What is the principle of superposition in the context of wave interference?

The principle of superposition states that when two or more waves overlap in space, the resultant wave displacement at any point is equal to the sum of the displacements of the individual waves at that point. This principle is fundamental in understanding interference patterns, where constructive interference occurs when waves are in phase, and destructive interference occurs when they are out of phase.

How do stationary waves form and what are their characteristics?

Stationary waves, or standing waves, form when two waves of the same frequency and amplitude travel in opposite directions and interfere with each other. This results in a wave pattern that appears to be standing still, characterized by nodes (points of zero amplitude) and antinodes (points of maximum amplitude). Stationary waves are commonly observed in musical instruments and are crucial in understanding resonance.

What is the difference between constructive and destructive interference?

Constructive interference occurs when two waves meet in phase, meaning their crests and troughs align, resulting in a wave with greater amplitude. Destructive interference happens when two waves meet out of phase, meaning the crest of one wave aligns with the trough of another, leading to a reduction or cancellation of amplitude. These concepts are essential in analyzing wave patterns and applications such as noise-canceling headphones.

How does the double-slit experiment demonstrate wave interference?

The double-slit experiment demonstrates wave interference by showing how light behaves as a wave. When light passes through two closely spaced slits, it creates an interference pattern of bright and dark fringes on a screen. This pattern results from constructive interference at the bright fringes and destructive interference at the dark fringes, providing evidence for the wave nature of light.

What role do stationary waves play in musical instruments?

Stationary waves are crucial in musical instruments as they determine the pitch and tone of the sound produced. In string instruments, stationary waves form along the strings, with nodes at fixed points and antinodes at varying points depending on the frequency. In wind instruments, stationary waves form in the air column, with the length of the column affecting the frequency and pitch of the sound. Understanding stationary waves helps in designing and playing musical instruments effectively.

Why is "Taking the distance between adjacent nodes (or antinodes) as a whole wavelength instead of half a wavelength" a common mistake in Interference and Stationary Waves?

Why it happens: Students picture one 'loop' as a full wave, forgetting that a complete wavelength spans two loops (node–antinode–node–antinode–node). How to avoid it: Remember that **adjacent nodes are 2 apart** (so are adjacent antinodes). One loop (node to next node) is half a wavelength; the full wavelength is the distance between alternate nodes. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Defining coherence as 'same amplitude' or 'in phase', rather than a constant phase difference" a common mistake in Interference and Stationary Waves?

Why it happens: Students confuse the conditions for a clear pattern with the precise definition of coherence. How to avoid it: Coherence means a **constant phase difference** (which requires the same frequency). Equal amplitude makes the cancellation at minima complete but is **not part of the definition** of coherence. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Saying superposition adds the amplitudes of the waves rather than their displacements" a common mistake in Interference and Stationary Waves?

Why it happens: The words 'amplitude' and 'displacement' are used loosely in everyday speech. How to avoid it: State that the **resultant displacement** is the **sum of the individual displacements** at that instant. Displacements can be positive or negative, which is why destructive interference is possible — amplitudes (always positive) cannot simply add. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Using $\lambda = L$ instead of $\lambda = 2L$ for the fundamental of a string fixed at both ends" a common mistake in Interference and Stationary Waves?

Why it happens: Students see one loop spanning the whole string and assume that is one wavelength. How to avoid it: For the fundamental there is a node at each end and **one antinode** in the middle, so the string holds **half a wavelength**: L = 2, giving ** = 2L**. Sketch the loop and mark the nodes to check. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Forgetting the square root in $v=\sqrt{T/\mu}$, or not squaring $f$ before plotting against $T$" a common mistake in Interference and Stationary Waves?

Why it happens: The relationship v=√(T/) is square-root, not linear, and is easy to misremember under time pressure. How to avoid it: Always **square-root T/** to get the speed. To linearise the CP5 data, **plot f^2 against T** (a straight line through the origin), not f against T (a curve). Source: Edexcel IAL Physics Unit 2 examiner reports (Core Practical 5)

Why is "Claiming a stationary wave transfers energy, or that all points have the same amplitude" a common mistake in Interference and Stationary Waves?

Why it happens: Students apply the properties of a progressive wave to a stationary one without thinking about the differences. How to avoid it: A stationary wave **stores energy and does not transfer it** along the medium, and its **amplitude varies with position** (zero at nodes, maximum at antinodes). Contrast this explicitly with a progressive wave in 'differences' questions. Source: Edexcel IAL Physics Unit 2 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Interference and Stationary Waves

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Short Notes - Interference and Stationary Waves

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Detailed Study Notes

Detailed notes on Waves and Electricity for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Interference and Stationary Waves — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 2)

The principle of superposition; constructive and destructive interference with the path-difference conditions $n\lambda$ and $\left(n+\tfrac12\right)\lambda$ ; phase difference and coherence; two-source (Young's double-slit) interference; stationary waves with nodes and antinodes; harmonics on a stretched string with $v=\sqrt{T/\mu}$ and $f=\tfrac{n}{2L}\sqrt{T/\mu}$ ; and Core Practical 5 — investigating the vibrating string.

At a glance

Superposition: when waves meet, the resultant displacement = the sum of the individual displacements (with signs). Displacements add, not amplitudes.
Constructive interference where waves arrive in phase — path difference $= n\lambda$ . Destructive interference where they arrive in antiphase — path difference $= \left(n+\tfrac{1}{2}\right)\lambda$ .
Coherence = constant phase difference (so the same frequency). It is needed for a stable interference pattern — not 'same amplitude'.
Stationary wave = two progressive waves of the same frequency and equal amplitude travelling in opposite directions superposing. It stores energy; it does not transfer it.
Nodes = amplitude always zero; antinodes = amplitude maximum. Adjacent nodes (or antinodes) are $\tfrac{\lambda}{2}$ apart.
Wave speed on a string: $v=\sqrt{\dfrac{T}{\mu}}$ ( $T$ = tension, $\mu$ = mass per unit length).
Harmonics (string fixed at both ends): $f=\dfrac{n}{2L}\sqrt{\dfrac{T}{\mu}}$ ; the fundamental ( $n=1$ ) has one antinode and $\lambda=2L$ ; $f_n = n f_1$ .
Core Practical 5: vary $L$ , $T$ or $\mu$ and find the fundamental frequency; plot $f^2$ against $T$ (straight line through the origin, gradient $=\tfrac{1}{4L^2\mu}$ ) to find $\mu$ .

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

2.1 — State and apply the principle of superposition of waves.
2.2 — Explain constructive and destructive interference in terms of path difference ( $n\lambda$ and $\left(n+\tfrac{1}{2}\right)\lambda$ ) and phase difference.
2.3 — Explain the meaning of coherence and why it is required to observe a stable interference pattern; describe two-source (Young's double-slit) interference.
2.4 — Explain how stationary waves are formed by superposition of two waves travelling in opposite directions, and describe nodes and antinodes (adjacent nodes $\tfrac{\lambda}{2}$ apart).
2.5 — Use $v=\sqrt{T/\mu}$ and the harmonic relation $f=\tfrac{n}{2L}\sqrt{T/\mu}$ for a string fixed at both ends; distinguish progressive and stationary waves.
CP5 — Investigate the effects of length, tension and mass per unit length on the frequency of a stationary wave on a vibrating string, and determine a quantity (e.g. $\mu$ ) from a suitable linearised graph.

Superposition — how waves combine

Where waves overlap, add their displacements (with signs) at every instant.

