach their maximum displacements at the same instant. Points in neighbouring loops are in antiphase.
End correction
A small extra length added to a pipe's measured length because the displacement antinode lies slightly beyond an open end. It cancels when the difference between two successive resonance lengths is used.

Common Mistakes and Misconceptions — Standing waves and resonance

The traps other students keep falling into on standing waves and resonance questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Applying the string/open-pipe harmonic series to a pipe closed at one end
IB Physics Theme C subject reports
Why it happens
Students memorise fₙ = nv/2L and use it for every pipe, forgetting that a closed pipe has a different boundary condition.
How to avoid it
A closed pipe (node at the closed end, antinode at the open end) fits L = (2n−1)λ/4 and produces ONLY odd harmonics (1st, 3rd, 5th…). Its fundamental is v/4L, and there is no 2nd, 4th… harmonic. Check the ends before choosing a formula.
✕Confusing where nodes and antinodes sit at the ends of a pipe
IB Physics Theme C subject reports
Why it happens
Nodes and antinodes look symmetric on a diagram, so students place them at the wrong ends.
How to avoid it
Learn the rule: an OPEN end is a displacement ANTINODE (air free to move); a CLOSED end (or a fixed string end) is a NODE (air/string cannot move). Draw the antinode as a bulge at every open end before doing any calculation.
✕Using λ = L instead of the correct fraction (L = nλ/2 or L = (2n−1)λ/4)
IB Physics Theme C subject reports
Why it happens
Students assume 'one wave fits the length', ignoring that the fundamental is a half-wavelength (string/open) or a quarter-wavelength (closed).
How to avoid it
Always sketch the standing wave first. Fundamental of a string/open pipe = ½λ (so λ = 2L); fundamental of a closed pipe = ¼λ (so λ = 4L). Never assume a whole wavelength fits.
✕Taking the distance between adjacent nodes to be a whole wavelength
IB Physics Theme C subject reports
Why it happens
Students see one 'loop' between nodes and assume it is a full wave.
How to avoid it
The distance between two adjacent nodes (or two adjacent antinodes) is HALF a wavelength (λ/2); a node to the nearest antinode is a QUARTER (λ/4). Multiply the node spacing by 2 to get λ.
✕Thinking a standing wave transfers energy along the medium
IB Physics Theme C subject reports
Why it happens
It is made of travelling waves, so students assume it must carry energy like them.
How to avoid it
A standing wave transfers NO NET energy — the two travelling waves carry equal energy in opposite directions, so it cancels. Energy is stored locally, oscillating between kinetic and potential within each loop.
✕Confusing the resonance (natural) frequency with the driving frequency in general
IB Physics Theme C subject reports
Why it happens
At resonance the two are equal, so students wrongly treat them as always the same.
How to avoid it
The driving frequency is set by the external oscillator and can be anything; the natural frequency is a fixed property of the system. Resonance is the special case where the driving frequency EQUALS a natural frequency — elsewhere they differ and the amplitude is small.

Standing waves and resonance

Detailed Study Notes

At a glance

What you’ll learn

How a standing wave forms: superposition of two waves

Standing wave vs travelling wave: the three differences

Harmonics on a string fixed at both ends

Deriving the harmonic frequencies fₙ = nv/2L (first principles)

Pipes: open at both ends vs closed at one end

Resonance and damping

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Node spacing gives the wavelength

2Fundamental frequency of a string

3Fundamental of a closed pipe

4First three harmonics of an open pipe

5Same tube, open vs closed

6Identifying a harmonic from the pattern

7Two consecutive closed-pipe resonances

8Measuring the speed of sound with a resonance tube

Harmonics on a string (or open pipe)

Wavelengths of a string / open pipe

Harmonics of a closed pipe (odd only)

Spacing of nodes and antinodes

Wave equation

Standing (stationary) wave

Travelling (progressive) wave

Principle of superposition

Node

Antinode

Harmonic

Fundamental frequency (first harmonic)

Overtone

Resonance

Natural frequency

Driving (forcing) frequency

Damping

Boundary condition

In phase (within a loop)

End correction

✕Applying the string/open-pipe harmonic series to a pipe closed at one end

✕Confusing where nodes and antinodes sit at the ends of a pipe

✕Using λ = L instead of the correct fraction (L = nλ/2 or L = (2n−1)λ/4)

✕Taking the distance between adjacent nodes to be a whole wavelength

✕Thinking a standing wave transfers energy along the medium

✕Confusing the resonance (natural) frequency with the driving frequency in general