Waves Revision Notes — 9702 Paper 1

What a progressive wave is

Syllabus requirement · §7.1

“

describe what is meant by wave motion as illustrated by vibration in ropes, springs and ripple tanks; understand and use the terms displacement, amplitude, phase difference, period, frequency, wavelength and speed; understand that energy is transferred by a progressive wave

”

The pattern moves; the water stays

Drop a stone into a still pond. A circular ripple spreads outward across the surface — and if you watch a floating leaf where the ripple passes, the leaf simply bobs up and down and settles back where it started. Nothing travelled from the stone to the edge of the pond except the disturbance: the shape of the wave moved outwards, while each particle of water only oscillated about its own fixed point.

Flick one end of a rope and watch the hump run down its length; strike a tuning fork and hear sound reach your ear though no air molecule made the whole journey; float a cork in a ripple-tank water wave and see it rise and fall without drifting shoreward. In each case something is transferred, and it is not the material. What transfers is energy.

A progressive wave is a wave that transfers energy without any net transfer of matter (B1).

The mark scheme for 9702/24 M/J 2025 Q4(a)(i) gave its mark for "(a wave that) transfers / propagates energy". "Progressive" means the wave travels along. It separates these waves from stationary waves, which do not carry energy along — you meet those in the AS Superposition note.

The vocabulary, built on one snapshot

To talk precisely about waves we need words for what the pond picture shows. Freeze the wave at one instant — a snapshot — and label its parts on the figure below. Every term here is examined by name, so learn each one as an exact sentence.

A transverse wave photographed at one instant. The dashed line is the equilibrium position; a is the amplitude; λ is the wavelength between adjacent crests; points P and Q are a quarter of a cycle apart.

term	exact meaning	watch out for
displacement ( $x$ )	the distance of an oscillating point from its equilibrium position, at a given instant, with a sign	measured FROM the equilibrium line, not from a crest or from zero on the page
amplitude ( $a$ )	the MAXIMUM displacement from equilibrium	amplitude is one number per wave; displacement varies continuously between $+a$ and $-a$
crests and troughs	the points of maximum positive and maximum negative displacement	labels, not definitions — amplitude is measured to either
wavelength ( $\lambda$ )	the distance between two adjacent points that are in phase — e.g. crest to next crest	"adjacent" matters: crest to crest-two-along is $2\lambda$
period ( $T$ )	the time for one complete oscillation of a point on the wave	a property of a POINT, read from a displacement–time graph (see “Two graphs, one wave”)
frequency ( $f$ )	the number of complete oscillations per unit time: $f = 1/T$	unit hertz (Hz); a frequency of 50 Hz means 50 cycles each second
speed ( $v$ )	how fast the wave PATTERN travels through the medium	not how fast the particles move — particles only oscillate about their fixed positions
phase difference	the fraction of a cycle by which one oscillation leads or lags another, expressed in degrees (or radians)	$360^{\circ}$ is a full cycle; see the phase rules below

The terms the syllabus lists, plus crest and trough. The middle column is close to mark-scheme wording.

Phase difference: measuring "out of step"

Two points on a wave (or two separate oscillations) are compared by phase difference — how far through their cycles one is ahead of the other. Take one full cycle as $360^{\circ}$ . Then:

in phase: the oscillations move exactly together, crest matching crest — phase difference $= 360^{\circ}\times k$ for any whole number $k$ (including $0^{\circ}$ );
antiphase (exactly out of phase): one is at a crest whenever the other is in a trough — phase difference $= 180^{\circ}\times(2k+1)$ , i.e. an odd multiple of $180^{\circ}$ ;
anything between: quote the angle, e.g. " $90^{\circ}$ out of phase".

A phase difference of $90^{\circ}$ means a quarter of a cycle. In “Two graphs, one wave” you will turn this into a formula that gives phase differences straight from a wave graph.

Why energy moves but matter does not: each oscillating particle pushes or pulls on its neighbour and passes energy on to it. The particles themselves stay in place: after any number of wave cycles, a floating cork is back where it began. When a question asks what a wave transfers, the answer is energy — never water, rope, air or matter.

A demonstration: flicking a rope past a post

Tie a bright ribbon at one point of a long rope and flick the far end rhythmically, sending identical crests down the rope. Suppose you count 5 crests passing the post in 10 s, and ask for the frequency and period of the wave.

