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General properties of waves

Key idea: Describe wave motion, quantities, energy transfer and transverse and longitudinal waves.

  • SEC G2 Science Physics component 2027
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Learning objectives

  • Describe wave generation by vibrating sources, ropes and springs
  • Describe ripple-tank waves using wavefronts
  • Explain that waves transfer energy
  • Explain that wave energy transfer does not transfer matter
  • Use amplitude, frequency and wavelength to describe wave motion
  • Define and use wave speed and period and interpret wave graphs
  • Recall and apply wave speed = frequency × wavelength
  • Compare transverse and longitudinal waves and give examples

Wave motion and quantities

Wave motion, wavefronts and energy transfer

A vibrating source produces a disturbance. The disturbance travels and transfers energy, while particles of the medium oscillate about equilibrium: there is no net transfer of matter with the wave.

Imagine a coloured mark on a stretched rope. Shake one end up and down: the disturbance moves along the rope, but the mark moves up and down near its original position. Each part pulls on its neighbour, passing on energy. The travelling crest is a pattern of displacement, not one piece of rope travelling from the source to the far end. This ideal travelling-wave model describes oscillation without a steady flow of the medium.

Wavefronts from line and point sourcesTwo ripple-tank plans compare parallel straight wavefronts from a vibrating line source with expanding circular wavefronts from a point source. Propagation arrows are perpendicular to each wavefront.Line sourcevibrating barenergy propagationstraight wavefrontsPoint sourcedipperwavefronts
Scroll diagram horizontally to read all labels.
A line source produces approximately straight wavefronts; a point source produces circular wavefronts. Each wavefront joins points in the same phase, and the arrows show energy propagation perpendicular to the wavefronts.

These ripple-tank views are snapshots from above. Each drawn line joins points at the same phase, such as neighbouring parts of one crest. Successive crest lines are one wavelength apart. Energy travels across the lines, perpendicular to them: away from the bar for straight wavefronts and radially outwards from the point source for circular ones. The drawn arrows describe propagation, not the path of a water particle.

Try first: a crest passes a floating marker. Must the marker travel with it to the far end of the tank?

No. In the taught wave model the marker oscillates locally as successive crests pass. Energy reaches new positions while the medium has no net transport with the wave. Saying “nothing moves” is also wrong: the local oscillation is what passes the disturbance on.

Wave quantities, frequency and period

Wave quantities in transverse and longitudinal representationsA transverse wave profile labels amplitude and wavelength. A longitudinal particle model labels compressions, rarefactions, vibration direction, and wavelength.Transverse wave profileequilibriumAwavelength λLongitudinal wave particle modelcompressionrarefactionwavelength λparticle vibration
Amplitude is measured from the equilibrium line; wavelength is measured between consecutive points in phase. In a longitudinal wave, one wavelength is the spacing between neighbouring compressions.
Amplitude
Maximum displacement from equilibrium.
Wavelength, λ
Shortest distance between two points in phase.
Frequency, f
Number of complete oscillations each second, in hertz.
Period, T
Time for one complete oscillation, with T = 1/f.

Read the axes before choosing an interval

A displacement–distance graph is a snapshot of many points at one instant. Its horizontal axis tells you where a point is. A displacement–time graph follows one fixed point as time passes. The curves can look alike, but a horizontal interval has a different meaning on each one. On either graph, read amplitude vertically from zero displacement to a peak, not from trough to crest.

Rope displacement at one instant

A snapshot at t = 0 s; the wave travels towards increasing distance.

Scroll across the graph to read all labels.

