You are viewing a free preview of this lesson.
Subscribe to unlock all 10 lessons in this course and every other course on LearningBro.
A wave is a disturbance that transfers energy from one place to another without transferring matter. Understanding the fundamental properties of waves — and the distinction between transverse and longitudinal types — is essential to every topic in this course.
Spec mapping: This lesson sits under AQA 7408 section 3.3.1 (progressive waves) and addresses the foundational descriptors of wave behaviour: amplitude, wavelength, frequency, period, wave speed, phase, phase difference, path difference, and the classification of transverse vs longitudinal waves. The wave equation v = fλ, the relation Δφ = (2π/λ)Δx, the inverse-square intensity law for a point source, and I ∝ A² are also introduced here. (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- Section 3.3.1 (stationary waves): the nodal spacing λ/2 and antinodal positions cannot be derived without first owning the phase-difference framework introduced here.
- Section 3.2.1 (mechanics and oscillations): simple harmonic motion is the kinematic engine driving every point on a progressive wave — the amplitude, period and angular frequency definitions used here recur in SHM.
- Section 3.4 (mechanics — energy): intensity I = P/A is the time-averaged power flux, linking wave concepts to mechanical-energy bookkeeping.
- Section 3.7 (fields and their consequences): inverse-square intensity for a point source is the same geometric scaling that governs Coulomb's and Newton's laws of gravitation — a common synoptic comparison point on Paper 2.
There are two fundamental categories of wave, classified by the direction of oscillation relative to the direction of energy transfer.
In a transverse wave, the oscillations are perpendicular to the direction of energy transfer (the direction of wave propagation).
Examples include:
A key property of transverse waves is that they can be polarised — the oscillations can be restricted to a single plane. This is explored in detail in Lesson 9.
In a longitudinal wave, the oscillations are parallel to the direction of energy transfer. The medium undergoes alternating compressions (regions of high pressure/density) and rarefactions (regions of low pressure/density).
Examples include:
Longitudinal waves cannot be polarised, because the oscillation direction is already constrained to be along the propagation direction.
Exam Tip: If asked how you can distinguish transverse from longitudinal waves experimentally, state that only transverse waves can be polarised. This is the definitive test.
Every periodic wave is described by a set of measurable quantities. These must be defined precisely for full marks in an exam.
| Quantity | Symbol | Definition | SI Unit |
|---|---|---|---|
| Amplitude | A | Maximum displacement of a point on the wave from its equilibrium (rest) position | m |
| Wavelength | λ | Minimum distance between two points oscillating exactly in phase (e.g., crest to crest) | m |
| Frequency | f | Number of complete oscillations passing a given point per unit time | Hz (s⁻¹) |
| Period | T | Time taken for one complete oscillation | s |
| Wave speed | v | Distance travelled by the wave per unit time | m s⁻¹ |
Frequency and period are reciprocals of each other:
f = 1/T and equivalently T = 1/f
Worked Example 1 — A sound wave has a period of 2.5 ms. Calculate its frequency.
T = 2.5 ms = 2.5 × 10⁻³ s
f = 1/T = 1/(2.5 × 10⁻³) = 400 Hz
The fundamental equation relating wave speed, frequency, and wavelength is:
v = f λ
This can be derived from first principles. In one complete period T, the wave advances by exactly one wavelength λ. Since speed = distance/time:
v = λ/T = λ × (1/T) = f λ
This equation applies to all waves — mechanical and electromagnetic, transverse and longitudinal.
Worked Example 2 — A radio wave has a frequency of 98.5 MHz. Calculate its wavelength.
v = c = 3.00 × 10⁸ m s⁻¹ (radio waves are electromagnetic)
f = 98.5 MHz = 98.5 × 10⁶ Hz
λ = v/f = (3.00 × 10⁸)/(98.5 × 10⁶) = 3.05 m
Worked Example 3 — A guitar string vibrates at 440 Hz and produces a sound wave in air with wavelength 0.773 m. Calculate the speed of sound in air.
v = f λ = 440 × 0.773 = 340 m s⁻¹ (to 3 s.f.)
Worked Example 4 — Ultrasound of frequency 2.0 MHz travels through soft tissue at 1540 m s⁻¹. Calculate the wavelength.