Whenever two or more waves pass through the same point at the same time, they superpose. The principle of superposition states:

When two or more waves meet at a point, the resultant displacement at that point equals the vector sum of the individual displacements of the waves at that instant.

The crucial word is displacement, not amplitude. Displacements carry a sign (a crest is positive, a trough is negative), so two waves can either reinforce or cancel:

If two crests (both positive) coincide, they add to a larger displacement — constructive.
If a crest (positive) coincides with a trough (negative), they partly or fully cancel — destructive.

After the waves have crossed, each one carries on unchanged — superposition is only what happens while they overlap.

Adding two in-phase waves point by point gives a resultant of double the amplitude — constructive interference. Two waves in antiphase would instead cancel to give zero.

The everyday consequence of superposition is interference: where waves repeatedly meet in step they build a strong signal, and where they repeatedly meet out of step they cancel to a weak (or zero) signal.

Resultant displacement = sum of the individual displacements (with signs).
Crest + crest → bigger (constructive); crest + trough → cancels (destructive).
Waves emerge unchanged after overlapping.
It is displacements that add, not amplitudes.

See the full worked example for interference and stationary waves →

Interference, path difference, phase and coherence

Constructive at $n\lambda$ , destructive at $\left(n+\tfrac12\right)\lambda$ — but only for coherent sources.

When two sets of waves from two sources overlap, whether you get a maximum or a minimum at a point depends on the path difference — the difference between the distances the two waves have travelled to reach that point.

The two conditions (learn these exactly):

At a point you get…	when the path difference is…	because the waves are…
Constructive (maximum)	$n\lambda$ (a whole number of wavelengths)	in phase
Destructive (minimum)	$\left(n+\tfrac{1}{2}\right)\lambda$ (odd number of half-wavelengths)	in antiphase

where $n = 0, 1, 2, 3, \ldots$

Phase difference is the partner idea. One whole wavelength of path difference corresponds to a full cycle, i.e. a phase difference of $2\pi$ rad (360°): $\Delta\phi = \frac{2\pi}{\lambda}\times(\text{path difference})$ So a path difference of $\tfrac{\lambda}{2}$ is a phase difference of $\pi$ rad (180°) — antiphase — and gives cancellation.

Coherence. A clear, stable interference pattern is only seen if the two sources are coherent:

Two sources are coherent if they maintain a constant phase difference (which requires them to have the same frequency).

If the phase difference between the sources kept changing randomly (incoherent sources, e.g. two separate lamps), the positions of the maxima and minima would jump about too fast to see, and the pattern would average out to uniform brightness. In practice coherence is achieved by deriving the two sources from a single source — e.g. by passing one beam of monochromatic light through two slits (Young's double-slit) so the slits act as two coherent sources.

Two coherent sources S₁ and S₂ produce overlapping waves. Solid green lines reach bright fringes (path difference = nλ, constructive); dashed red lines reach dark fringes (path difference = (n + ½)λ, destructive). The central fringe is bright because the two paths are equal.

Constructive: path difference $= n\lambda$ (in phase).
Destructive: path difference $= \left(n+\tfrac{1}{2}\right)\lambda$ (antiphase).
Phase difference $= \tfrac{2\pi}{\lambda}\times$ path difference; $\tfrac{\lambda}{2}$ ⇒ $\pi$ rad.
Coherent = constant phase difference (same frequency) ⇒ stable pattern.

See the full worked example for interference and stationary waves →

Stationary waves — nodes and antinodes

Two equal waves going opposite ways superpose into a fixed pattern of nodes and antinodes, $\tfrac{\lambda}{2}$ apart.

A stationary (or standing) wave is formed when two progressive waves of the same frequency and equal amplitude travel in opposite directions and superpose. On a string this happens when a wave is reflected at a fixed end and travels back to overlap the incoming wave.

The superposition produces a pattern that does not travel — it just oscillates in place:

Nodes — points where the two waves are always in antiphase, so they continually cancel: the amplitude is always zero.
Antinodes — points where the two waves are always in phase, so they continually reinforce: the amplitude is a maximum.

The geometry to memorise: adjacent nodes are $\tfrac{\lambda}{2}$ apart, and so are adjacent antinodes; a node and its neighbouring antinode are $\tfrac{\lambda}{4}$ apart. One "loop" (node to next node) is therefore half a wavelength.

A stationary wave (third harmonic) on a string fixed at both ends. The two faint curves show the extreme positions; nodes (red) stay still while antinodes (green) oscillate with maximum amplitude. Adjacent nodes are half a wavelength apart.

Stationary versus progressive — the differences examiners test:

	Progressive wave	Stationary wave
Energy	Transferred along the wave	Stored — not transferred along
Amplitude	Same at every point	Varies: zero at nodes, max at antinodes
Phase	Neighbouring points slightly out of phase	All points between two nodes oscillate in phase
Waveform	Profile moves along	Profile stays in a fixed position

Formed by two waves of equal frequency/amplitude going in OPPOSITE directions.
Node = amplitude always zero; antinode = amplitude maximum.
Adjacent nodes (or antinodes) are $\tfrac{\lambda}{2}$ apart; node-to-antinode $\tfrac{\lambda}{4}$ .
Stationary wave stores energy; amplitude varies with position.

See the full worked example for interference and stationary waves →

Harmonics on a stretched string

Fundamental: $\lambda=2L$ ; speed $v=\sqrt{T/\mu}$ ; $f=\tfrac{n}{2L}\sqrt{T/\mu}$ , with $f_n = n f_1$ .

A string fixed at both ends must have a node at each end, so only certain wavelengths "fit" and form a stationary wave. These special frequencies are the harmonics.

Fundamental (1st harmonic, $n=1$ ): one antinode in the middle, a node at each end. The string holds half a wavelength, so $L = \tfrac{\lambda}{2}$ and $\lambda = 2L$ .
2nd harmonic ( $n=2$ ): two antinodes, $\lambda = L$ .
$n$ th harmonic: $n$ antinodes, $L = n\tfrac{\lambda}{2}$ , so $\lambda = \dfrac{2L}{n}$ .

The first three harmonics of a string fixed at both ends. The number of loops (antinodes) equals the harmonic number n; red dots mark the internal nodes. Each higher harmonic fits more half-wavelengths into the same length.

The speed of a transverse wave on a string depends on how tight and how heavy it is: $v = \sqrt{\frac{T}{\mu}}$ where $T$ is the tension (N) and $\mu$ is the mass per unit length (kg m⁻¹).

Combine this with $v = f\lambda$ and $\lambda = \dfrac{2L}{n}$ to get the frequency of any harmonic: $f = \frac{v}{\lambda} = \frac{n}{2L}\sqrt{\frac{T}{\mu}}$

So the harmonics are simple multiples of the fundamental: $f_n = n f_1$ . Increasing the tension raises the frequency (tighter strings sound higher); a longer or heavier string lowers it. This is exactly how a guitarist tunes a string (tension) and why the thick (high- $\mu$ ) strings sound low.

Fundamental: one loop, $\lambda = 2L$ , lowest frequency.
$n$ th harmonic: $n$ loops, $\lambda = \tfrac{2L}{n}$ , $f_n = n f_1$ .
Wave speed: $v=\sqrt{T/\mu}$ .
Frequency: $f=\tfrac{n}{2L}\sqrt{T/\mu}$ — raise $T$ ⇒ higher $f$ ; raise $L$ or $\mu$ ⇒ lower $f$ .

See the full worked example for interference and stationary waves →

Core Practical 5 — investigating the vibrating string

Vary L, T or μ, find the fundamental, and plot a linear graph (e.g. f² vs T) to extract μ.