A snapshot at t = 0 s; the wave travels towards increasing distance.A snapshot at t = 0 s; the wave travels towards increasing distance.
A snapshot at t = 0 s; the wave travels towards increasing distance.
Open full-size graph
View figure data
Values for Rope displacement at one instant
Distance along rope (m)Rope
00
0.1250.585271
0.251.14805
0.3751.666711
0.52.12132
0.6252.494409
0.752.771639
0.8752.942356
13
1.1252.942356
1.252.771639
1.3752.494409
1.52.12132
1.6251.666711
1.751.14805
1.8750.585271
20
2.125-0.585271
2.25-1.14805
2.375-1.666711
2.5-2.12132
2.625-2.494409
2.75-2.771639
2.875-2.942356
3-3
3.125-2.942356
3.25-2.771639
3.375-2.494409
3.5-2.12132
3.625-1.666711
3.75-1.14805
3.875-0.585271
40
4.1250.585271
4.251.14805
4.3751.666711
4.52.12132
4.6252.494409
4.752.771639
4.8752.942356
53
5.1252.942356
5.252.771639
5.3752.494409
5.52.12132
5.6251.666711
5.751.14805
5.8750.585271
60
6.125-0.585271
6.25-1.14805
6.375-1.666711
6.5-2.12132
6.625-2.494409
6.75-2.771639
6.875-2.942356
7-3
7.125-2.942356
7.25-2.771639
7.375-2.494409
7.5-2.12132
7.625-1.666711
7.75-1.14805
7.875-0.585271
80

In this snapshot, the crest at 1 m and the next crest at 5 m are at equivalent phases. Their separation is 5 − 1 = 4 m, so the wavelength is 4 m. The trough at 3 m is only halfway through that cycle: crest-to-trough gives half a wavelength. The peak displacement is 3 cm, so the amplitude is 3 cm even though the full crest-to-trough height is 6 cm.

Displacement of one marked rope point

The record of a fixed point during the same steady travelling wave.

Scroll across the graph to read all labels.

The record of a fixed point during the same steady travelling wave.The record of a fixed point during the same steady travelling wave.
The record of a fixed point during the same steady travelling wave.
Open full-size graph
View figure data
Values for Displacement of one marked rope point
Time (s)Marked point
00
0.01250.585271
0.0251.14805
0.03751.666711
0.052.12132
0.06252.494409
0.0752.771639
0.08752.942356
0.13
0.11252.942356
0.1252.771639
0.13752.494409
0.152.12132
0.16251.666711
0.1751.14805
0.18750.585271
0.20
0.2125-0.585271
0.225-1.14805
0.2375-1.666711
0.25-2.12132
0.2625-2.494409
0.275-2.771639
0.2875-2.942356
0.3-3
0.3125-2.942356
0.325-2.771639
0.3375-2.494409
0.35-2.12132
0.3625-1.666711
0.375-1.14805
0.3875-0.585271
0.40
0.41250.585271
0.4251.14805
0.43751.666711
0.452.12132
0.46252.494409
0.4752.771639
0.48752.942356
0.53
0.51252.942356
0.5252.771639
0.53752.494409
0.552.12132
0.56251.666711
0.5751.14805
0.58750.585271
0.60
0.6125-0.585271
0.625-1.14805
0.6375-1.666711
0.65-2.12132
0.6625-2.494409
0.675-2.771639
0.6875-2.942356
0.7-3
0.7125-2.942356
0.725-2.771639
0.7375-2.494409
0.75-2.12132
0.7625-1.666711
0.775-1.14805
0.7875-0.585271
0.80

Now follow the marked point. Successive maxima occur at 0.10 s and 0.50 s, so one complete oscillation takes 0.40 s. This is the period, not a wavelength: the horizontal unit is seconds. Frequency counts complete cycles per second, so f = 1/T = 1/0.40 = 2.5 Hz. A longer period means fewer cycles per second.

Try first: a learner calls the 0.10 s to 0.30 s interval “one period” and calculates 5 Hz. Which step fails first?

The chosen endpoints are a maximum and a minimum, not equivalent phases. That 0.20 s interval is half a cycle. Use maximum to next maximum, or a zero crossing to the next crossing in the same direction: T = 0.40 s and f = 2.5 Hz. Two successive zero crossings in opposite directions are also only half a cycle apart.

Another rope at one instant

Displacement observations at t = 0 s; the wave travels right.

Scroll across the graph to read all labels.