λ = v/f = 1540/(2.0 × 10⁶) = 7.7 × 10⁻⁴ m = 0.77 mm
Exam Tip: Always convert units before substituting into the wave equation. Frequencies in MHz or kHz must be converted to Hz; wavelengths in nm or mm must be converted to m.
The phase of a point on a wave describes its position within the oscillation cycle. Phase is measured in radians (rad) or degrees (°), where one complete cycle = 2π rad = 360°.
The phase difference between two points on a wave (or between two waves) describes how far one oscillation is ahead of or behind the other.
| Phase difference | In degrees | In radians | Meaning |
|---|---|---|---|
| In phase | 0° (or 360°) | 0 (or 2π) | Points oscillate together — same displacement at all times |
| Antiphase | 180° | π | Points have equal but opposite displacements at all times |
| Quarter cycle ahead | 90° | π/2 | One point leads the other by a quarter of a wavelength |
If two points on a wave are separated by a distance Δx along the direction of propagation, their phase difference Δφ is:
Δφ = (2π/λ) × Δx
Equivalently, the path difference in terms of wavelength determines the phase relationship:
Worked Example 5 — Two points on a wave of wavelength 0.60 m are separated by 0.45 m. What is their phase difference?
Δφ = (2π/λ) × Δx = (2π/0.60) × 0.45 = (2π × 0.75) = 1.5π rad
Convert: 1.5π rad = 270° (or equivalently 3π/2 rad)
The path difference is 0.75λ, which is neither a whole number nor a half-integer multiple of λ, so the points are neither in phase nor in antiphase.
Worked Example 6 — Two loudspeakers emit sound of wavelength 0.50 m in phase. A listener is 3.00 m from one speaker and 3.75 m from the other. What does the listener hear?
Path difference = 3.75 − 3.00 = 0.75 m
Path difference in wavelengths = 0.75/0.50 = 1.5λ
Since 1.5λ = (1 + ½)λ, this is a half-integer multiple of λ. The waves arrive in antiphase, producing destructive interference — the listener hears a quiet sound (a minimum).
Waves can be represented graphically in two important ways:
This is a snapshot of the wave at one moment in time, plotting displacement (y-axis) against position along the wave (x-axis). From this graph you can read:
This shows how the displacement of one point varies with time. From this graph you can read:
Common Misconception: Students often confuse wavelength and period. Wavelength is a spatial quantity (metres) read from a displacement-distance graph; period is a temporal quantity (seconds) read from a displacement-time graph.
The intensity of a wave is the power transmitted per unit area, measured perpendicular to the direction of energy transfer:
I = P/A
where I is intensity (W m⁻²), P is power (W), and A is the area (m²) through which the wave passes.
For a point source radiating equally in all directions, the power spreads over the surface of a sphere of radius r:
I = P/(4πr²)
This gives the inverse square law: intensity is inversely proportional to the square of the distance from the source.
Intensity is proportional to the square of the amplitude:
I ∝ A²
This means that doubling the amplitude quadruples the intensity.
Worked Example 7 — A point source emits sound with a power of 0.50 W. Calculate the intensity at a distance of 4.0 m from the source.
I = P/(4πr²) = 0.50/(4π × 4.0²) = 0.50/(4π × 16) = 0.50/201.1 = 2.5 × 10⁻³ W m⁻²
Worked Example 8 — At a distance of 2.0 m from a source, the intensity of a sound wave is 0.080 W m⁻². What is the intensity at 8.0 m?
Using the inverse square law: I₁r₁² = I₂r₂²
I₂ = I₁ × (r₁/r₂)² = 0.080 × (2.0/8.0)² = 0.080 × (0.25)² = 0.080 × 0.0625 = 5.0 × 10⁻³ W m⁻²
The distance quadrupled, so the intensity decreased by a factor of 16.
A wavefront is a surface (a curve in two dimensions, a surface in three) connecting all points on a wave that are in phase — for example, the locus of all crests at a given instant. For a wave from a point source, the wavefronts are concentric spheres expanding at speed v; far from the source, a small patch of the spherical wavefront looks approximately flat and is called a plane wave. A ray is a line perpendicular to the wavefronts indicating the direction of energy transfer; rays and wavefronts together form an orthogonal grid that completely characterises the geometry of wave propagation.
| Wave type | Wavefront shape | Intensity decay with distance | Typical example |
|---|---|---|---|
| Plane wave | Flat (parallel planes) | No decay (in non-absorbing medium) | Distant starlight, laser beam |
| Spherical wave | Concentric spheres | I ∝ 1/r² (inverse-square law) | Light from a point source, sound from a small loudspeaker |
| Cylindrical wave | Concentric cylinders | I ∝ 1/r | Sound from a long line source, light from a fluorescent tube viewed nearby |
The inverse-square law is therefore a consequence of three-dimensional geometric spreading from a point source; it is not a universal law of all wave propagation. Plane waves do not decay; cylindrical waves decay as 1/r.