Frequency means oscillations (here, crests) per unit time:

f = \frac{5}{10\ \text{s}} = 0.50\ \text{Hz}

Period is the time for ONE crest — one full cycle — which follows from $f = 1/T$ rearranged:

T = \frac{1}{f} = \frac{1}{0.50} = 2.0\ \text{s}

Check it against the raw count: 5 crests in 10 s must give 2 s per crest. It does. Notice what the rope's ribbon did throughout: oscillated vertically, never travelling — the crests moved, the rope did not.

Common mistakes

"A progressive wave transfers the particles of the medium from source to detector."
It transfers ENERGY; there is no net transfer of matter. Each particle oscillates about a fixed point.
The B1 of 9702/24 M/J 2025 Q4(a)(i) is for 'transfers / propagates energy'. Saying the wave carries the material loses the mark.
Amplitude quoted as the crest-to-trough height of a snapshot graph.
Amplitude is measured from the equilibrium line to a crest (or trough): half the crest-to-trough distance.
In 9702/22 M/J 2025 Q3(a) the A1 was for 10.0 cm, the height from the axis to a peak. The 20 cm crest-to-trough height is the trap.
"Frequency is the time for one oscillation."
That is the PERIOD. Frequency is the NUMBER of oscillations per unit time; they are reciprocals, $f = 1/T$ .
Mixing them up spoils every later calculation: a period of 2.0 s means a frequency of 0.50 Hz, not 2.0 Hz.
Calling two crests "in antiphase".
Adjacent crests are IN PHASE ( $360^{\circ}$ apart ≡ $0^{\circ}$ ); antiphase pairs a crest with the adjacent TROUGH ( $180^{\circ}$ ).
Phase compares states of oscillation, not shapes: same state = in phase whatever the separation.

Two recall marks: define the wave, then express its progress

9702/24 M/J 2025 Q4(a)(i)–(ii)3 marks

A source oscillates with frequency $f$ to produce a progressive wave of wavelength $\lambda$ . The source takes time $t$ to produce $n$ complete oscillations.

(a)(i) State what is meant by a progressive wave. [1]

(a)(ii) State expressions, in terms of some or all of $f$ , $\lambda$ and $n$ , for:

the distance moved by a wavefront in time $t$
time $t$ . [2]

Show full working

1
(a)(i) A progressive wave is a wave that transfers energy (B1) — with no net transfer of matter.
The scheme credits '(wave that) transfers / propagates energy'. 'Energy' is the word that earns the mark.
2
(a)(ii), distance: in ONE oscillation of the source the wave moves on by one wavelength, so in $n$ oscillations a wavefront moves:
$\text{distance} = n\lambda \quad (B1)$
Build it from the definition: one oscillation moves the wave one λ, so n oscillations move it nλ.
3
(a)(ii), time: the source makes $f$ oscillations every second, so $n$ oscillations take:
$t = \frac{n}{f} \quad (B1)$
f is oscillations per second. Dividing the number of oscillations by oscillations-per-second leaves seconds. Students often write nf, which has the wrong unit (s⁻¹).

Answer

(a)(i) A wave that transfers energy (with no net transfer of matter). (a)(ii) distance = $n\lambda$ ; time $t = n/f$ .

Distance ÷ time here is nλ ÷ (n/f) = fλ — the wave equation of the next section.

Your turn

Vocabulary first, then a Paper 1 period–frequency conversion with prefixes.

1
Define phase difference between two points on a wave, and state the phase difference between:

(i) two adjacent crests of the same wave
(ii) one crest and the adjacent trough.
Stuck? Show hint
Take one full cycle to be 360° and count fractions of a cycle between the two points.
Show solution
1. 1
  Phase difference is the fraction of a cycle by which the oscillations of the two points are out of step, expressed in degrees (or radians).
  The definition needs the fraction-of-a-cycle idea; 'how far apart they are' is too vague to earn the mark.
2. 2
  (i) Adjacent crests are at the same stage of their cycles — separated by one full cycle:
  $\Delta\varphi = 360^{\circ}$
  In phase. A whole cycle apart is the same state of motion as 0° apart, so 360° and 0° both mean 'in phase'.
3. 3
  (ii) Crest to adjacent trough is half a cycle:
  $\Delta\varphi = 180^{\circ}$
  Antiphase — the odd-multiple-of-180° family from the section text.
Answer
Fraction of a cycle between two oscillations, in degrees (or radians). (i) 360° — in phase. (ii) 180° — antiphase.
29702/11 O/N 2024 Q261 mark
A wave has a frequency of $5\ \text{GHz}$ .

