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Interference & Stationary Waves

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  Edexcel IAL Physics — Waves

Interference & Stationary Waves

Superposition Path Difference Nodes & Antinodes Wave Speed on a String Core Practical 5
The Big Idea: When two waves of the same frequency overlap, they add together point-by-point — and depending on exactly how "in step" they are when they meet, they can either boost each other into a bigger wave (constructive) or cancel each other out (destructive). When this happens continuously between a wave and its own reflection, you get a special pattern that looks like it's standing still — a stationary wave.

Quick Summary

  • Superposition is what happens when two or more waves meet — their displacements add together to give one resultant wave.
  • Constructive interference = waves add up to a bigger amplitude. Destructive interference = waves cancel to a smaller (or zero) amplitude.
  • Interference is only visible if the sources are coherent — same frequency, constant phase difference.
  • Phase difference is about how "out of step" two wave cycles are (measured in degrees/radians). Path difference is about how much extra distance one wave has travelled (measured in wavelengths).
  • Constructive interference happens when path difference = (whole number of wavelengths). Destructive happens when path difference = (n + ½)λ.
  • A stationary wave forms when two waves of the same frequency and amplitude travel in opposite directions and superpose — usually a wave and its own reflection.
  • Stationary waves have fixed nodes (zero displacement, always) and antinodes (maximum displacement, oscillating up and down).
  • Wave speed on a stretched string: v = √(T/μ). Fundamental frequency: f₀ = (1/2L)√(T/μ).
  • Core Practical 5 investigates how the frequency of the first harmonic depends on string length, tension, or mass per unit length.

1. Interference & Superposition of Waves

Picture two ripples on a pond heading toward each other. What happens the instant they cross paths? They don't bounce off each other or get destroyed — instead, at every point where they overlap, the water surface height is simply the sum of what each wave would have done on its own. Once they've fully passed through each other, both ripples carry on completely unchanged, as if the meeting never happened.

This "just add the displacements together" rule is called the principle of superposition, and the process itself is called interference. It applies to any type of wave — water, sound, light, waves on a string — as long as they're travelling through the same medium at the same time.

Principle of Superposition
Resultant displacement = Sum of individual displacements
In plain English: at any point where waves overlap, just add up (with correct sign) what each wave is doing there.

Constructive vs Destructive Interference

There are two extreme outcomes of superposition, depending on how the waves line up:

CONSTRUCTIVE INTERFERENCE (crest meets crest) Wave 1: /\ /\ /\ / \ / \ / \ Wave 2: /\ /\ /\ / \ / \ / \ | v Result: /‾‾\ <- amplitude = 2A (bigger!) / \ / \ DESTRUCTIVE INTERFERENCE (crest meets trough) Wave 1: /\ /\ /\ / \ / \ / \ Wave 2: \ / \ / \ / \ / \ / \ / | v Result: ------------------ <- amplitude = 0 (cancelled out!)

Constructive interference happens when the two waves meet "in phase" — crest lines up with crest, trough lines up with trough. The resultant amplitude is larger than either individual wave (if both waves have amplitude A, the resultant can be as big as 2A).

Destructive interference happens when the waves meet "in antiphase" — crest lines up with trough. The resultant amplitude is smaller than the individual waves, and if both amplitudes are equal, it cancels to exactly zero.

Quick Way to Decide

If two waves meet at the same point on each wave (e.g. both at a crest, or both at a trough) → constructive. If one is at a crest and the other at a trough → destructive. Anything in between gives something in between.

Coherence — the Essential Requirement

Here's the catch: you only get a stable, observable interference pattern if the two sources are coherent. Coherent waves must have:

  • The same frequency
  • A constant phase difference (it doesn't have to be zero, it just can't keep randomly changing)

Why does this matter so much? Imagine two sources whose phase relationship keeps jumping around randomly. One instant they might add constructively, the next instant destructively — the interference pattern would flicker so fast (and randomly) that you'd never actually see fringes or a fixed pattern. It would just look like an average, blurred-out mess.