Core Practical 5 investigates how the length $L$ , the tension $T$ and the mass per unit length $\mu$ of a stretched string affect the frequency of a stationary wave on it.

Typical set-up. A string runs over two fixed bridges, with one end attached to a vibration generator (driven by a signal generator) and the other passing over a pulley to a hanging mass, which sets the tension $T = mg$ . The length between the bridges is $L$ .

Finding the fundamental. Slowly increase the driving frequency until the string vibrates with the largest amplitude in a single loop (one antinode, a node at each end). This resonance is the fundamental; read the frequency from the signal generator. (A paper rider or fine sand can confirm a node; a stroboscope helps freeze the motion.)

What you vary, and the graph that linearises it. Starting from $f = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$ (fundamental), choose one variable, hold the others constant, and plot a straight-line graph:

Tension: square the equation → $f^2 = \dfrac{1}{4L^2\mu}\,T$ . Plot $f^2$ (y) against $T$ (x): a straight line through the origin with gradient $=\dfrac{1}{4L^2\mu}$ .
Length: with $T$ , $\mu$ fixed, $f = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}} \propto \dfrac{1}{L}$ . Plot $f$ against $\dfrac{1}{L}$ : a straight line through the origin.

Plotting f² against T gives a straight line through the origin whose gradient is 1/(4L²μ). Reading the gradient from a large triangle and rearranging gives the string's mass per unit length μ.

Reducing and judging uncertainty

Repeat the resonance for each setting and take a mean frequency to reduce random error.
Use a range of tensions (or lengths) and a best-fit line so a single anomalous reading does not distort the gradient.
Measure $L$ with a metre rule and the masses on a balance; find $\mu$ independently (mass ÷ length of a measured sample) to check the value from the gradient.
The line passing through the origin confirms the expected proportionality (e.g. $f^2 \propto T$ ).

Tension set by a hanging mass: $T = mg$ ; drive with a vibration generator.
Fundamental = increase frequency to the single-loop resonance; read $f$ .
Plot $f^2$ vs $T$ : straight line through origin, gradient $=\tfrac{1}{4L^2\mu}$ .
Get $\mu = \dfrac{1}{4L^2 \times \text{gradient}}$ ; repeat and use a best-fit line.

See the full worked example for interference and stationary waves →

Quick recap

Superposition: the resultant displacement is the sum of the individual displacements (with signs) — displacements add, not amplitudes.
Constructive interference at path difference $= n\lambda$ (in phase); destructive at $\left(n+\tfrac{1}{2}\right)\lambda$ (antiphase).
Coherence means a constant phase difference (same frequency) and is required for a stable interference pattern.
A stationary wave is two equal waves travelling in opposite directions; nodes have zero amplitude, antinodes maximum, and adjacent nodes are $\tfrac{\lambda}{2}$ apart.
Wave speed on a string $v=\sqrt{T/\mu}$ ; for a string fixed at both ends $f=\tfrac{n}{2L}\sqrt{T/\mu}$ with the fundamental $\lambda=2L$ and $f_n = n f_1$ .
Core Practical 5: find the fundamental at each setting and linearise (e.g. plot $f^2$ vs $T$ , gradient $=\tfrac{1}{4L^2\mu}$ ) to determine $\mu$ .
Progressive waves transfer energy with constant amplitude; stationary waves store energy with amplitude varying from node to antinode.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Superposition: resultant displacement = sum of the individual displacements.
Constructive: path difference $= n\lambda$ (in phase). Destructive: path difference $= \left(n+\tfrac{1}{2}\right)\lambda$ (antiphase).
Coherent = constant phase difference (same frequency) — needed for a stable pattern.
Stationary wave = two waves, same frequency, equal amplitude, OPPOSITE directions; nodes = zero amplitude, antinodes = maximum.
Adjacent nodes (or antinodes) are $\tfrac{\lambda}{2}$ apart; node-to-antinode $\tfrac{\lambda}{4}$ .
$v=\sqrt{\dfrac{T}{\mu}}$ ; fundamental $\lambda = 2L$ ; $f=\dfrac{n}{2L}\sqrt{\dfrac{T}{\mu}}$ , $f_n = n f_1$ .
Stationary wave stores energy (no transfer along it); amplitude varies — progressive wave transfers energy with constant amplitude.
CP5: plot $f^2$ against $T$ → straight line through origin, gradient $=\dfrac{1}{4L^2\mu}$ .

How it’s examined

Interference and stationary waves are core Unit 2 content and appear on the Unit 2 paper (WPH12, Waves and Electricity), which is structured with short calculations, 'show that' items, explanation questions and a longer multi-part question. Typical formats: state the principle of superposition or define coherence/node/antinode (1–2 marks, AO1); decide constructive vs destructive from a path difference using $n\lambda$ or $\left(n+\tfrac12\right)\lambda$ (2–3 marks, AO2); calculate a wave speed or harmonic frequency with $v=\sqrt{T/\mu}$ and $f=\tfrac{n}{2L}\sqrt{T/\mu}$ , showing equation, substitution with units and a 3 s.f. answer (3–4 marks, M and A); explain how a stationary wave forms or contrast progressive and stationary waves in linked prose (4–6 marks); and a CP5 data question (linearise to $f^2$ vs $T$ , find a gradient from a large triangle, determine $\mu$ , evaluate uncertainty). ALWAYS show the equation, substitution with units, and a 3 s.f. answer — 'no working = risk zero'. For 'show that' work fully to the value (one extra s.f.); for definitions quote the exact mark-scheme phrase (e.g. 'constant phase difference' for coherence, 'amplitude always zero' for a node).

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 2 examiner reports; Edexcel IAL Physics Unit 2 sample assessment materials and mark schemes. Last reviewed 2026-06-26.

Take this whole topic with you

Step-by-step worked examples — Interference and Stationary Waves

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

Question type:

1Resultant displacement by superposition (2 marks)
Getting startedDirect calculation• superposition, AO2, Unit 2
Question
Two waves arrive at the same point at the same instant. One has a displacement of +6.0 mm and the other a displacement of −2.5 mm at that point. Use the principle of superposition to find the resultant displacement. (2 marks)
Step-by-step solution
1. Step 1
  The principle of superposition says the resultant displacement is the vector sum of the individual displacements — here just add them with their signs.
  $y = y_1 + y_2$
2. Step 2
  Substitute the two displacements (keep the signs).
  $y = (+6.0) + (-2.5) = +3.5\ \text{mm}$
Answer
Resultant displacement $= +3.5\ \text{mm}$ (3.5 mm in the positive direction).
Examiner tip
Mark scheme: (1) state/use superposition = add displacements; (2) $+3.5\ \text{mm}$ . The displacements are partly out of phase, so this is partial destructive interference — the resultant (3.5 mm) is less than the larger individual displacement.
2Path difference for constructive interference (2 marks)
Getting startedDirect calculation• path difference, interference, Unit 2
Question
Two coherent sources emit waves of wavelength 0.60 m. At a point P the path difference between the waves from the two sources is 1.8 m. State whether constructive or destructive interference occurs at P, and justify your answer. (2 marks)
Step-by-step solution
1. Step 1
  Find the path difference as a number of wavelengths by dividing by $\lambda$ .
  $\frac{\text{path difference}}{\lambda} = \frac{1.8}{0.60} = 3.0$
2. Step 2
  The path difference is a whole number of wavelengths ( $3\lambda$ , i.e. $n=3$ ). For constructive interference the path difference $= n\lambda$ .
  $\text{path difference} = n\lambda \;\Rightarrow\; \text{constructive}$
Answer
Constructive interference occurs at P, because the path difference is $3\lambda$ (a whole number of wavelengths), so the waves arrive in phase.
Examiner tip
Mark scheme: (1) path difference $= 3\lambda$ (or $n\lambda$ with $n=3$ ); (2) therefore constructive (waves arrive in phase, reinforce). Quoting the rule path difference $= n\lambda$ for constructive secures the justification mark.
3Wavelength from node spacing (2 marks)
Getting startedDirect calculation• stationary waves, nodes, Unit 2
Question
On a stationary wave on a stretched string, the distance between two adjacent nodes is measured to be 0.24 m. Calculate the wavelength of the waves that form this stationary wave. (2 marks)
Step-by-step solution
1. Step 1
  On a stationary wave the distance between adjacent nodes is half a wavelength, $\tfrac{\lambda}{2}$ .
  $\text{node-to-node distance} = \frac{\lambda}{2}$
2. Step 2
  Rearrange and substitute to find the wavelength.
  $\lambda = 2 \times 0.24 = 0.48\ \text{m}$
Answer
$\lambda = 0.48\ \text{m}$ .
Examiner tip
Mark scheme: (1) node-to-node $= \lambda/2$ ; (2) $\lambda = 0.48\ \text{m}$ . The commonest slip is taking the node spacing to be a whole wavelength — adjacent nodes (and adjacent antinodes) are always half a wavelength apart.
4Speed of a wave on a string from tension and mass per unit length (3 marks)
Building confidenceDirect calculation• wave speed, string, AO2, Unit 2
Question
A stretched string has a tension of 36 N and a mass per unit length of 2.5 × 10⁻³ kg m⁻¹. Calculate the speed of a transverse wave travelling along the string. (3 marks)
Step-by-step solution
1. Step 1
  The speed of a transverse wave on a string depends on the tension $T$ and the mass per unit length $\mu$ .
  $v = \sqrt{\frac{T}{\mu}}$
2. Step 2
  Substitute the values with units.
  $v = \sqrt{\frac{36}{2.5 \times 10^{-3}}} = \sqrt{14400}$
3. Step 3
  Take the square root.
  $v = 120\ \text{m s}^{-1}$
Answer
$v = 120\ \text{m s}^{-1}$ .
Examiner tip
Mark scheme: (1) correct equation $v=\sqrt{T/\mu}$ ; (2) correct substitution → $\sqrt{14400}$ (M); (3) $120\ \text{m s}^{-1}$ (A). A common error is forgetting the square root, giving 14400 m s⁻¹.
5Fundamental frequency of a vibrating string (4 marks)
Building confidenceMulti-step problem• Adapted from Edexcel IAL Physics Unit 2• harmonics, fundamental, AO2, Unit 2
Question
A guitar string of length 0.65 m has a tension of 80 N and a mass per unit length of 5.0 × 10⁻³ kg m⁻¹. The string is fixed at both ends. Calculate the fundamental frequency (first harmonic) of the string. (4 marks)
Step-by-step solution
1. Step 1
  First find the wave speed from the tension and mass per unit length.
  $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{80}{5.0 \times 10^{-3}}} = \sqrt{16000} = 126.5\ \text{m s}^{-1}$
2. Step 2
  For the fundamental (first harmonic) of a string fixed at both ends there is one antinode and a node at each end, so the string length is half a wavelength: $L = \tfrac{\lambda}{2}$ , giving $\lambda = 2L$ .
  $\lambda = 2L = 2 \times 0.65 = 1.30\ \text{m}$
3. Step 3
  Use the wave equation $v = f\lambda$ rearranged for frequency.
  $f = \frac{v}{\lambda} = \frac{126.5}{1.30}$
4. Step 4
  Evaluate the fundamental frequency.
  $f = 97.3\ \text{Hz}$
Answer
$f = 97.3\ \text{Hz}$ (to 3 s.f.).
Examiner tip
Mark scheme: (1) $v=\sqrt{T/\mu}=126.5\ \text{m s}^{-1}$ ; (2) $\lambda=2L=1.30\ \text{m}$ for the fundamental; (3) $f=v/\lambda$ (M); (4) $97.3\ \text{Hz}$ (A). Equivalently $f=\tfrac{1}{2L}\sqrt{T/\mu}$ in one step. Forgetting that the fundamental wavelength is $2L$ (not $L$ ) is the usual error.
6Identifying a harmonic from the number of antinodes (4 marks)
Building confidenceGraph or diagram• harmonics, stationary waves, AO2, Unit 2
Question
A string of length 1.2 m is fixed at both ends and made to vibrate. A stationary wave is set up with three antinodes along its length. (a) State which harmonic this is. (b) Calculate the wavelength. (c) If the wave speed on the string is 90 m s⁻¹, calculate the frequency. (4 marks)
Step-by-step solution
1. Step 1
  For a string fixed at both ends, the number of antinodes equals the harmonic number $n$ . Three antinodes ⇒ the third harmonic ( $n=3$ ).
2. Step 2
  For the $n$ th harmonic, $L = n\tfrac{\lambda}{2}$ , so $\lambda = \dfrac{2L}{n}$ . With $n=3$ :
  $\lambda = \frac{2L}{n} = \frac{2 \times 1.2}{3} = 0.80\ \text{m}$
3. Step 3
  Use $v = f\lambda$ rearranged for frequency.
  $f = \frac{v}{\lambda} = \frac{90}{0.80} = 112.5\ \text{Hz}$
Answer
(a) Third harmonic ( $n=3$ ). (b) $\lambda = 0.80\ \text{m}$ . (c) $f = 113\ \text{Hz}$ (to 3 s.f.).
Examiner tip
Mark scheme: (a) third harmonic (number of antinodes = $n$ ); (b) $\lambda=2L/n=0.80\ \text{m}$ ; (c) $f=v/\lambda$ (M); $113\ \text{Hz}$ (A). Counting loops (each loop = one antinode = $\lambda/2$ ) is a reliable way to get the wavelength.
7Show that the fundamental frequency formula gives a stated value (5 marks)
StretchShow that / prove• Adapted from Edexcel IAL Physics Unit 2• harmonics, show that, AO2, Unit 2
Question
A string of length 0.50 m is fixed at both ends. It has a mass per unit length of 1.6 × 10⁻³ kg m⁻¹ and is under a tension of 64 N. Show that the fundamental frequency of the string is about 200 Hz. (5 marks)
Step-by-step solution
1. Step 1
  The fundamental frequency of a string fixed at both ends is given by combining $v=\sqrt{T/\mu}$ with $f=v/\lambda$ and $\lambda=2L$ .
  $f_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$
2. Step 2
  Evaluate the wave speed first.
  $v = \sqrt{\frac{64}{1.6 \times 10^{-3}}} = \sqrt{40000} = 200\ \text{m s}^{-1}$
3. Step 3
  The fundamental wavelength is twice the length.
  $\lambda = 2L = 2 \times 0.50 = 1.00\ \text{m}$
4. Step 4
  Use $f = v/\lambda$ .
  $f_1 = \frac{200}{1.00} = 200\ \text{Hz}$
5. Step 5
  To one extra significant figure the result is $200.0\ \text{Hz}$ , confirming the stated value of about 200 Hz.
  $f_1 \approx 200\ \text{Hz}$
Answer
$f_1 = 200\ \text{Hz}$ , as required.
Examiner tip
Mark scheme: (1) correct combined relation $f_1=\tfrac{1}{2L}\sqrt{T/\mu}$ ; (2) $v=200\ \text{m s}^{-1}$ ; (3) $\lambda=2L=1.00\ \text{m}$ ; (4) $f=v/\lambda$ ; (5) $200\ \text{Hz}$ shown. For a 'show that' you must work fully to the value (to one extra s.f.), never just quote it.
8Determining mass per unit length from a CP5 graph (6 marks)
StretchGraph or diagram• Adapted from Edexcel IAL Physics Unit 2 (Core Practical 5)• Core Practical 5, graph, AO3, Unit 2
Question
In an investigation a string of fixed length 0.40 m is fixed at both ends and vibrated in its fundamental mode. The tension T is varied and the fundamental frequency f recorded. A graph of f² (y-axis) against T (x-axis) is a straight line through the origin with gradient 1.0 × 10⁴ Hz² N⁻¹. (a) Show that f² = T / (4L²μ). (b) Use the gradient to determine the mass per unit length μ of the string. (6 marks)
Step-by-step solution
1. Step 1
  Start from the fundamental frequency relation and square it to linearise.
  $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} \;\Rightarrow\; f^2 = \frac{1}{4L^2}\cdot\frac{T}{\mu} = \frac{T}{4L^2\mu}$
2. Step 2
  Compare $f^2 = \dfrac{1}{4L^2\mu}\,T$ with $y = mx$ (with $y=f^2$ , $x=T$ ). The gradient is $\dfrac{1}{4L^2\mu}$ .
  $\text{gradient} = \frac{1}{4L^2\mu}$
3. Step 3
  Rearrange to make $\mu$ the subject.
  $\mu = \frac{1}{4L^2 \times \text{gradient}}$
4. Step 4
  Substitute $L = 0.40\ \text{m}$ and the gradient.
  $\mu = \frac{1}{4 \times (0.40)^2 \times 1.0 \times 10^4}$
5. Step 5
  Evaluate the denominator: $4 \times 0.16 \times 1.0\times10^4 = 6400$ .
  $\mu = \frac{1}{6400} = 1.5625 \times 10^{-4}\ \text{kg m}^{-1}$
6. Step 6
  Quote the final value to 3 significant figures with units.
  $\mu = 1.56 \times 10^{-4}\ \text{kg m}^{-1}$
Answer
(a) $f^2 = \dfrac{T}{4L^2\mu}$ , a straight line through the origin. (b) $\mu = 1.56 \times 10^{-4}\ \text{kg m}^{-1}$ .
Examiner tip
Mark scheme: (a) 2 marks — square $f=\tfrac{1}{2L}\sqrt{T/\mu}$ to reach $f^2=T/(4L^2\mu)$ ; (b) gradient $=1/(4L^2\mu)$ (1), rearrange for $\mu$ (1), substitute (1), $\mu=1.56\times10^{-4}\ \text{kg m}^{-1}$ (1). Linearising by plotting $f^2$ against $T$ is the key skill; the line through the origin confirms $f^2 \propto T$ .
9Two-source interference: explaining a bright fringe (6 marks)
StretchMulti-step problem• Young's double slit, coherence, AO1, AO2, Unit 2
Question
Monochromatic light of wavelength 600 nm passes through two narrow, closely spaced slits and falls on a screen, producing a pattern of bright and dark fringes. (a) State the condition on coherence needed for a stable fringe pattern. (b) Explain, in terms of path difference, why a bright fringe forms at the centre of the screen and at certain other positions. (c) At a particular bright fringe the path difference from the two slits is 1.8 × 10⁻⁶ m. Determine the order n of this fringe. (6 marks)
Step-by-step solution
1. Step 1
  (a) For a stable, observable pattern the two sources must be coherent — they must have a constant phase difference (and therefore the same frequency/wavelength). In practice this is achieved by using two slits illuminated by the same monochromatic source.
2. Step 2
  (b) At the centre the two paths are equal (path difference = 0), so the waves arrive in phase and reinforce — constructive interference gives a bright (central) fringe.
3. Step 3
  Bright fringes occur wherever the path difference is a whole number of wavelengths, $n\lambda$ , because the waves then arrive in phase.
  $\text{path difference} = n\lambda \quad (n = 0, 1, 2, \ldots)$
4. Step 4
  (c) Divide the given path difference by the wavelength to find $n$ . Convert: $\lambda = 600\ \text{nm} = 600 \times 10^{-9}\ \text{m}$ .
  $n = \frac{\text{path difference}}{\lambda} = \frac{1.8 \times 10^{-6}}{600 \times 10^{-9}}$
5. Step 5
  Evaluate.
  $n = \frac{1.8 \times 10^{-6}}{6.0 \times 10^{-7}} = 3$
Answer
(a) The sources must be coherent — constant phase difference / same frequency. (b) A bright fringe forms where the path difference is $n\lambda$ (in phase ⇒ constructive); the centre has zero path difference. (c) $n = 3$ (the third-order bright fringe).
Examiner tip
Mark scheme: (a) coherent = constant phase difference (1); (b) path difference $= n\lambda$ → in phase → constructive (1), centre has equal paths/zero path difference (1); (c) $n = 1.8\times10^{-6}/600\times10^{-9}$ (M), $n=3$ (A) (2 marks), plus the link that $n$ is a whole number (1). Always convert nm to m before dividing.