I = P/(4πr²) = 1.2/(4π × 25) = 1.2/314.16 = 3.82 × 10⁻³ W m⁻²
Intensity level (dB) = 10 log₁₀(I/I₀) = 10 log₁₀((3.82 × 10⁻³)/(1.0 × 10⁻¹²)) = 10 × log₁₀(3.82 × 10⁹) = 95.8 dB
This is comparable to a noisy nightclub or a heavy lorry passing at close range — well above the 85 dB threshold for hearing damage with prolonged exposure.
(a) E₀ = √(2I/(ε₀c)) = √(2 × 250 / (8.85 × 10⁻¹² × 3.00 × 10⁸)) = √(2 × 250 / 2.655 × 10⁻³) = √(1.883 × 10⁵) = 434 V m⁻¹
(b) Energy per photon: E_photon = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(500 × 10⁻⁹) = 3.98 × 10⁻¹⁹ J
Number per second per cm² = I × A / E_photon = (250 × 10⁻⁴) / 3.98 × 10⁻¹⁹ = 6.3 × 10¹⁶ photons s⁻¹ cm⁻²
This shows why the photon picture is rarely needed for everyday illumination — the photon flux is so high that the granularity of the energy delivery is undetectable.
A progressive wave can be written mathematically as:
y(x, t) = A sin(ωt − kx + φ₀)
where ω = 2πf is the angular frequency, k = 2π/λ is the angular wavenumber, and φ₀ is the initial phase at x = 0, t = 0. The argument of the sine function (ωt − kx + φ₀) is the phase at position x and time t; the wave at any point oscillates with angular frequency ω, and at any instant the spatial pattern repeats with wavenumber k.
The wave speed v is the rate at which a point of constant phase moves through space. Setting (ωt − kx) = constant and differentiating gives v = dx/dt = ω/k. Substituting ω = 2πf and k = 2π/λ recovers v = fλ. This formal derivation underpins all the qualitative arguments made so far.
Amplitude: A = 0.020 m = 20 mm
Angular frequency: ω = 40π rad s⁻¹, so f = ω/(2π) = 20 Hz
Angular wavenumber: k = 4π m⁻¹, so λ = 2π/k = 0.50 m
Period: T = 1/f = 0.050 s = 50 ms
Phase velocity: v = ω/k = 40π/4π = 10 m s⁻¹
Check with v = fλ: 20 × 0.50 = 10 m s⁻¹ ✓
The wave travels in the +x direction because the sign of kx is negative inside the sine. Had the equation been written y = A sin(ωt + kx), the wave would travel in the −x direction.
ω = 2200 rad s⁻¹, k = 6.47 m⁻¹
v = ω/k = 2200/6.47 = 340 m s⁻¹ (consistent with the speed of sound in air at 20°C)
λ = 2π/k = 2π/6.47 = 0.971 m
f = v/λ = 340/0.971 = 350 Hz
This is a low-frequency sound, well within the speech-and-music register.
(a) k = 2.5 m⁻¹, λ = 2π/k = 2.513 m. Phase difference Δφ = k Δx = 2.5 × 0.40 = 1.0 rad (≈ 57.3°).
(b) Wave speed v = ω/k = 50/2.5 = 20 m s⁻¹. Time delay = Δx/v = 0.40/20 = 0.020 s = 20 ms. The further point lags behind the closer point by this amount.