This is exactly why a laser is coherent (it emits monochromatic light — a single frequency — with waves locked in step), while a filament lamp is not coherent (it emits a jumble of different frequencies and random phases from millions of independent atoms).

Practice Question 1.1
Explain why you cannot see a stable interference pattern using light from two separate torches, even if they are identical models pointed at the same wall.
Practice Question 1.2
Two waves of equal amplitude 3 cm meet at a point. State the resultant amplitude if (a) they arrive in phase, and (b) they arrive in antiphase.

2. Phase & Path Difference

These two terms sound alike, and students mix them up constantly on exams — but they describe two genuinely different things. Let's separate them clearly.

Phase Difference

Two points are "in phase" if they are at the exact same stage of their wave cycle at the same moment — same displacement, moving in the same direction. Think of a wave cycle as a full circle (0° to 360°): phase difference is simply the angle between where two points sit on that circle.

One full wave cycle = 360° = 2π radians Amplitude | * 90° (peak) | * * | * * 0°---*--------------*----180°------> Phase Angle | * * | * * | * 270° (trough) 360° = back to 0°

So if one wave is at its peak (phase = 90°) and another is at zero going upward (phase = 0°), their phase difference is 90°, or a quarter of a cycle.

Remember

Phase difference is measured in degrees or radians — it's about comparing the "stage" of the cycle each wave is at, like comparing their peaks and troughs directly.

Path Difference

Path difference is a completely different idea. It's about distance travelled, not angle. Formally:

Definition
Path difference = the difference in distance travelled by two waves from their sources to the point where they meet
It's expressed in multiples of wavelength (λ), e.g. "2λ" or "half a wavelength".

Why does path difference matter? Because if one wave has travelled exactly one whole wavelength further than the other, it will have completed one extra full cycle — meaning it arrives back "in step" with the other wave, and you get constructive interference again. But if it's travelled half a wavelength further, it arrives exactly out of step, giving destructive interference.

Interference Conditions (Path Difference)
Constructive: path difference = nλ
Destructive: path difference = (n + ½)λ
n = an integer (0, 1, 2, 3, ...)
λ = wavelength of the waves

Worked Example — Two Sources

Two coherent sources S₁ and S₂ emit identical waves. At point P₁, the wave from S₁ has travelled 6λ, and the wave from S₂ has travelled 6.5λ. At point P₂, S₁'s wave has travelled 7λ and S₂'s has travelled 6λ.

Path difference at P1 = 6.5λ - 6λ = λ/2 → (n + ½)λ form → DESTRUCTIVE Path difference at P2 = 7λ - 6λ = λ → nλ form (n=1) → CONSTRUCTIVE

This matches the general rule perfectly: a path difference that's a whole number of wavelengths → constructive; a path difference that's a whole number plus a half → destructive.

Practice Question 2.1 (based on the "XYZ" style exam question)
At point X, two coherent waves have travelled 5.5λ and 4.5λ from their sources. At point Y, they have travelled 3.5λ and 3.5λ. At point Z, they have travelled 4λ and 3.5λ. Identify the type of interference at each point.
Practice Question 2.2
Explain the difference between phase difference and path difference in your own words.

3. Stationary Waves

Now for the main event of this chapter. A stationary wave (also called a standing wave) is a special, very particular result of superposition: it happens when two waves of the same frequency and the same amplitude travel in opposite directions and overlap continuously.

In practice, this is almost always created by taking a single travelling wave and reflecting it back on itself — for example, a wave sent down a string that's fixed at the far end. The original outward wave and its reflected wave then superpose continuously, and something remarkable happens: the resulting pattern's peaks and troughs stop moving along the string. They just oscillate up and down in place. That's why it's called "stationary" — even though the two waves that make it up are very much still travelling (in opposite directions), the interference pattern itself appears frozen in space.