Model Answers — Interference and Stationary Waves

High-scoring sample answers for interference and stationary waves 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). State the principle of superposition of waves.
Model answer
When two (or more) waves meet at a point, the resultant displacement at that point is equal to the vector sum of the individual displacements of the waves at that instant.

After they have passed, each wave continues unchanged.
Why this scores
Why this scores 2/2. (1) resultant displacement = sum (vector sum) of the individual displacements; (2) 'at the same point/instant', or that the waves emerge unchanged. The key phrase examiners credit is 'sum of the displacements' — not 'sum of the amplitudes' or 'the waves add together' (too vague).
Question 2
2 marks
Q (2 marks). Explain what is meant by two sources being coherent, and state why coherence is needed to observe a stable interference pattern.
Model answer
Two sources are coherent if they have a constant phase difference (which requires them to have the same frequency).

Coherence is needed because only then is the path-difference condition at each point on the screen constant in time, so the maxima and minima stay in fixed positions and a stable (steady, observable) interference pattern is seen. Incoherent sources have a randomly changing phase difference, so the pattern would shift too rapidly to observe.
Why this scores
Why this scores 2/2. (1) coherent = constant phase difference (same frequency); (2) constant phase difference keeps the pattern fixed/stable. 'Same amplitude' is NOT part of the definition of coherence — the discriminator is 'constant phase difference'.
Question 3
3 marks
Q (3 marks). A stationary wave is formed on a string. (a) State what is meant by a node and by an antinode. (b) State the distance, in terms of wavelength, between two adjacent nodes.
Model answer
(a) A node is a point on the stationary wave where the amplitude is always zero (the two waves always cancel by destructive interference). An antinode is a point where the amplitude is a maximum (the two waves reinforce by constructive interference).

