Wave Properties

A wave is a disturbance that transfers energy from one place to another without transferring matter. Understanding the fundamental properties of waves — and being able to describe them precisely using mathematical language — is essential for everything that follows in this topic.

Transverse and Longitudinal Waves

Waves come in two fundamental types, defined by the relationship between the direction of oscillation and the direction of energy transfer.

In a transverse wave, the oscillations are perpendicular to the direction of energy transfer. If you flick a rope up and down, the wave travels horizontally along the rope while the particles of the rope move vertically. Light, water surface waves, and all electromagnetic waves are transverse.

In a longitudinal wave, the oscillations are parallel to the direction of energy transfer. Sound is the most important example. When a loudspeaker cone pushes forward, it compresses the air molecules in front of it, creating a region of compression (high pressure). As the cone pulls back, it creates a region of rarefaction (low pressure). These compressions and rarefactions propagate through the air, carrying energy to your ear.

A useful way to remember: in a longitudinal wave, the particles oscillate along the direction the wave travels (longitudinal — along).

graph LR
    A["Wave Type?"] -->|"Oscillation ⊥ energy transfer"| B["Transverse"]
    A -->|"Oscillation ∥ energy transfer"| C["Longitudinal"]
    B --> D["Examples: light, EM waves,\nwater surface waves,\nS-waves (seismic)"]
    C --> E["Examples: sound,\nultrasound,\nP-waves (seismic)"]
    B --> F["Can be polarised ✓"]
    C --> G["Cannot be polarised ✗"]

Key Wave Quantities

Every wave can be described using a small set of measurable quantities.

Amplitude (A) is the maximum displacement of a particle from its equilibrium (rest) position. It is measured in metres (m). For a sound wave, greater amplitude means a louder sound. For a light wave, greater amplitude means greater intensity. Crucially, the intensity of a wave is proportional to the square of its amplitude: I ∝ A². This means doubling the amplitude quadruples the intensity.

Wavelength (λ) is the distance between two consecutive points in phase — for example, from one crest to the next crest, or from one compression to the next compression. It is measured in metres (m).

Frequency (f) is the number of complete oscillations per second. It is measured in hertz (Hz), where 1 Hz = 1 oscillation per second. A tuning fork vibrating at 440 Hz completes 440 full cycles every second.

Period (T) is the time taken for one complete oscillation. It is measured in seconds (s). Frequency and period are reciprocals:

$T = \frac{1}{f} \quad \text{and} \quad f = \frac{1}{T}$

For example, if a wave has a frequency of 500 Hz, its period is T = 1/500 = 0.002 s = 2 ms.

Wave speed (v) is the distance travelled by the wave per unit time. It is measured in m s⁻¹. The fundamental wave equation links speed, frequency, and wavelength:

$v = f\lambda$

This equation applies to all waves. If you know any two of the three quantities, you can calculate the third.

Quantity	Symbol	SI Unit	Key relationship
Amplitude	A	m	I ∝ A²
Wavelength	λ	m	λ = v / f
Frequency	f	Hz (s⁻¹)	f = 1 / T
Period	T	s	T = 1 / f
Wave speed	v	m s⁻¹	v = fλ
Phase difference	Δφ	rad (or °)	Δφ = 2πΔx / λ

Worked Calculations

Worked example 1: A sound wave in air has a frequency of 680 Hz and a wavelength of 0.50 m. Calculate the speed of the wave.

v = fλ = 680 × 0.50 = 340 m s⁻¹

This is consistent with the known speed of sound in air at room temperature (~340 m s⁻¹ at 20 °C).

Worked example 2: A radio station broadcasts at 98.5 MHz. Calculate the wavelength of the broadcast signal (speed of EM waves = 3.00 × 10⁸ m s⁻¹).

First convert the frequency: f = 98.5 MHz = 98.5 × 10⁶ Hz

λ = v / f = 3.00 × 10⁸ / 98.5 × 10⁶ = 3.05 m

FM radio waves have wavelengths of the order of a few metres — this is why FM radio antennas are roughly 75 cm long (about λ/4).