Step 1: Outward pulse travels toward fixed end hand ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ o fixed end (pulse moving right -->) Step 2: Pulse reflects off the fixed end hand ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ o fixed end (<-- reflected pulse moving left) Step 3: Outward + reflected waves overlap continuously = a STATIONARY WAVE PATTERN (peaks/troughs don't travel)
Conditions for a Stationary Wave
  • Two waves with the same frequency
  • Two waves with the same amplitude
  • Travelling in opposite directions
  • (Usually achieved via a wave and its own reflection)

Where You See This In Real Life

This chapter covers three classic demonstrations, and it's genuinely useful to picture each one:

DemonstrationSet-upHow it shows nodes/antinodes
Stretched stringOscillator vibrates one end; other end fixed via a pulley + mass (keeps it taut)Adjust the oscillator's frequency until a clean standing wave pattern forms — you literally see the loops (antinodes) and still points (nodes)
MicrowavesMicrowave source facing a metal reflecting plate, with a probe detector in betweenMove the detector along the gap — the meter reading peaks at antinodes and drops to (near) zero at nodes
Air columnLoudspeaker at open end of a tube, fine powder scattered insideAt the right frequency, the powder gets shaken into neat heaps exactly at the nodes (zero disturbance points) — the powder stays still there while it's flung about everywhere else

These all work on the same underlying physics: a driven wave reflects off some kind of boundary and interferes with itself.

Nodes and Antinodes

Every stationary wave is built from two key kinds of points:

Nodes

Points of zero displacement — the medium never moves here, no matter what. Nodes stay completely fixed in position.

Antinodes

Points of maximum amplitude — the medium oscillates here with the biggest possible swing. Antinodes stay fixed in horizontal position, but move up and down vertically.

Snapshot of a stationary wave at one instant in time: N AN N AN N AN N |------/‾‾\------|------/‾‾\------|------/‾‾\------| | / \ | / \ | / \ | --*----+------+----*----+------+----*----+------+----*-- | \ / | \ / | \ / | |------\__/------|------\__/------|------\__/------| N = Node (always zero displacement) AN = Antinode (max displacement, oscillates up & down) Distance between adjacent nodes = λ/2
Memory Trick

Can't remember which is which? Nodes = NO Disturbance. Both start with "N" — nodes are where nothing happens.

One more subtle but important fact: between two adjacent nodes, every point on the stationary wave is in phase with every other point in that same "loop" (they all reach max displacement together, and all pass through zero together) — they just have different amplitudes depending on how close they are to the antinode. But cross over a node into the next loop, and everything flips into antiphase (exactly opposite direction) compared to the loop before it.

Practice Question 3.1
A stretched string forms a stationary wave pattern with 5 complete loops between its two fixed ends, over a total length L. State (a) how many nodes and how many antinodes there are, and (b) how many wavelengths fit into length L.
Practice Question 3.2
A stationary wave is set up in a stretched string of length L with just one loop (the fundamental / first harmonic — a single antinode in the middle, nodes at both ends). Express the wavelength λ in terms of L.

4. Wave Speed on a Stretched String

How fast does a wave actually travel along a taut string, like a guitar string? It turns out this depends on just two physical properties: how tightly the string is stretched (tension), and how "heavy" the string is per unit length. Makes intuitive sense — a tighter string snaps back faster (higher wave speed), and a heavier/thicker string is more sluggish to move (lower wave speed).

Wave Speed on a Stretched String
v = √(T / μ)
v = wave speed (m s⁻¹)
T = tension in the string (N)
μ = mass per unit length of the string (kg m⁻¹) — i.e. mass ÷ length
More tension → faster wave. Heavier string (more mass per metre) → slower wave.

Getting to the Fundamental Frequency

Now combine this with what we know about stationary waves. At the fundamental frequency (also called the first harmonic) of a string of length L fixed at both ends, we established that λ = 2L (one loop, nodes at both ends, antinode in the middle).