(b) The distance between two adjacent nodes is half a wavelength, $\dfrac{\lambda}{2}$ .
Why this scores
Why this scores 3/3. (a) node = zero amplitude (1), antinode = maximum amplitude (1); (b) $\lambda/2$ (1). Say 'amplitude' (not 'displacement') — at a node the displacement is momentarily zero everywhere when the string passes through the straight position, so it is the amplitude that is permanently zero at a node.
Question 4
4 marks
Q (4 marks). Two loudspeakers are connected to the same signal generator and emit a sound of wavelength 0.25 m. A listener stands at a point where the distance from one speaker is 4.50 m and from the other is 4.875 m. (a) Calculate the path difference. (b) State and explain whether the listener hears a loud sound or a quiet sound at this point.
Model answer
(a) The path difference is the difference between the two distances:

$\text{path difference} = 4.875 - 4.50 = 0.375\ \text{m}$

(b) Express this as a number of wavelengths:

$\frac{0.375}{0.25} = 1.5 = \left(1 + \tfrac{1}{2}\right)\lambda$

The path difference is an odd number of half-wavelengths $\left(n + \tfrac{1}{2}\right)\lambda$ , so the two sounds arrive in antiphase and undergo destructive interference. The listener therefore hears a quiet sound (a minimum).
Why this scores
Why this scores 4/4. (a) path difference $= 0.375\ \text{m}$ (1); (b) $= 1.5\lambda$ (1), recognise odd number of half-wavelengths / antiphase (1), conclude destructive → quiet (1). The loudspeakers are driven by the same generator, so they are coherent — a necessary condition for this reasoning to apply.
Question 5
5 marks
Q (5 marks). A wire of length 0.80 m is fixed at both ends and stretched to a tension of 50 N. The wire has a mass per unit length of 8.0 × 10⁻⁴ kg m⁻¹. (a) Calculate the speed of a transverse wave on the wire. (b) Calculate the frequency of the fundamental (first harmonic). (c) State the frequency of the third harmonic.
Model answer
(a) The wave speed depends on tension and mass per unit length:

$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{50}{8.0 \times 10^{-4}}} = \sqrt{62500} = 250\ \text{m s}^{-1}$

(b) For the fundamental of a string fixed at both ends, $\lambda = 2L = 2 \times 0.80 = 1.60\ \text{m}$ . Using $v = f\lambda$ :

$f_1 = \frac{v}{\lambda} = \frac{250}{1.60} = 156\ \text{Hz}$

(c) The harmonics are integer multiples of the fundamental, so the third harmonic has frequency:

$f_3 = 3 f_1 = 3 \times 156 = 469\ \text{Hz}$
Why this scores
Why this scores 5/5. (a) 2 marks: $v=\sqrt{T/\mu}$ and $250\ \text{m s}^{-1}$ . (b) 2 marks: $\lambda=2L=1.60\ \text{m}$ and $f_1=156\ \text{Hz}$ . (c) 1 mark: $f_3=3f_1=469\ \text{Hz}$ (harmonics are integer multiples of the fundamental). Keeping the full value $f_1=156.25\ \text{Hz}$ before tripling gives $468.75 \approx 469\ \text{Hz}$ .
Question 6
4 marks
Q (4 marks). State two differences between a progressive (travelling) wave and a stationary (standing) wave.
Model answer
Difference 1 — energy transfer. A progressive wave transfers energy in the direction of travel, whereas a stationary wave stores energy and does not transfer it along the medium (no net energy is moved).

Difference 2 — amplitude. In a progressive wave every point oscillates with the same amplitude. In a stationary wave the amplitude varies with position — it is zero at nodes and maximum at antinodes.

(A further valid difference: in a progressive wave neighbouring points are slightly out of phase, so a phase 'lag' travels along the wave; in a stationary wave all points between two adjacent nodes oscillate in phase with one another.)
Why this scores
Why this scores 4/4. 2 marks each for two clear, contrasted differences. Accept: energy transferred vs not transferred; constant amplitude vs amplitude varying (nodes/antinodes); points in phase between nodes vs progressive phase lag; profile moves vs profile fixed. Marks require the contrast for BOTH wave types, not a one-sided statement.
Question 7
5 marks
Q (5 marks). Explain how a stationary wave is formed on a string fixed at both ends, and explain why nodes and antinodes occur at fixed positions.
Model answer
A stationary wave is formed when two progressive waves of the same frequency and (ideally) equal amplitude travel in opposite directions along the string and superpose. On a string fixed at both ends, a wave sent along the string is reflected at the fixed end and travels back, so the original and reflected waves continually overlap.

By the principle of superposition the two waves add at every point. At certain fixed positions the two waves are always in antiphase, so they continually cancel — these are the nodes, where the amplitude is always zero. Half a wavelength away the two waves are always in phase, so they continually reinforce — these are the antinodes, where the amplitude is a maximum.

The nodes and antinodes stay in fixed positions because the two interfering waves have the same frequency, so the path-difference (and hence the phase relationship) at each point does not change with time. Adjacent nodes are $\tfrac{\lambda}{2}$ apart.
Why this scores
Why this scores 5/5. (1) two waves, same frequency, opposite directions; (2) superposition / reflection at the fixed end; (3) nodes where waves are always in antiphase → cancel → zero amplitude; (4) antinodes where always in phase → reinforce → maximum amplitude; (5) fixed positions because same frequency keeps the phase relationship constant. Examiners want 'opposite directions' and 'superposition' explicitly.
Question 8
Adapted from Edexcel IAL Physics Unit 2 (Core Practical 5)6 marks
Q (6 marks). A student investigates how the fundamental frequency f of a stretched string fixed at both ends depends on its tension T. The length and mass per unit length are kept constant. (a) State how the apparatus could be used to find the fundamental frequency for a given tension. (b) The student plots a graph of f² against T. Explain why this graph is a straight line through the origin, and describe how its gradient could be used to find the mass per unit length μ. (6 marks)
Model answer
(a) Fix the string at both ends over two bridges with one end passing over a pulley to a hanging mass that sets the tension $T = mg$ . Drive the string with a vibration generator (signal generator) and slowly increase the driving frequency until the string vibrates with the largest amplitude in a single loop (one antinode, a node at each end) — this is the fundamental. Read the frequency from the signal generator (a strobe or fine sand/paper rider can confirm the resonance). Repeat for each tension.

(b) The fundamental frequency is $f = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$ . Squaring gives:

$f^2 = \frac{1}{4L^2\mu}\,T$

Since $L$ and $\mu$ are constant, $f^2$ is directly proportional to $T$ , so a graph of $f^2$ against $T$ is a straight line through the origin (the $y = mx$ form with $x = T$ ).

The gradient equals $\dfrac{1}{4L^2\mu}$ . Measuring the gradient from a large triangle and rearranging gives:

$\mu = \frac{1}{4L^2 \times \text{gradient}}$

so $\mu$ is found from the measured gradient and the (constant) length $L$ .
Why this scores
Why this scores 6/6. (a) 3 marks: tension via hanging mass ( $T=mg$ ), drive with vibration generator and increase frequency to resonance/fundamental (single loop), read the frequency. (b) 3 marks: square the equation to get $f^2 \propto T$ (straight line through origin), gradient $=1/(4L^2\mu)$ , rearrange to $\mu = 1/(4L^2 \times \text{gradient})$ . 'Through the origin' must be justified by the proportionality.
Question 9
Adapted from Edexcel IAL Physics Unit 26 marks
Q (6 marks). In a demonstration, microwaves from a single source are passed through two narrow gaps in a metal sheet to act as two sources. A detector moved along a line parallel to the sheet records a series of maxima and minima. (a) Explain, in terms of path difference and phase, why maxima and minima are detected. (b) Explain why the two gaps act as coherent sources. (c) State and explain what happens to the spacing of the maxima if the two gaps are moved closer together. (6 marks)
Model answer
(a) Microwaves spread out (diffract) from each gap and overlap, so the principle of superposition applies. At positions where the path difference is a whole number of wavelengths ( $n\lambda$ ) the two waves arrive in phase and reinforce — constructive interference gives a maximum. At positions where the path difference is an odd number of half-wavelengths $\left(n+\tfrac{1}{2}\right)\lambda$ the waves arrive in antiphase and cancel — destructive interference gives a minimum.