Worked example 3: An ultrasound scanner emits pulses at 3.5 MHz. The speed of ultrasound in soft tissue is approximately 1540 m s⁻¹. Calculate the wavelength and explain why this frequency is chosen.

λ = v / f = 1540 / (3.5 × 10⁶) = 4.4 × 10⁻⁴ m ≈ 0.44 mm

This short wavelength provides good resolution in medical imaging because the ultrasound can distinguish structures that are about 1 mm or larger. Higher frequencies give shorter wavelengths and better resolution but are absorbed more quickly, reducing penetration depth.

Displacement–Distance and Displacement–Time Graphs

Waves are commonly represented using two types of graph, and it is vital to distinguish between them.

A displacement–distance graph is a snapshot of the wave at a single instant. The x-axis shows position along the wave (in metres), and the y-axis shows displacement. From this graph you can read off the amplitude (maximum displacement) and the wavelength (distance between consecutive points in phase).

A displacement–time graph shows how a single point on the wave oscillates over time. The x-axis shows time (in seconds), and the y-axis shows displacement. From this graph you can read off the amplitude and the period (time for one complete cycle). From the period, you can calculate the frequency using f = 1/T.

Common exam mistake: A very frequent error is reading a "wavelength" from a displacement–time graph. The repeat distance on a displacement–time graph gives the period, not the wavelength. Wavelength can only be determined from a displacement–distance graph. Similarly, you cannot read the period from a displacement–distance graph.

Phase and Phase Difference

Phase describes the stage a point on a wave has reached in its cycle. It is measured in degrees (0° to 360°) or radians (0 to 2π).

Two points on a wave are in phase if they have the same displacement and are moving in the same direction at all times. They have a phase difference of 0° (or 360°, or any multiple of 360°). For example, two adjacent crests are in phase, separated by exactly one wavelength.

Two points are in antiphase (or completely out of phase) if they have a phase difference of 180° (π radians). When one is at maximum positive displacement, the other is at maximum negative displacement.

The phase difference between two points separated by a distance Δx along a wave of wavelength λ is:

$\Delta\phi = \frac{2\pi \, \Delta x}{\lambda}$

Worked example 4: Two points on a wave of wavelength 0.40 m are separated by 0.10 m. Calculate the phase difference.

Δφ = (2π × 0.10) / 0.40 = π/2 radians = 90°

The points are a quarter of a cycle apart.

Worked example 5: On a wave of wavelength 12 cm, how far apart are two points with a phase difference of 120° (2π/3 radians)?

Rearranging: Δx = Δφ × λ / (2π) = (2π/3) × 0.12 / (2π) = 0.12/3 = 0.040 m = 4.0 cm

Intensity and the Inverse Square Law

For a point source radiating uniformly in all directions, the wave energy spreads out over ever-larger spheres. At a distance r from the source, the intensity (power per unit area) is:

$I = \frac{P}{4\pi r^2}$

This gives the inverse square law: intensity is inversely proportional to the square of the distance from the source.

Since I ∝ A², this also means A ∝ 1/r — the amplitude halves when you double the distance from the source.

Worked example 6: A loudspeaker emits sound at a power of 0.50 W. Calculate the intensity at a distance of 4.0 m, assuming the sound radiates uniformly.

I = P / (4πr²) = 0.50 / (4π × 4.0²) = 0.50 / (4π × 16) = 0.50 / 201.1 = 2.5 × 10⁻³ W m⁻²

Real-world application: The inverse square law explains why sound becomes quieter as you move away from a source, why we need more powerful transmitters to communicate over greater distances, and why the intensity of sunlight on Mars (~1.5 AU from the Sun) is only about 44% of that on Earth (~1 AU).

Wave Speed in Different Media

The speed of a wave depends on the medium through which it travels, not on its frequency or amplitude (for most waves at normal amplitudes).