The universal wave equation says v = fλ, so at the fundamental:

Step-by-step derivation
v = f × λ = f × 2L
Combine with v = √(T/μ) ...
f₀ = (1/2L) × √(T/μ)
f₀ = fundamental frequency (Hz)
L = length of the string (m)
T = tension (N)
μ = mass per unit length (kg m⁻¹)

Harmonics — Higher Modes of Vibration

The fundamental (first harmonic) isn't the only possible standing wave pattern a string can support. If you drive the string at higher frequencies, you can get 2, 3, or more loops — these are called the second harmonic, third harmonic, and so on.

HarmonicPatternWavelengthFrequency
1st (fundamental)1 loop, 1 antinodeλ = 2Lf = v / 2L
2nd2 loops, 2 antinodesλ = Lf = v / L (= 2f₀)
3rd3 loops, 3 antinodesλ = 2L/3f = 3v / 2L (= 3f₀)

Notice the pattern: each successive harmonic's frequency is just a whole-number multiple of the fundamental frequency. This is exactly why musical instruments have such a rich, characteristic sound — the string vibrates in a mix of all these harmonics at once, not just the fundamental.

Worked Example

A guitar string of mass 3.2 g and length 90 cm is fixed onto a guitar. It's tightened to a tension of 65 N between two bridges 75 cm apart. Find (a) the wave speed, and (b) the fundamental frequency.

PART (A) — Wave speed Step 1: Convert to SI units T = 65 N m = 3.2 x 10^-3 kg (mass of the WHOLE string, 90 cm, used to find mu) L(whole string) = 0.90 m Step 2: mass per unit length mu = m / L = (3.2x10^-3) / 0.90 = 3.56 x 10^-3 kg/m Step 3: v = sqrt(T / mu) v = sqrt(65 / 3.56x10^-3) = sqrt(18258) = 135.1 m/s v ~= 140 m/s (2 s.f.) PART (B) — Fundamental frequency Step 1: This time use the VIBRATING length between the bridges L(vibrating) = 0.75 m <-- NOT the same L as part (a)! Step 2: f0 = v / (2L) = 135.1 / (2 x 0.75) = 90.1 Hz f0 ~= 90 Hz (2 s.f.)
Watch Out!

Notice how the string's total length (90 cm, used to find μ) is different from the vibrating length between the bridges (75 cm, used to find f₀). This is a classic trap — always check whether you need the whole string's length or just the vibrating section's length for each part of a calculation.

Practice Question 4.1
A wire has mass 0.50 g and length 1.2 m. It is stretched between two fixed points 1.0 m apart under a tension of 40 N. Calculate (a) the mass per unit length, (b) the wave speed, and (c) the fundamental frequency.
Practice Question 4.2
If the tension in a stretched string is doubled (everything else unchanged), what happens to the fundamental frequency?

5. Core Practical 5: Investigating Stationary Waves

This is a required practical, so exam boards love asking about the method, the apparatus, the graph, and — especially — the sources of error. Let's build the whole picture.

Aim & Variables

The overall aim is to measure how the frequency of the first harmonic depends on one of three things (you only vary one at a time, keeping the others constant):

Independent variableWhat you keep constant (control variables)
Length of string, LSame masses attached (tension), same string (μ)
Tension, TSame length of string, same string (μ)
Mass per unit length, μ (different strings)Same masses attached (tension), same length

Dependent variable in every case: the frequency of the first harmonic, f.