(b) Both gaps are illuminated by the same single source, so the waves leaving them have the same frequency and a constant phase difference — that is the definition of coherent sources. This is why a steady pattern of maxima and minima is observed rather than a blur.

(c) Moving the gaps closer together (decreasing the slit separation) increases the spacing of the maxima. The fringe spacing is inversely proportional to the slit separation (for a fixed wavelength and source-to-detector distance), so a smaller separation spreads the maxima further apart.
Why this scores
Why this scores 6/6. (a) 3 marks: superposition, path difference $n\lambda$ → in phase → maximum, $\left(n+\tfrac12\right)\lambda$ → antiphase → minimum. (b) 2 marks: same single source → same frequency and constant phase difference → coherent. (c) 1 mark: closer gaps → wider fringe spacing (inverse relationship). The reasoning must link path difference to phase to constructive/destructive throughout.
Question 10
Adapted from Edexcel IAL Physics Unit 26 marks
Q (6 marks). A wire of length 0.75 m is fixed at both ends and vibrates in its fundamental mode at a frequency of 220 Hz when the tension is 90 N. (a) Calculate the mass per unit length of the wire. (b) The tension is now increased. Explain, with a calculation, what new tension is needed to raise the fundamental frequency to 330 Hz, the length being unchanged. (c) State one change, other than tension, that would raise the fundamental frequency, and explain why.
Model answer
(a) For the fundamental of a string fixed at both ends, $f = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$ . Rearranging for $\mu$ :

$\mu = \frac{T}{4L^2 f^2} = \frac{90}{4 \times (0.75)^2 \times (220)^2} = \frac{90}{4 \times 0.5625 \times 48400}$

$\mu = \frac{90}{108900} = 8.26 \times 10^{-4}\ \text{kg m}^{-1}$

(b) Because $L$ and $\mu$ are constant, $f \propto \sqrt{T}$ , so $\dfrac{f_2}{f_1} = \sqrt{\dfrac{T_2}{T_1}}$ , which gives $T_2 = T_1\left(\dfrac{f_2}{f_1}\right)^2$ :

$T_2 = 90 \times \left(\frac{330}{220}\right)^2 = 90 \times (1.5)^2 = 90 \times 2.25 = 203\ \text{N}$

So the tension must be raised to about 203 N (a $1.5\times$ rise in frequency needs a $2.25\times$ rise in tension, because $f \propto \sqrt{T}$ ).

(c) Any one of: shortening the wire (decreasing $L$ ) raises the fundamental frequency because $f \propto \dfrac{1}{L}$ (the same wave speed now fits into a shorter half-wavelength); OR using a thinner/lighter wire (smaller $\mu$ ) raises it because $f \propto \dfrac{1}{\sqrt{\mu}}$ (a lower mass per unit length gives a higher wave speed). Either change increases the fundamental frequency for the same tension.
Why this scores
Why this scores 6/6. (a) 2 marks: rearrange to $\mu = T/(4L^2f^2)$ (1) and $8.26\times10^{-4}\ \text{kg m}^{-1}$ (1). (b) 3 marks: use $f\propto\sqrt{T}$ so $T_2 = T_1(f_2/f_1)^2$ (1), substitute $90\times2.25$ (1), $203\ \text{N}$ (1). (c) 1 mark: shorten the wire OR reduce $\mu$ , WITH the reason ( $f\propto 1/L$ or $f\propto 1/\sqrt{\mu}$ ). The dominant error is treating $f\propto T$ (giving $135\ \text{N}$ ) — the square-root relationship means tension must rise by the square of the frequency ratio.

Key Formulae — Interference and Stationary Waves

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

Speed of a transverse wave on a string
$v = \sqrt{\frac{T}{\mu}}$
$v$
wave speed / m s⁻¹
$T$
tension in the string / N
$\mu$
mass per unit length / kg m⁻¹
When to use
Finding the speed of a transverse wave on a stretched string from its tension and mass per unit length. In the Edexcel formulae booklet (Appendix 7).
Example
$T=36\ \text{N}$ , $\mu=2.5\times10^{-3}$ : $v=\sqrt{14400}=120\ \text{m s}^{-1}$ .
Fundamental frequency of a string fixed at both ends
$f = \frac{n}{2L}\sqrt{\frac{T}{\mu}}$
$f$
frequency of the $n$ th harmonic / Hz
$n$
harmonic number (1, 2, 3, …)
$L$
length of the string / m
$T$
tension / N
$\mu$
mass per unit length / kg m⁻¹
When to use
Finding the frequency of any harmonic of a string fixed at both ends. With $n=1$ it gives the fundamental (first harmonic). Combine $v=\sqrt{T/\mu}$ with $f=v/\lambda$ and $\lambda=2L/n$ .
Example
$n=1$ , $L=0.50$ , $T=64$ , $\mu=1.6\times10^{-3}$ : $f=200\ \text{Hz}$ .
The wave equation
$v = f\lambda$
$v$
wave speed / m s⁻¹
$f$
frequency / Hz
$\lambda$
wavelength / m
When to use
Linking speed, frequency and wavelength for any progressive wave; used to convert a wave speed into a frequency once the wavelength is known. Must be recalled.
Example
$v=126.5\ \text{m s}^{-1}$ , $\lambda=1.30\ \text{m}$ : $f=v/\lambda=97.3\ \text{Hz}$ .
Constructive interference (path difference condition)
$\text{path difference} = n\lambda$
$n$
any integer (0, 1, 2, …)
$\lambda$
wavelength / m
When to use
Deciding whether two coherent waves reinforce. A whole number of wavelengths means the waves arrive in phase, giving a maximum (bright fringe / loud point). Must be recalled.
Example
Path difference $1.8\ \text{m}$ , $\lambda=0.60\ \text{m}$ : $1.8/0.60=3$ → constructive.
Destructive interference (path difference condition)
$\text{path difference} = \left(n + \tfrac{1}{2}\right)\lambda$
$n$
any integer (0, 1, 2, …)
$\lambda$
wavelength / m
When to use
Deciding whether two coherent waves cancel. An odd number of half-wavelengths means the waves arrive in antiphase, giving a minimum (dark fringe / quiet point). Must be recalled.
Example
Path difference $\tfrac{1}{2}\lambda$ , $\tfrac{3}{2}\lambda$ , … → destructive.
Node (and antinode) spacing on a stationary wave
$d_{\text{node-node}} = \frac{\lambda}{2}$
$d_{\text{node-node}}$
distance between adjacent nodes / m
$\lambda$
wavelength of the waves forming the stationary wave / m
When to use
Finding the wavelength of a stationary wave by measuring the distance between adjacent nodes (or adjacent antinodes), which is half a wavelength. Must be recalled.
Example
Adjacent nodes $0.24\ \text{m}$ apart: $\lambda = 2 \times 0.24 = 0.48\ \text{m}$ .
Phase difference from path difference
$\Delta\phi = \frac{2\pi}{\lambda} \times (\text{path difference})$
$\Delta\phi$
phase difference / rad
$\lambda$
wavelength / m
When to use
Converting a path difference into a phase difference. One whole wavelength of path difference corresponds to $2\pi$ rad (360°) of phase. Must be recalled.
Example
Path difference $\tfrac{\lambda}{2}$ : $\Delta\phi = \pi$ rad (180°, antiphase).