Medium	Speed of sound / m s⁻¹	Notes
Air (20 °C)	343	Increases ~0.6 m s⁻¹ per °C
Water (25 °C)	1498	About 4× faster than air
Soft tissue	~1540	Used in medical ultrasound
Steel	~5960	Waves travel fastest in stiff materials
Vacuum	0	Sound cannot travel in a vacuum

For electromagnetic waves in a vacuum, v = c = 3.00 × 10⁸ m s⁻¹ regardless of frequency. In a medium, EM waves slow down (v = c/n, where n is the refractive index).

When a wave passes from one medium to another, its frequency stays constant (determined by the source), but its speed and wavelength change. Since v = fλ and f is fixed, a decrease in speed means a decrease in wavelength, and vice versa.

Summary of Key Equations

Quantity	Symbol	Unit	Equation
Period	T	s	T = 1/f
Frequency	f	Hz	f = 1/T
Wave speed	v	m s⁻¹	v = fλ
Phase difference	Δφ	rad	Δφ = 2πΔx / λ
Intensity	I	W m⁻²	I = P/(4πr²), I ∝ A²

These relationships form the foundation for every wave calculation in A-Level Physics. Make sure you can rearrange each equation confidently and select the correct one for a given problem.

A-Level Deep Dive: Wave Properties

Spec mapping

Edexcel 9PH0 specification Topic 5 — Waves and the Particle Nature of Light, in its opening sub-strands, requires students to describe and distinguish transverse and longitudinal progressive waves; define amplitude, wavelength, frequency, period and phase; use the relationship $v = f\lambda$ ; and apply the inverse-square relationship for intensity from a point source (refer to the official Pearson Edexcel specification document for exact wording). Although these statements appear at the start of Topic 5, the underlying ideas thread through every later sub-strand: superposition and standing-wave conditions are written in the language of wavelength and phase difference; diffraction grating geometry uses $d\sin\theta = n\lambda$ , which collapses without confidence in $v = f\lambda$ ; and photon energy $E = hf$ is examined alongside wavelength via $E = hc/\lambda$ . Wave-property fluency is also load-bearing in Topic 7 (electric and magnetic fields) when discussing electromagnetic radiation, and in Topic 12 (turning points / nuclear) for de Broglie wavelengths. The Edexcel data-and-formulae booklet lists $v = f\lambda$ , $I = P/(4\pi r^2)$ for a point source, and $E = hf$ , but does not list the phase-difference formula $\Delta\phi = 2\pi\Delta x/\lambda$ — that one must be memorised.

Worked example with full mark scheme

Question (8 marks):

A small loudspeaker emits sound uniformly in all directions at a constant power. A microphone records a frequency of 1.20 kHz and a sound intensity of $4.0 \times 10^{-3}\ \text{W m}^{-2}$ at a distance of 2.0 m from the loudspeaker. The speed of sound in air is $340\ \text{m s}^{-1}$ .

(a) Calculate the wavelength of the sound. (2)

(b) Calculate the power output of the loudspeaker. (3)

(c) The microphone is moved to a new position. The intensity is now $1.0 \times 10^{-3}\ \text{W m}^{-2}$ . Calculate the new distance from the loudspeaker, and state the new amplitude as a fraction of the original amplitude. (3)

Solution with mark scheme:

(a) Step 1 — apply the wave equation.

$\lambda = \frac{v}{f} = \frac{340}{1.20 \times 10^3} = 0.283\ \text{m}$

M1 — correct rearrangement of $v = f\lambda$ to $\lambda = v/f$ with values substituted (frequency in Hz, not kHz). The conversion $1.20\ \text{kHz} = 1200\ \text{Hz}$ is the most common slip; leaving frequency in kHz produces $\lambda = 283\ \text{m}$ , off by a factor of 1000.

A1 — correct value, $\lambda = 0.28\ \text{m}$ (2 sf is acceptable; $0.283\ \text{m}$ to 3 sf is preferred).

(b) Step 1 — recognise the inverse-square geometry.

The loudspeaker is treated as a point source radiating into a full sphere, so the intensity at radius $r$ is $I = P/(4\pi r^2)$ . C1 — selecting the correct relationship.

Step 2 — rearrange and substitute.

$P = I \cdot 4\pi r^2 = (4.0 \times 10^{-3}) \times 4\pi \times (2.0)^2$

M1 — algebraically correct rearrangement and substitution.

