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Year 11 Physics · Waves · Wave properties · 11 min read

Wave properties

A wave carries energy from one place to another without carrying the material along with it. A ripple crosses a pond, but the water just bobs up and down; a sound reaches your ear, but the air near the speaker stays where it is.

This guide covers the two kinds of mechanical wave, the parts of a wave (amplitude, wavelength, period and frequency), the wave equation v = fλ, and the two wave graphs that look the same but aren't. Then there is a quick quiz and a full worked example.

What you need to know

The equations you will use:

f = 1/TFrequency in hertz (Hz = s−1) from the period in seconds. It works the other way too: T = 1/f.
v = fλWave speed (m s−1) = frequency (Hz) × wavelength (m). Equivalent to v = λ/T.

Dot points: PY11-WV-01, PY11-WV-02, PY11-WV-03, PY11-WV-04, PY11-WV-05, PY11-WV-06 (2025 Physics syllabus, Year 11), in our own words.

Mini lesson

Energy moves, the medium doesn't

A mechanical wave is a disturbance passed from particle to particle through a medium (a rope, water, air). Each particle oscillates about its own rest position and ends up where it started. It's the energy that travels.

Top: a transverse wave on a rope, with one particle moving up and down while the energy travels to the right. Bottom: a longitudinal wave as a row of coils with compressions C and rarefactions R, with one particle moving back and forth along the direction the energy travels.transverseparticleenergylongitudinalCCCRRparticleenergy
Transverse: particles move at right angles to the energy's direction. Longitudinal: particles move back and forth along it, forming compressions (C) and rarefactions (R).

The parts of a wave

A transverse wave drawn as displacement y against distance x. The wavelength lambda is marked from one crest to the next. The amplitude A is marked from the rest line down to a trough. A crest and a trough are labelled.yxλAcresttrough
The x-axis is the rest position. Amplitude A runs from the rest line to a crest or trough; wavelength λ runs from a crest to the next crest.

The wave equation: v = fλ

In one period, a wave moves forward exactly one wavelength. So its speed is v = λ/T, and since f = 1/T, v = fλ.

Two graphs that look the same

Wave questions use two graphs that look alike but mean different things. Always read the horizontal axis first.

Two graphs side by side that look alike. Left: displacement y against distance x, where crest to crest is the wavelength lambda. Right: displacement y against time t, where crest to crest is the period T.yxλsnapshot: y vs distanceytTone particle: y vs time
Same shape, different meaning. On a distance axis, crest to crest is λ (in m); on a time axis, it is T (in s).

To find which way a particle on a snapshot is moving, picture the wave shifted a little in its direction of travel. If the curve at that particle's position is now higher, the particle is moving up; if lower, it is moving down.

Longitudinal waves on a displacement graph

A sound wave can be drawn on the same axes, but now the displacement is along the direction of travel. Take positive to mean a particle has moved forward (the way the wave is going) and negative to mean it has moved back.

At a compression, the particles just behind have moved forward and the particles just ahead have moved back, so they crowd together: the graph crosses zero going from positive to negative. At a rarefaction, the graph crosses zero going from negative to positive. The peaks are not compressions; they are where particles have moved furthest forward.

Compression to compression is still one wavelength, and v = fλ works in exactly the same way.

A displacement-distance graph for a longitudinal wave, with a row of particles drawn underneath. Where the graph crosses zero going from positive to negative the particles are bunched into a compression C; where it crosses going from negative to positive they are spread into a rarefaction R.sxCCR+−
s is each particle's displacement along the line of travel (+ = moved forward, the way the wave travels: to the right here). Compressions (C) sit where the graph crosses zero going down; rarefactions (R) where it crosses going up.

Common mistakes

Mini quiz

6 quick questions. Pick an answer to see if you're right and why.

  1. As a sound wave travels through air, how do the air particles move?
    Show answer

    B. Sound is longitudinal: the particles oscillate back and forth along the direction the energy travels, forming compressions and rarefactions. They don't travel with the wave.

  2. A wave has a period of 0.020 s. What is its frequency?
    Show answer

    C. f = 1/T = 1 ÷ 0.020 = 50 Hz. Taking the period itself as the frequency gives 0.020 Hz.

  3. A wave on a rope has a frequency of 4.0 Hz and a wavelength of 0.75 m. What is its speed?
    Show answer

    B. v = fλ = 4.0 × 0.75 = 3.0 m s−1. 0.19 comes from λ ÷ f and 5.3 from f ÷ λ.

  4. On a displacement–time graph for one particle, what is the horizontal distance between two neighbouring crests?
    Show answer

    B. The horizontal axis is time, so crest to crest is the time for one oscillation: the period. The wavelength comes from a displacement–distance graph.

  5. The frequency of a wave on a string is doubled, with the tension unchanged. What happens?
    Show answer

    B. The string (the medium) sets the speed, so v doesn't change. From v = fλ, doubling f halves λ.

  6. A displacement–distance graph shows crests 1.2 m apart. A displacement–time graph for one point on the same wave shows crests 0.30 s apart. What is the wave speed?
    Show answer

    D. λ = 1.2 m and T = 0.30 s, so v = λ/T = 1.2 ÷ 0.30 = 4.0 m s−1. 0.36 multiplies them and 0.25 divides the wrong way.

Worked example

Waves travel across a lake towards a jetty. A buoy floating on the water moves from its highest point to its lowest point, a vertical distance of 0.60 m, in 1.25 s. A photo taken from a drone shows that the horizontal distance from a crest to the next trough is 3.0 m.