Apparatus & Its Purpose

ApparatusPurpose
Signal generatorDrives the vibration generator and measures the frequency
Vibration generatorPhysically shakes one end of the string to produce the wave
Retort stand + G-clamp/2 kg massProvides a stable, fixed end on the table
PulleyLets masses hang vertically with less friction than the table edge
Wooden bridgeProvides the other fixed end; can be moved to vary L
Mass hanger + 100 g massesHangs from the pulley to vary tension in the string
Metre rulerMeasures the length of the string (resolution: 1 mm)
Top-pan balanceMeasures the mass of the string (resolution: 0.005 g)

Method (varying length, as an example)

  1. Attach one end of the string to the vibration generator, pass the other end over the pulley, and secure it to the mass hanger.
  2. Position the wooden bridge so the length L (from vibration generator to bridge) can be measured with a metre ruler.
  3. Turn on the signal generator to set the string oscillating.
  4. Increase the frequency until the first harmonic is observed (nodes at both ends, single antinode in the middle) — read off this frequency.
  5. Repeat with different values of L (a good spread, e.g. 0.2 m intervals over at least 1.0–1.6 m range).
  6. Repeat each frequency reading at least twice more and average.
  7. Calculate tension using T = mg (m = mass on hanger, g = 9.81 N kg⁻¹).
  8. Find μ by weighing a known length of string (1 m is ideal) on the balance: μ = mass ÷ length.

Analysing the Results — the Linear Graph Trick

Here's the elegant bit. We know f = v/(2L), which can be rewritten as:

f = (v/2) × (1/L)

Compare this to the straight-line equation y = mx:

y = f (Hz)
x = 1/L (m⁻¹)

gradient = v/2 (m s⁻¹)

So if you plot f against 1/L, you should get a straight line through the origin. The gradient of that line, multiplied by 2, gives you the wave speed v — which you can then cross-check against v = √(T/μ) calculated independently. This is a really elegant way to test the theory using real data rather than a single calculation.

f/Hz ^ | * | * | * | * gradient = v/2 | * |* +----------------------------> 1/L (m^-1) 0

Evaluating: Errors & How to Reduce Them

Systematic Errors
  • Use an oscilloscope to verify the signal generator's frequency readings independently.
  • Let the signal generator run for ~20 minutes to stabilise before taking readings.
  • Use as large a range of L as possible (e.g. 20 cm intervals over at least 1.0 m) — this improves resolution/percentage uncertainty in your gradient.
Random Errors
  • The biggest issue is the "sharpness of resonance" — it can be hard to judge exactly which frequency gives the "true" first harmonic.
  • Fix: watch a node closely while adjusting the frequency, rather than judging by eye from the amplitude (which moves too fast to track reliably).
  • Best repeat procedure: find the frequency giving max vibration, note it → increase frequency then gradually reduce until the harmonic reappears, note it → repeat for a third reading → average all three.
Safety
  • Use a rubber string rather than metal wire, in case it snaps under tension.
  • Wear goggles if using metal wire.
  • Stand well away from the hanging masses in case they fall.
  • Place a crash mat / soft surface beneath the masses.

Uncertainty Calculation Example

A student measures a wire: mass = 0.16 g, length weighed = 1.0 m, distance between fixed ends L = 0.4 m. Resolution: metre ruler = 1 mm (but real set-up errors up to 1 cm), top-pan balance = 0.005 g.

Percentage uncertainty in mass on hanger (0.20 kg, +/- 0.005 kg): (0.005 / 0.20) x 100% = 2.5% Percentage uncertainty in length (0.4 m, +/- 0.01 m): (0.01 / 0.4) x 100% = 2.5% Percentage uncertainty in mass per unit length (mu): Delta(mu)/mu = Delta(m)/m + Delta(l)/l = (0.005/0.16) + (0.001/1.0) -> Delta(mu) = 0.005 g/m (0.005 / 0.16) x 100% = 3% Since T and mu are under a SQUARE ROOT in f = (1/2L)*sqrt(T/mu), their % uncertainties are HALVED before adding: Total % uncertainty in f = 2.5% (length, full) + 1.25% (half of T's 2.5%) + 1.5% (half of mu's 3%) = 5.25%
Key Uncertainty Rule

When a quantity appears under a square root in a formula (like T and μ do in f = (1/2L)√(T/μ)), its percentage uncertainty is halved before you combine it with the others by addition. This trips up a lot of students!