Step 3 — evaluate.

$P = (4.0 \times 10^{-3}) \times 50.27 = 0.20\ \text{W}$

A1 — final answer $P = 0.20\ \text{W}$ (or $0.201\ \text{W}$ ). A unit error here (" $0.20\ \text{W m}^{-2}$ ") loses the A1.

$\frac{I_2}{I_1} = \frac{r_1^2}{r_2^2} \implies r_2 = r_1\sqrt{\frac{I_1}{I_2}} = 2.0 \times \sqrt{\frac{4.0 \times 10^{-3}}{1.0 \times 10^{-3}}} = 2.0 \times 2 = 4.0\ \text{m}$

M1 — applying the inverse-square law as a ratio (avoids re-computing $P$ ).

A1 — $r_2 = 4.0\ \text{m}$ .

Step 2 — relate amplitude to intensity.

Since $I \propto A^2$ , $A \propto \sqrt{I}$ . The intensity has dropped by a factor of 4, so $A_2/A_1 = \sqrt{1/4} = 1/2$ .

B1 — the new amplitude is half the original.

Total: 8 marks (M2 A2 C1 M1 A1 B1).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A student investigates a wave on a stretched rope. A displacement–time graph for one point on the rope shows that the period is $0.40\ \text{s}$ and the maximum displacement is $5.0\ \text{cm}$ . A separate displacement–distance graph (taken at one instant) shows that two adjacent crests are separated by $1.20\ \text{m}$ .

(a) Calculate the speed of the wave. (2)

(b) Two points on the rope are $0.30\ \text{m}$ apart. Calculate, in radians, the phase difference between them. (2)

(c) State and explain how the answer to (b) would change if the wavelength were doubled while the frequency remained the same. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO2) — recognising that frequency comes from the displacement–time graph and wavelength from the displacement–distance graph; computing $f = 1/T = 2.5\ \text{Hz}$ .
A1 (AO1) — $v = f\lambda = 2.5 \times 1.20 = 3.0\ \text{m s}^{-1}$ .

(b)

M1 (AO1) — substitution into $\Delta\phi = 2\pi\Delta x/\lambda = 2\pi \times 0.30 / 1.20$ .
A1 (AO1) — $\Delta\phi = \pi/2\ \text{rad}$ (or $1.57\ \text{rad}$ ).

(c)

M1 (AO2) — recognising that $\Delta\phi$ is inversely proportional to $\lambda$ for a fixed separation, so doubling $\lambda$ halves $\Delta\phi$ .
A1 (AO3) — new phase difference is $\pi/4\ \text{rad}$ , with explicit reasoning that the proportion of one cycle covered by the same physical separation has halved.

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. This question is typical of the Edexcel pattern of weaving graph-reading (AO2) and proportional reasoning (AO3) into questions whose surface looks like pure substitution.

Synoptic links

Connects to:

Topic 5, sub-strand on superposition and interference: the constructive/destructive interference conditions $\Delta x = n\lambda$ (constructive) and $\Delta x = (n + \tfrac{1}{2})\lambda$ (destructive) are direct rephrasings of the phase-difference formula $\Delta\phi = 2\pi\Delta x/\lambda$ at $\Delta\phi = 2n\pi$ and $(2n+1)\pi$ respectively. Confidence with phase difference is the prerequisite for Young's double-slit and thin-film analysis.
Topic 5, sub-strand on standing waves: a stationary wave on a string fixed at both ends has length $L = n\lambda/2$ . Resonance frequencies follow from $f_n = nv/(2L)$ — pure $v = f\lambda$ applied at fixed wavelengths set by the boundary conditions. Without secure handling of $v = f\lambda$ , harmonic analysis is impossible.
Topic 5, sub-strand on the photon model: $E = hf$ links the wave property of frequency to the particle property of energy. Combining with $c = f\lambda$ gives $E = hc/\lambda$ — the equation used for every photoelectric and energy-level question.
Topic 6 (further mechanics — circular motion and SHM): simple harmonic motion is the time-domain description of a single oscillating point on a wave; the period $T$ in the wave equation is identical to the SHM period $T = 2\pi/\omega$ . The displacement–time graph of a point on a transverse wave is a sinusoid governed by $x(t) = A\sin(\omega t)$ .
Topic 7 (electric and magnetic fields — EM radiation context): all electromagnetic waves obey $c = f\lambda$ in vacuum. The inverse-square law for intensity carries straight across to radio-wave power flux, gamma-ray dosimetry, and the solar constant calculation.