Key Definitions and Keywords — Interference and Stationary Waves

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

Principle of superposition
Examiner keyword
When two or more waves meet at a point, the resultant displacement at that point equals the vector sum of the individual displacements of the waves at that instant.
Interference
Examiner keyword
The superposition of two or more coherent waves to give a resultant wave whose amplitude varies from point to point, producing a pattern of reinforcement (maxima) and cancellation (minima).
Constructive interference
Examiner keyword
Reinforcement that occurs where waves meet in phase. The path difference is a whole number of wavelengths, path difference $= n\lambda$ , giving a maximum amplitude.
Destructive interference
Examiner keyword
Cancellation that occurs where waves meet in antiphase. The path difference is an odd number of half-wavelengths, path difference $= \left(n+\tfrac{1}{2}\right)\lambda$ , giving a minimum amplitude.
Path difference
Examiner keyword
The difference in the distances travelled by two waves from their sources to a given point, usually expressed as a number of wavelengths.
Phase difference
Examiner keyword
A measure of how far one oscillation lags or leads another, expressed in radians or degrees. A path difference of one wavelength corresponds to a phase difference of $2\pi$ rad (360°).
Coherence
Examiner keyword
Two sources are coherent if they have a constant phase difference (which requires the same frequency). Coherence is required for a stable, observable interference pattern.
In phase / antiphase
Two oscillations are in phase when their phase difference is zero (or a multiple of $2\pi$ rad); they are in antiphase when the phase difference is $\pi$ rad (180°), i.e. half a cycle apart.
Stationary (standing) wave
Examiner keyword
A wave pattern formed by the superposition of two progressive waves of the same frequency and equal amplitude travelling in opposite directions. It stores energy and does not transfer it along the medium.
Node
Examiner keyword
A point on a stationary wave where the amplitude is always zero, because the two interfering waves are permanently in antiphase there and cancel.
Antinode
Examiner keyword
A point on a stationary wave where the amplitude is a maximum, because the two interfering waves are permanently in phase there and reinforce.
Progressive (travelling) wave
A wave that transfers energy in the direction of travel; every point oscillates with the same amplitude, and there is a progressive phase difference between neighbouring points.
Fundamental frequency (first harmonic)
Examiner keyword
The lowest frequency at which a string fixed at both ends forms a stationary wave — one antinode with a node at each end, so the string length equals $\tfrac{\lambda}{2}$ ( $\lambda = 2L$ ).
Harmonic
Examiner keyword
A frequency at which a stationary wave is set up on a string fixed at both ends; the $n$ th harmonic has $n$ antinodes and a frequency $n$ times the fundamental, $f_n = n f_1$ .
Tension (T)
The stretching force in the string (in newtons). Increasing the tension increases the wave speed and therefore the natural (harmonic) frequencies, since $v=\sqrt{T/\mu}$ .
Mass per unit length (μ)
Examiner keyword
The mass of the string per metre of its length (kg m⁻¹), equal to mass ÷ length. A heavier string (larger $\mu$ ) has a lower wave speed and lower harmonic frequencies.

Common Mistakes and Misconceptions — Interference and Stationary Waves

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

✕Taking the distance between adjacent nodes (or antinodes) as a whole wavelength instead of half a wavelength
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students picture one 'loop' as a full wave, forgetting that a complete wavelength spans two loops (node–antinode–node–antinode–node).
How to avoid it
Remember that adjacent nodes are $\tfrac{\lambda}{2}$ apart (so are adjacent antinodes). One loop (node to next node) is half a wavelength; the full wavelength is the distance between alternate nodes.
✕Defining coherence as 'same amplitude' or 'in phase', rather than a constant phase difference
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students confuse the conditions for a clear pattern with the precise definition of coherence.
How to avoid it
Coherence means a constant phase difference (which requires the same frequency). Equal amplitude makes the cancellation at minima complete but is not part of the definition of coherence.
✕Saying superposition adds the amplitudes of the waves rather than their displacements
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
The words 'amplitude' and 'displacement' are used loosely in everyday speech.
How to avoid it
State that the resultant displacement is the sum of the individual displacements at that instant. Displacements can be positive or negative, which is why destructive interference is possible — amplitudes (always positive) cannot simply add.
✕Using $\lambda = L$ instead of $\lambda = 2L$ for the fundamental of a string fixed at both ends
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students see one loop spanning the whole string and assume that is one wavelength.
How to avoid it
For the fundamental there is a node at each end and one antinode in the middle, so the string holds half a wavelength: $L = \tfrac{\lambda}{2}$ , giving $\lambda = 2L$ . Sketch the loop and mark the nodes to check.
✕Forgetting the square root in $v=\sqrt{T/\mu}$ , or not squaring $f$ before plotting against $T$
Edexcel IAL Physics Unit 2 examiner reports (Core Practical 5)
Why it happens
The relationship $v=\sqrt{T/\mu}$ is square-root, not linear, and is easy to misremember under time pressure.
How to avoid it
Always square-root $T/\mu$ to get the speed. To linearise the CP5 data, plot $f^2$ against $T$ (a straight line through the origin), not $f$ against $T$ (a curve).
✕Claiming a stationary wave transfers energy, or that all points have the same amplitude
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students apply the properties of a progressive wave to a stationary one without thinking about the differences.
How to avoid it
A stationary wave stores energy and does not transfer it along the medium, and its amplitude varies with position (zero at nodes, maximum at antinodes). Contrast this explicitly with a progressive wave in 'differences' questions.

Interference and Stationary Waves — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Interference and Stationary Waves

Short Notes - Interference and Stationary Waves

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Resultant displacement by superposition (2 marks)

2Path difference for constructive interference (2 marks)

3Wavelength from node spacing (2 marks)

4Speed of a wave on a string from tension and mass per unit length (3 marks)

5Fundamental frequency of a vibrating string (4 marks)

6Identifying a harmonic from the number of antinodes (4 marks)

7Show that the fundamental frequency formula gives a stated value (5 marks)

8Determining mass per unit length from a CP5 graph (6 marks)

9Two-source interference: explaining a bright fringe (6 marks)

Speed of a transverse wave on a string

Fundamental frequency of a string fixed at both ends

The wave equation

Constructive interference (path difference condition)

Destructive interference (path difference condition)

Node (and antinode) spacing on a stationary wave

Phase difference from path difference

Principle of superposition

Interference

Constructive interference

Destructive interference

Path difference

Phase difference

Coherence

In phase / antiphase

Stationary (standing) wave

Node

Antinode

Progressive (travelling) wave

Fundamental frequency (first harmonic)

Harmonic

Tension (T)

Mass per unit length (μ)

✕Taking the distance between adjacent nodes (or antinodes) as a whole wavelength instead of half a wavelength

✕Defining coherence as 'same amplitude' or 'in phase', rather than a constant phase difference

✕Saying superposition adds the amplitudes of the waves rather than their displacements

✕Using λ=L\lambda = Lλ=L instead of λ=2L\lambda = 2Lλ=2L for the fundamental of a string fixed at both ends

✕Forgetting the square root in v=T/μv=\sqrt{T/\mu}v=T/μ​, or not squaring fff before plotting against TTT

✕Claiming a stationary wave transfers energy, or that all points have the same amplitude

Interference and Stationary Waves — frequently asked questions

✕Using $\lambda = L$ instead of $\lambda = 2L$ for the fundamental of a string fixed at both ends

✕Forgetting the square root in $v=\sqrt{T/\mu}$ , or not squaring $f$ before plotting against $T$