(a) State the amplitude of the waves. (1 mark)

(b) Calculate the frequency and the speed of the waves. (3 marks)

(c) The buoy is 45 m from the jetty. Calculate how long a crest takes to travel from the buoy to the jetty. (1 mark)

(5 marks)

I'll think out loud, the way you should in the exam, using KNOW-UD: Knowns, Not known, Organise, Work it out, Units, Direction.

Prediction first. Highest to lowest is half an oscillation, and crest to the next trough is half a wavelength, so I expect T ≈ 2.5 s and λ ≈ 6 m. That makes the speed a bit over 2 m s−1, so 45 m should take roughly 20 s. The amplitude should be half of 0.60 m.

K · Knowns

I sketch the wave with the numbers on it. Both measurements are halves: high to low is half a period, and a crest to the next trough is half a wavelength. Writing that down stops me using them as whole ones.

  • crest-to-trough height = 0.60 m
  • highest → lowest: 1.25 s = ½T
  • crest → next trough: 3.0 m = ½λ
  • distance to the jetty: d = 45 m
Sketch of the lake waves. A buoy sits on a crest. The height from crest to trough is 0.60 m, and the buoy goes from highest to lowest in 1.25 s. The horizontal distance from a crest to the next trough is 3.0 m. The jetty is 45 m from the buoy.buoy: highest to lowest in 1.25 s0.60 m3.0 mjetty45 m
Knowns on the sketch: crest-to-trough height 0.60 m, highest to lowest in 1.25 s (half a period), crest to next trough 3.0 m (half a wavelength), 45 m to the jetty. Not to scale.

N · Not known

There are four answers: the amplitude in metres, the frequency in hertz, the speed in m s−1, and a time in seconds.

  • A = ? m, f = ? Hz, v = ? m s−1, t = ? s

O · Organise

The amplitude is half the crest-to-trough height. T and λ are double the halves I was given. Then f = 1/T and v = fλ. A crest moves at the wave speed, so the travel time is distance ÷ speed.

  • A = ½ × (crest-to-trough height)
  • T = 2 × 1.25 s, λ = 2 × 3.0 m
  • f = 1/T, v = fλ, t = d ÷ v

W · Work it out

I go line by line and keep the full time value until the end.

  • A = ½ × 0.60 = 0.30 m
  • T = 2 × 1.25 = 2.5 s → f = 1 ÷ 2.5 = 0.40 Hz
  • λ = 2 × 3.0 = 6.0 m → v = fλ = 0.40 × 6.0 = 2.4 m s−1
  • t = d ÷ v = 45 ÷ 2.4 = 18.75 s

U · Units

Metres, hertz, metres per second and seconds. The data have two significant figures, so I round the time to two.

  • A = 0.30 m, f = 0.40 Hz, v = 2.4 m s−1, t ≈ 19 s

D · Direction

The waves move towards the jetty, so I say that with the speed. As a sense check I find the speed the other way, from v = λ/T, and it should agree.

  • v = 2.4 m s−1 towards the jetty
  • check: λ/T = 6.0 ÷ 2.5 = 2.4 m s−1, the same

(a) A = 0.30 m. (b) f = 0.40 Hz and v = 2.4 m s−1. (c) About 19 s.

Check against the prediction. I predicted T ≈ 2.5 s and λ ≈ 6 m, a speed a bit over 2 m s−1 and roughly 20 s to the jetty. 2.4 m s−1 and 19 s fit.

Exam advice

What markers look for

Summaries in our own words. Read NESA's own feedback with the exam packs: 2024 · 2025.

Where this leads in Year 12

In Year 12 the wave model is applied to light: v = fλ becomes c = fλ for electromagnetic waves, and superposition explains the bright and dark bands of double-slit and diffraction-grating patterns (d sin θ = mλ). The wave model of light is then weighed against the particle model, where each photon has energy E = hf.

FAQ

What is the difference between transverse and longitudinal waves?
In a transverse wave the particles oscillate at right angles to the direction the energy travels (a wave on a rope). In a longitudinal wave they oscillate parallel to it, making compressions and rarefactions (sound). Both carry energy without carrying the medium along.
What is the difference between a displacement–time graph and a displacement–distance graph?
A displacement–distance graph is a snapshot of the whole wave at one instant: crest to crest is the wavelength. A displacement–time graph follows one particle over time: crest to crest is the period. Check the horizontal axis to tell them apart.
Is amplitude measured from crest to trough?
No. Amplitude is the maximum displacement from the rest position, so it is measured from the middle line to a crest. Crest to trough is twice the amplitude.
Does the frequency of a wave change when it enters a new medium?
No. The frequency is set by the source. The speed changes in the new medium, so the wavelength changes too (λ = v/f). This is what happens in refraction.

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