Mark-scheme literacy

Wave-property questions on 9PH0 split AO marks broadly as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Selecting and substituting correct formulae ( $v = f\lambda$ , $T = 1/f$ , $I = P/(4\pi r^2)$ , phase-difference formula); converting units (kHz to Hz, MHz to Hz, cm to m); reading graph values
AO2 (application / analysis)	25–35%	Distinguishing displacement–time from displacement–distance graphs; recognising which formula applies to a non-routine context; ratio-style intensity calculations
AO3 (problem-solving / evaluation)	10–15%	Multi-stage problems combining wave equation with intensity, or explaining proportional changes (e.g. how amplitude scales with distance)

Examiner-rewarded phrasing: "the wavelength is read from the displacement–distance graph because…"; "since intensity is inversely proportional to the square of distance…"; "the frequency is determined by the source and is unchanged on entering a new medium". Phrases that lose marks: "the wave gets quieter" (no quantification); "the wavelength changes" without stating why; mixing " $\propto$ " with " $=$ " carelessly.

A specific Edexcel pattern to watch: questions with wording like "calculate" demand a numerical answer with correct unit and appropriate significant figures; "show that" demands derivation with explicit substitution; "explain" demands a causal sentence linking physics to outcome. Answering an "explain" question with a calculation typically loses the explanation marks.

Grade-band model answers

3-mark question

Question: A water wave has a frequency of $0.50\ \text{Hz}$ and a wavelength of $3.0\ \text{m}$ . State its period and calculate its speed.

Grade C response (~150 words):

The period is $T = 1/f = 1/0.50 = 2.0\ \text{s}$ .

The speed is $v = f\lambda = 0.50 \times 3.0 = 1.5\ \text{m s}^{-1}$ .

Examiner commentary: Full marks (3/3). The candidate selects both formulae correctly, substitutes accurately, and includes units. The presentation is brief but every step is verifiable. A common Grade C pattern is to drop units on the period ("T = 2.0") — that would lose one mark even with the rest correct. The order of operations here (period first, then speed) is sensible because period is a one-step rearrangement. Note that $v = f\lambda$ requires frequency in Hz — had the question given frequency in kHz, conversion would be needed before substitution.

Grade A response (~210 words):*

Using $T = 1/f$ :

$T = \frac{1}{0.50\ \text{Hz}} = 2.0\ \text{s}$

Using the wave equation $v = f\lambda$ :

$v = (0.50\ \text{Hz}) \times (3.0\ \text{m}) = 1.5\ \text{m s}^{-1}$

The period of $2.0\ \text{s}$ means each point on the water surface completes one full oscillation every two seconds, while the wave profile advances $1.5\ \text{m}$ along the surface in the same time — consistent with the wavelength of $3.0\ \text{m}$ being covered in two periods (one wavelength per period).

Examiner commentary: Full marks (3/3). The candidate goes beyond procedural correctness by adding a one-sentence physical interpretation that links period, wavelength and speed. The cross-check ("one wavelength per period") is the kind of consistency reasoning that signals examiner-aware understanding. On a 3-mark question this extra is unnecessary for the marks, but the same disciplined habit on a 6-mark question is the difference between A and A*.

6-mark question

Question: A loudspeaker mounted on a tall pole emits sound uniformly. At a distance of $5.0\ \text{m}$ the sound intensity is measured as $1.6 \times 10^{-3}\ \text{W m}^{-2}$ . (a) Calculate the power output of the loudspeaker. (b) Calculate the intensity at $20\ \text{m}$ . (c) State, with reasoning, the ratio of amplitudes at $5.0\ \text{m}$ and $20\ \text{m}$ .