What is the period of the wave?

Options

A $200\ \text{ps}$
B $2\ \text{ns}$
C $20\ \text{ns}$
D $20\,000\ \mu\text{s}$
Stuck? Show hint
Convert GHz to hertz FIRST, then invert — and check the prefix on each option before trusting it.
Show solution
1. 1
  Convert the frequency to hertz: $5\ \text{GHz} = 5\times10^{9}\ \text{Hz}$ .
  Giga means 10⁹. Convert before inverting, not after.
2. 2
  State the rule: $T = \dfrac{1}{f}$ .
  Period and frequency are reciprocals — the definition from the table above.
3. 3
  Substitute and evaluate:
  $T = \frac{1}{5\times10^{9}} = 0.2\times10^{-9} = 2\times10^{-10}\ \text{s}$
  1 ÷ 5 = 0.2, and 1 ÷ 10⁹ = 10⁻⁹. Then write 0.2×10⁻⁹ in standard form.
4. 4
  Express in picoseconds ( $1\ \text{ps} = 10^{-12}\ \text{s}$ ):
  $T = \frac{2\times10^{-10}}{10^{-12}} = 200\ \text{ps} \quad\rightarrow\ \textbf{A}$
  Dividing by 10⁻¹² is the same as multiplying by 10¹². Option B (2 ns = 2000 ps) is ten times too big; C and D are further out still.
Answer
A — $T = 1/(5\times10^{9}) = 2\times10^{-10}\ \text{s} = 200\ \text{ps}$ .

Paper 1 · MCQReal past-paper questions · Progressive waves and wave terminology Paper 2 · TheoryReal past-paper questions · Progressive waves and wave terminology

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Can you do all of these?

Define a progressive wave as one that transfers energy with no net transfer of matter
Convert period to frequency and back with f = 1/T, converting prefixes (GHz, ms, ps, nm) before substituting
Derive v = fλ: one wavelength is travelled in one period, so speed = λ/T = fλ
Identify which graph you are reading: wavelength from displacement–distance, period from displacement–time, amplitude from either
Find phase difference as (d/λ) × 360° or (Δt/T) × 360°; in phase = multiples of 360°, antiphase = odd multiples of 180°
Find a point's direction of motion by shifting the snapshot slightly in the direction of travel; points are at rest at crests and troughs and fastest at zero displacement
Read a CRO trace: T = divisions × time-base, then f = 1/T; amplitude = vertical divisions × y-gain
Calculate intensity as I = P/A with the area facing the wave: side² for squares, πr² for circles, lengths converted to metres first
Apply I ∝ A² in both directions: amplitude ×k gives intensity ×k²; intensity ÷n gives amplitude ÷√n
Compare transverse and longitudinal waves with the mark-scheme words: oscillations perpendicular (transverse) or parallel (longitudinal) to the direction of energy transfer
Locate compressions and rarefactions on a longitudinal wave's displacement–position graph
Choose the Doppler sign from the physics: approaching (minus) raises the observed frequency, receding (plus) lowers it; then solve for whichever quantity is unknown
Describe the observed frequency for a passing source (high, then equal to f_s and falling as it passes, then low) and a rotating source (smoothly above and below f_s)
Recall the EM spectrum in order radio → microwaves → infrared → visible → ultraviolet → X-rays → gamma; convert wavelengths to metres before naming the region; visible = 400–700 nm
State the meaning of polarisation and explain that sound, being longitudinal, cannot be polarised
Use the angle BETWEEN the plane of polarisation and the transmission axis in Malus's law — if the filter starts at right angles, use 90° − α; for a chain, work one filter at a time
Work cos² problems in DEGREE mode, and take a square root to go from an intensity ratio to an amplitude ratio

Paper 1 · MCQ

Real past-paper questions, sorted by difficulty, with mark schemes

Paper 2 · Theory

Real past-paper questions, sorted by difficulty, with mark schemes

Deformation of Solids

Superposition