(a) Duration: τ = L/c = 1.50/(3.00 × 10⁸) = 5.00 × 10⁻⁹ s = 5.0 ns
(b) Average power: P = E/τ = 0.40/(5.0 × 10⁻⁹) = 8.0 × 10⁷ W = 80 MW (this is the peak power of a typical Q-switched pulse laser)
(c) Photon energy: E_photon = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(6.33 × 10⁻⁷) = 3.14 × 10⁻¹⁹ J
Number of photons: N = E/E_photon = 0.40/(3.14 × 10⁻¹⁹) = 1.27 × 10¹⁸ photons
A short, energy-rich light pulse can therefore deliver an enormous photon count even though each individual photon carries a minute energy.
v = λ/T = 2.4/1.5 = 1.6 m s⁻¹
Time to cross 60 m: t = distance/v = 60/1.6 = 37.5 s
This is a slow wave compared with EM radiation: ocean surface waves typically travel at metres per second, sound at hundreds, and light at hundreds of millions — a separation of seven orders of magnitude. The fact that the same v = fλ equation applies across this range is a testament to the universality of wave kinematics.
Specimen question modelled on the AQA paper format. Six marks.
A teacher demonstrates a longitudinal sound wave in a slinky and, on a separate bench, a transverse pulse on a long rubber rope. A microphone-and-oscilloscope arrangement is used to inspect the air pressure produced by a 256 Hz tuning fork held above a closed organ pipe. The teacher then arranges two loudspeakers, driven in phase from a single signal generator at frequency 680 Hz. Student A stands such that the path from speaker 1 is 4.00 m and the path from speaker 2 is 4.50 m. The speed of sound is 340 m s⁻¹.
(a) State two observable features by which the slinky and rope waves could be classified as longitudinal or transverse without any equipment beyond visual inspection. (2 marks)
(b) Calculate the phase difference, in radians, between the sound arriving at Student A from each speaker. (3 marks)
(c) Hence state and justify whether Student A hears a maximum, a minimum, or an intermediate intensity. (1 mark)
The six available marks split conventionally across the three AO categories. AO1 (knowledge and recall) accounts for the two marks in part (a), which test direct recall of the transverse/longitudinal distinguishing features. AO2 (application of knowledge to a familiar context) accounts for the three calculation marks in part (b) — one for λ = v/f, one for converting path difference to a fraction of λ, and one for the final phase-difference value. AO3 (analysis and evaluation) accounts for the single mark in part (c), where the student must combine the numerical result with the interference-condition framework and reach a justified conclusion. Generic AO descriptors only — the precise wording of mark-scheme bullets is not reproduced here.
Grade C response (6/6 attempted, expected 3–4 awarded):
(a) The slinky wave is longitudinal because the coils move backwards and forwards along the direction the wave travels. The rope is transverse because the rope moves up and down while the wave goes sideways.
(b) λ = v/f = 340/680 = 0.50 m. Path difference = 4.50 − 4.00 = 0.50 m, which is one wavelength. Phase difference = 2π × (0.50/0.50) = 2π rad.
(c) The waves arrive in phase, so it is a maximum.
Examiner commentary: M1 awarded for naming the longitudinal slinky behaviour and M2 for the transverse rope behaviour, though the language is loose ("goes sideways" rather than "perpendicular to the direction of energy transfer"). M3 awarded for correct λ. M4 awarded for the correct path difference. M5 awarded for the correct phase difference of 2π rad. M6 awarded for stating "maximum" with the in-phase justification. Borderline C: arithmetic is right but the physics vocabulary is GCSE-tier; a stronger response would name "compression" and "rarefaction" and reference path-difference as nλ.
Grade A response (6/6):*
(a) In the slinky the particles oscillate parallel to the direction of energy transfer, generating visible compressions and rarefactions; this is the defining feature of a longitudinal wave. In the rope each element moves perpendicular to the direction of energy transfer, producing a transverse displacement profile. A second confirmatory test is that the rope wave can be plane-polarised by passing it through a slit aligned with the oscillation plane, whereas the slinky wave cannot — only transverse waves can be polarised.
(b) The wavelength of the sound is λ = v/f = 340 / 680 = 0.500 m. The geometric path difference between the two paths to Student A is Δx = 4.50 − 4.00 = 0.500 m. Applying Δφ = (2π/λ)·Δx gives Δφ = (2π/0.500) × 0.500 = 2π rad.
(c) A phase difference of 2π rad is equivalent to a phase difference of zero (full-cycle wrap-around). The waves arrive in phase from coherent in-phase sources, so the path difference satisfies the constructive-interference condition Δx = nλ with n = 1. Student A therefore hears an intensity maximum.