Grade B response (~250 words):

(a) $P = I \times 4\pi r^2 = 1.6 \times 10^{-3} \times 4\pi \times 25 = 0.50\ \text{W}$ .

(b) Distance has gone from $5\ \text{m}$ to $20\ \text{m}$ , a factor of 4. Intensity goes down by a factor of $4^2 = 16$ . So $I_2 = 1.6 \times 10^{-3} / 16 = 1.0 \times 10^{-4}\ \text{W m}^{-2}$ .

Examiner commentary: Method correct throughout, secured (a) with full marks (2/2) and (b) with full marks (2/2). In (c), the candidate states the right magnitude but inverts the ratio: $A_5 : A_{20}$ should be $4 : 1$ (the closer point has the larger amplitude), and the reasoning is too thin to recover the lost mark. Total: 5/6. The slip is characteristic of Grade B: the physics is right, the direction is wrong.

Grade A response (~290 words):*

(a) Treat the loudspeaker as a point source radiating uniformly into a sphere of radius $r$ . Then $I = P/(4\pi r^2)$ , so:

$P = I \cdot 4\pi r^2 = (1.6 \times 10^{-3}) \times 4\pi \times (5.0)^2 = 0.50\ \text{W}$

(b) Using the inverse-square law as a ratio (no need to recompute $P$ ):

$I_2 = I_1 \left(\frac{r_1}{r_2}\right)^2 = 1.6 \times 10^{-3} \times \left(\frac{5.0}{20}\right)^2 = 1.6 \times 10^{-3} \times \frac{1}{16} = 1.0 \times 10^{-4}\ \text{W m}^{-2}$

(c) Since $I \propto A^2$ , $A \propto \sqrt{I}$ . The intensity at $5.0\ \text{m}$ is 16 times the intensity at $20\ \text{m}$ , so the amplitude is $\sqrt{16} = 4$ times larger. Hence $A_5 : A_{20} = 4 : 1$ .

Examiner commentary: Full marks (6/6). The ratio approach in (b) is more elegant than recomputing $P$ and demonstrates the AO2 reasoning Edexcel rewards. In (c) the direction of the ratio is stated unambiguously and the underlying $I \propto A^2$ relationship is named explicitly. The "no need to recompute $P$ " remark is the kind of self-aware methodological commentary characteristic of A* writing.

9-mark question

Question: A microphone is connected to an oscilloscope and used to record sound from a tuning fork at a distance of $1.0\ \text{m}$ . The trace shows a sinusoidal signal of period $2.5\ \text{ms}$ and peak displacement $0.80\ \text{cm}$ on the screen.

(a) Calculate the frequency of the tuning fork. (1)

(b) The speed of sound in air is $340\ \text{m s}^{-1}$ . Calculate the wavelength. (2)

(c) Two microphones are now placed in line with the tuning fork. Microphone 1 is at $1.0\ \text{m}$ and microphone 2 is at $1.34\ \text{m}$ . Calculate the phase difference, in radians, between the signals at the two microphones. (3)

(d) Microphone 2 is then moved further from the source until the intensity it records is one ninth of the intensity at microphone 1. Calculate its new distance from the source, and explain in terms of energy why intensity falls with distance even though no energy is absorbed. (3)

Grade A response (~340 words):*

(a) $f = 1/T = 1/(2.5 \times 10^{-3}) = 400\ \text{Hz}$ .

(b) $\lambda = v/f = 340/400 = 0.85\ \text{m}$ .

$\Delta\phi = \frac{2\pi \Delta x}{\lambda} = \frac{2\pi \times 0.34}{0.85} = 0.80\pi\ \text{rad} \approx 2.51\ \text{rad}$

(d) Using $I \propto 1/r^2$ , intensity falls by a factor of 9 means $r$ increases by a factor of $\sqrt{9} = 3$ . So the new distance is $r_2 = 3 \times 1.0 = 3.0\ \text{m}$ .