Practice Question 5.1
In Core Practical 5, explain why the student should plot a graph of f against 1/L rather than just calculating v directly from a single set of L and f values.
Practice Question 5.2
Why is it important to determine the frequency of the first harmonic by both increasing to it and then decreasing to it, and averaging?

What to Memorise

Key Terms

TermMeaning
SuperpositionAdding the displacements of overlapping waves at a point to find the resultant displacement
InterferenceThe result of superposition — waves combining to produce a resultant wave with a new amplitude
Constructive interferenceResultant amplitude is larger than the individual waves (waves meet in phase)
Destructive interferenceResultant amplitude is smaller than the individual waves (waves meet in antiphase)
Coherent sourcesSame frequency + constant phase difference — required for a stable interference pattern
MonochromaticLight of a single frequency (e.g. laser light)
Phase differenceThe angle between two waves' positions in their cycle (measured in degrees/radians)
Path differenceThe difference in distance travelled by two waves to reach the same point (measured in λ)
Stationary (standing) waveFormed by two waves of equal frequency & amplitude travelling in opposite directions superposing; the pattern doesn't travel
NodeA point of permanently zero displacement in a stationary wave
AntinodeA point of maximum displacement (oscillates) in a stationary wave
Fundamental frequency (1st harmonic)The lowest possible resonant frequency of a string — one loop, λ = 2L

Formulas

FormulaWhat it's for
Constructive: path difference = nλCondition for constructive interference (n = 0, 1, 2, ...)
Destructive: path difference = (n + ½)λCondition for destructive interference
Distance between adjacent nodes = λ/2Spacing rule for any stationary wave
v = √(T/μ)Wave speed on a stretched string
f₀ = (1/2L)√(T/μ)Fundamental frequency of a stretched string
λ = 2L (1st harmonic), λ = L (2nd), λ = 2L/3 (3rd)Harmonic wavelength patterns on a string
μ = mass ÷ lengthMass per unit length of a string

Concepts Checklist

Exam Tips & Common Mistakes

Trap 1: Confusing phase difference and path difference

These sound similar but measure completely different things — phase difference is an angle (comparing cycle position), path difference is a distance (in multiples of λ). Examiners deliberately test this confusion.

Trap 2: Forgetting n = 0 counts as constructive

A path difference of exactly 0 (waves travelling equal distances) still satisfies "nλ" with n = 0 — it's still constructive interference. Don't dismiss a zero path difference as "no interference."

Trap 3: Mixing up which length to use

In stretched-string calculations, always check whether you need the whole string's length (for finding μ) or the vibrating length between the fixed points (for finding f₀ using λ = 2L). These are often different numbers in the same question!

Trap 4: Assuming frequency is proportional to tension

f₀ ∝ √T, not f₀ ∝ T. Doubling the tension multiplies frequency by √2 ≈ 1.41, not by 2. This is a favourite "sneaky" calculation question.

Trap 5: Square-root uncertainties

When T or μ (both under a square root in the frequency formula) contribute to an uncertainty calculation, halve their percentage uncertainty before adding to the others. Forgetting to halve is one of the most common marks lost in this practical's write-up.

What Examiners Reward
  • Always show full working with units at every step, not just a final answer.
  • When identifying interference type, explicitly state the path difference calculation, not just the conclusion.
  • For practical-based questions, always tie your answer back to the actual experimental set-up described (e.g. "the reflected wave from the loudspeaker at the closed end...").
  • Use the correct number of significant figures matching the data given in the question.
Revision Guide — Interference & Stationary Waves · Edexcel International A Level Physics
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Also in the full note
  • 1. Interference & Superposition of Waves
  • 2. Phase & Path Difference
  • Exam Tips & Common Mistakes
  • Aim & Variables
  • Apparatus & Its Purpose
  • Evaluating: Errors & How to Reduce Them
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