Examiner commentary: All six marks awarded. M1–M2 for the rigorous parallel/perpendicular framing in (a) plus the polarisation confirmatory test — an A*-band move because it goes beyond the minimum mark-scheme entry. M3–M5 for full numerical working with units shown. M6 for the explicit nλ identification and recognition that 2π rad ≡ 0 phase. The synoptic move to polarisation in (a) is what distinguishes this from a Grade B-style response that would simply name the perpendicular/parallel distinction.
Many candidates lose marks by treating "longitudinal" and "transverse" as if the labels were obvious from the medium alone — students will sometimes assert that sound in solids is transverse because solids can support shear. The correct framing is to examine the oscillation direction of a particle relative to the energy-transfer direction in that propagation mode. A typical pitfall on phase-difference questions is forgetting that path differences exceeding λ wrap modulo 2π: a path difference of 1.5λ is in antiphase, but a path difference of 3.5λ is also in antiphase, and many candidates compute (2π × 3.5) = 7π rad and leave it there, when the marker expects either the bare 7π or the equivalent π modulo 2π. A further common pitfall is conflating path difference with phase difference — they are related by Δφ = (2π/λ)Δx, and writing "the path difference is 90°" loses a mark even if the arithmetic is right. On intensity questions, candidates routinely state I ∝ A (a linear relationship) rather than I ∝ A², which is incorrect and loses any subsequent reasoning marks. Finally, on inverse-square problems, students sometimes drop the 4π factor when the question specifies a point source radiating equally in all directions — they apply I = P/A with A as a flat area rather than the spherical surface 4πr².
For a more rigorous treatment of progressive waves, The Physics of Waves and Oscillations by N.K. Bajaj develops the wave equation from first principles and is a standard first-year undergraduate text. For the geometric optics underpinning intensity-versus-distance arguments, Optics by Eugene Hecht — even just chapter 1 — embeds intensity within the Poynting-vector framework, which is the rigorous root of the inverse-square law. Students considering physics at university should also browse the early chapters of Halliday, Resnick and Walker (Fundamentals of Physics), particularly the chapter on travelling waves and the wave equation y(x,t) = A sin(kx − ωt + φ), which formalises everything in this lesson with calculus. The Hyperphysics web resource (Georgia State University) is a free, well-curated cross-reference; its "Travelling Waves" pages connect this material to acoustic intensity in decibels (an admissions-interview favourite).
The most subtle error distinguishing A from A* is the treatment of phase difference for points separated by more than one wavelength. A path difference of 2.5λ does not represent a "more antiphase" configuration than 0.5λ — both are antiphase, and an A* candidate explicitly reduces Δφ modulo 2π before quoting a final value. A second subtle error is confusing the amplitude of a wave with its maximum displacement at a given point: for a progressive wave the two are equal, but for a stationary wave (next lesson) they are not. Students who carry forward a progressive-wave intuition mis-state stationary-wave amplitudes. A third error is asserting that the inverse-square law "applies to all waves" — strictly it applies to an isotropic point source in a non-absorbing medium with 3-dimensional propagation. A line source gives 1/r decay; a plane wave gives no decay. Naming the geometry that produces inverse-square behaviour, rather than parroting the law, is what an A* answer does.
In Paper 1 and Paper 2 of AQA 7408, foundational wave questions typically appear as part-marked openers in which a quick numerical answer is followed by a one-line justification. Always write units alongside symbols when defining quantities — "v = 340 m s⁻¹" not "v = 340" — because mark schemes routinely penalise unit omissions on the final answer line. When asked to convert between phase difference in degrees and in radians, show the conversion factor (×π/180 or ×180/π) explicitly to claim the method mark even if a slip occurs in the final value. For path-difference questions, draw a quick sketch of the geometry: with two sources S₁, S₂ and a point P, the path lengths S₁P and S₂P should be labelled directly on the figure — this both clarifies your thinking and earns presentational credit on the multi-step parts. When commenting on the type of interference (constructive vs destructive), use full physics vocabulary: "the waves arrive in phase satisfying the condition Δx = nλ, producing constructive interference and an intensity maximum" earns more marks than "they add up to make it loud". If a question asks for intensity at a new distance, use the ratio form I₂/I₁ = (r₁/r₂)² rather than recomputing P explicitly — it is faster and avoids arithmetic slips. Finally, for any wave-classification question, the polarisation test is the cleanest single discriminator and should be reached for first.