The wave energy emitted per second is constant (the loudspeaker output power is fixed), but at a distance $r$ this energy is spread over the surface of a sphere of area $4\pi r^2$ . Doubling $r$ quadruples the area, so the power per unit area — the intensity — drops by a factor of 4. No energy is lost; it is simply distributed over a larger surface.

Examiner commentary: Full marks (9/9). The candidate handles unit conversion (ms to s) cleanly in (a), keeps the wavelength to an appropriate number of significant figures in (b), and uses the path-difference shortcut in (c) without being distracted by the position of the source. In (d), the explanation is precisely what an examiner wants: the constancy of the source power, the geometric spreading over the sphere, and the explicit denial that energy has been "absorbed". This last sentence is the AO3 mark — many candidates score the calculation but lose the explanation by writing "the energy spreads out" without specifying what is geometrically constant and what is changing.

A-Level-depth misconceptions

The errors that distinguish A from A* on wave-property questions:

"Frequency changes when a wave enters a new medium." Frequency is set by the source and is unchanged on transmission. It is the speed and wavelength that change at a boundary. A* candidates state this explicitly and use it as a constraint when applying $v = f\lambda$ in two media.
"Wavelength can be read from a displacement–time graph." It cannot. A displacement–time graph gives the period (and hence frequency); a displacement–distance graph (a snapshot) gives the wavelength. The two graphs look identical to the eye but mean entirely different things. Always check the x-axis units.
Confusing intensity with loudness or brightness. Intensity has SI units of $\text{W m}^{-2}$ — it is a measurable physical quantity. Loudness (in phons) and brightness (perceived) are physiological responses that scale roughly logarithmically with intensity. On exam questions, "intensity" always refers to the physical $\text{W m}^{-2}$ quantity.
Applying $I = P/(4\pi r^2)$ to non-isotropic sources. The $4\pi r^2$ comes from a full spherical surface. For a hemispherical source (a loudspeaker on the ground radiating into the upper half-space), the area is $2\pi r^2$ ; for a directional beam, the formula does not apply at all. Read the question for the geometry.
Forgetting that $A \propto 1/r$ from $I \propto 1/r^2$ and $I \propto A^2$ . The combined chain gives amplitude falling as the first power of distance, not the second. Many candidates correctly halve the intensity and then incorrectly halve the amplitude as well.
Mixing degrees and radians for phase difference. $\Delta\phi = 2\pi\Delta x/\lambda$ produces a radian answer. If the question asks for degrees, multiply by $180/\pi$ . A common slip is to leave $\pi/2$ as "1.57°" — that's $1.57$ radians, equivalent to $90°$ .
"Longitudinal waves can be polarised." They cannot. Polarisation requires oscillation perpendicular to propagation, which is only possible for transverse waves. Stating that sound can be polarised is an automatic mark loss on any descriptive question.

Common errors and mark-loss patterns on wave-property calculations

Three patterns repeatedly cost candidates marks on Paper 1 wave-property questions. They are all about discipline, not difficulty.

Frequency unit slips. Frequencies given in $\text{kHz}$ ( $\times 10^3$ ), $\text{MHz}$ ( $\times 10^6$ ) or $\text{GHz}$ ( $\times 10^9$ ) must be converted to $\text{Hz}$ before substitution into $v = f\lambda$ . The cure: always rewrite the frequency in scientific notation in $\text{Hz}$ on the line before the substitution, never inside the substitution itself.
Wrong-graph readings. Reading "wavelength" from a displacement–time graph (or "period" from a displacement–distance graph) is endemic. The cure: always check the x-axis label before extracting any wave property. Train yourself to label the axes mentally as "TIME → period/frequency" or "DISTANCE → wavelength".
Inverse-square direction errors. Candidates know $I \propto 1/r^2$ but apply it the wrong way, multiplying when they should divide. The cure: always compute the ratio $r_2/r_1$ first, square it, and decide consciously whether intensity should rise or fall on physical grounds (further away → smaller intensity).

This pattern is endemic to Paper 1 wave questions: candidates know the equations, lose marks on bookkeeping.

Going further — university and academic signposting

Wave-property fluency points directly toward several undergraduate trajectories:

Optics and Fourier analysis (Year 1/2 physics): any wave can be decomposed into a sum of pure sinusoids of different frequencies — Fourier's theorem. Speech, music, and the spectrum of a star are all Fourier decompositions of complex waveforms into pure $f\lambda$ -obeying components.
Acoustics (Year 2/3 physics or engineering): the inverse-square law is modified in real rooms by reflections, absorption coefficients of surfaces, and reverberation time (Sabine's formula $T_{60} = 0.161 V/A$ ). Concert-hall acoustics is essentially the engineering of these deviations from the ideal point-source model.
Astrophysics (Year 1/2 physics): the apparent brightness of a star follows $F = L/(4\pi d^2)$ — the inverse-square law applied to luminosity. Combined with parallax distances and the distance modulus, this is how stellar luminosities (and, via Cepheid variables, the size of the universe) are measured.
Quantum mechanics (Year 2/3 physics): the de Broglie wavelength $\lambda = h/p$ extends the wave-particle equation $v = f\lambda$ to massive particles. A 1 kg ball moving at 1 m s⁻¹ has wavelength $\sim 10^{-34}\ \text{m}$ — undetectable; an electron at $10^6\ \text{m s}^{-1}$ has $\sim 10^{-9}\ \text{m}$ , which is comparable to atomic spacings and explains electron diffraction.

Oxbridge interview prompt: "A point source emits sound uniformly in all directions. Why does intensity fall as $1/r^2$ ? Now suppose the same source emits in a thin horizontal plane (a 2D world). How would intensity fall with distance? What about a wave on a 1D string?"

Required practical reference

Two Edexcel A-Level core practicals draw directly on wave-property fluency. Core Practical 3 — determination of the speed of sound in air using a resonance tube uses a tuning fork of known frequency held above a tube whose effective length can be varied. At resonance, the tube length corresponds to a quarter-wavelength (first resonance) and three-quarter-wavelengths (second resonance) of the standing wave, with end corrections accounted for. Subtracting consecutive resonance lengths gives a half-wavelength directly, which combined with $v = f\lambda$ yields the speed of sound. Typical results lie within 5% of the accepted value; the practical assesses unit handling and precision-of-measurement reasoning.

Core Practical 5 — determination of the refractive index of a transparent block is the second wave-property practical, in which a ray of light is traced through a glass block at varied angles of incidence and the angles of refraction are measured with a protractor. A graph of $\sin\theta_i$ against $\sin\theta_r$ yields a straight line through the origin whose gradient is the refractive index $n$ . This connects Topic 5's wave equation to the slowing of light in dense media via $n = c/v$ , and reinforces the constancy of frequency across a boundary.

Edexcel A-Level alignment footer

This content is aligned with the Pearson Edexcel GCE A Level Physics (9PH0) specification, Topic 5 — Waves and the Particle Nature of Light. For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.

Visual summary

graph TD
    A["Wave property<br/>question"] --> B{"What is<br/>given?"}
    B -->|"Frequency f<br/>and wavelength λ"| C["Use v = fλ<br/>directly"]
    B -->|"Period T"| D["Convert:<br/>f = 1/T"]
    B -->|"Displacement<br/>graph(s)"| E{"Which<br/>x-axis?"}
    E -->|"Time"| F["Read T,<br/>compute f = 1/T"]
    E -->|"Distance"| G["Read λ<br/>directly"]
    D --> C
    F --> C
    G --> C
    B -->|"Power P and<br/>distance r"| H["Use I = P / (4πr²)"]
    H --> I{"Need<br/>amplitude?"}
    I -->|"Yes"| J["Use I ∝ A²<br/>so A ∝ √I"]
    I -->|"No"| K["Stop"]
    B -->|"Two points,<br/>separation Δx"| L["Use Δφ = 2π Δx / λ"]
    C --> M["Check units:<br/>f in Hz, λ in m,<br/>v in m s⁻¹"]
    J --> M
    L --> M

    style C fill:#27ae60,color:#fff
    style M fill:#3498db,color:#fff