Displacement observations at t = 0 s; the wave travels right.Displacement observations at t = 0 s; the wave travels right.
Displacement observations at t = 0 s; the wave travels right.
Open full-size graph
View figure data
Values for Another rope at one instant
Distance along rope (m)Rope
00
0.018750.390181
0.03750.765367
0.056251.11114
0.0751.414214
0.093751.662939
0.11251.847759
0.131251.961571
0.152
0.168751.961571
0.18751.847759
0.206251.662939
0.2251.414214
0.243751.11114
0.26250.765367
0.281250.390181
0.30
0.31875-0.390181
0.3375-0.765367
0.35625-1.11114
0.375-1.414214
0.39375-1.662939
0.4125-1.847759
0.43125-1.961571
0.45-2
0.46875-1.961571
0.4875-1.847759
0.50625-1.662939
0.525-1.414214
0.54375-1.11114
0.5625-0.765367
0.58125-0.390181
0.60
0.618750.390181
0.63750.765367
0.656251.11114
0.6751.414214
0.693751.662939
0.71251.847759
0.731251.961571
0.752
0.768751.961571
0.78751.847759
0.806251.662939
0.8251.414214
0.843751.11114
0.86250.765367
0.881250.390181
0.90
0.91875-0.390181
0.9375-0.765367
0.95625-1.11114
0.975-1.414214
0.99375-1.662939
1.0125-1.847759
1.03125-1.961571
1.05-2
1.06875-1.961571
1.0875-1.847759
1.10625-1.662939
1.125-1.414214
1.14375-1.11114
1.1625-0.765367
1.18125-0.390181
1.20

Read this new snapshot before opening the feedback: give the amplitude and wavelength, then decide whether the period can be obtained from this graph alone.

Amplitude is 2 cm from equilibrium to a peak. Crests at 0.15 m and 0.75 m give a wavelength of 0.60 m. This single snapshot does not say how quickly the shape changes, so it cannot alone give a period or frequency. A time record, or speed information combined with the wavelength, is needed.

Wave speed and transverse/longitudinal models

Use

v = f λ

. Convert wavelength to metres before substitution and check that Hz × m gives m/s. For f = 5.0 Hz and λ = 0.40 m, the speed is 2.0 m/s.

In one period a crest advances one wavelength. Dividing that distance by the time gives v = λ/T; because f = 1/T, this is v = fλ. For the first pair of graphs, λ = 4 m and f = 2.5 Hz, giving v = 10 m/s. Do not multiply frequency by the 3 cm amplitude: amplitude describes local displacement, not the distance the pattern travels in one cycle.

Try first: the second rope snapshot is produced by a source oscillating at 5.0 Hz. Find the wave speed using the graph.

The wavelength read from the snapshot is 0.60 m, so v = 5.0 × 0.60 = 3.0 m/s. If using 60 cm, convert it to 0.60 m before reporting metres per second. The 2 cm amplitude is not needed.

Compare the two physical directions

In a transverse wave, vibration is perpendicular to travel. In a longitudinal wave, vibration is parallel to travel and produces compressions and rarefactions. Classify a wave from physical directions, not merely from a sinusoidal graph shape.

Try first: a rope wave travels right while a marked point oscillates vertically. Which motion determines the wave type?

Compare the vertical particle oscillation with the horizontal propagation: they are perpendicular, so the wave is transverse. Reversing propagation would leave that classification unchanged. In the longitudinal model above, the particles move back and forth along the propagation direction; the compression pattern travels while each particle stays near its equilibrium position.

Practise

Practise: General properties of waves

A text-first General Waves assessment with labelled controls and explicit motion, graph and equation data.

About 10 minutes

Practise

Questions are selected when you start. Use the feedback to decide what to practise next; this does not prove mastery.

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Practise

Practise after feedback: General properties of waves

A text-first General Waves assessment with labelled controls and explicit motion, graph and equation data.

About 10 minutes

Practise

Questions are selected when you start. Use the feedback to decide what to practise next; this does not prove mastery.

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Check what I know

Check what I know: General properties of waves

A text-first General Waves assessment with labelled controls and explicit motion, graph and equation data.

About 8 minutes

Check what I know

Answer 6 short questions. This starting check helps choose what to work on; it does not prove mastery.

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Check my progress

Check my progress: General properties of waves

A text-first General Waves assessment with labelled controls and explicit motion, graph and equation data.

About 10 minutes

Check my progress

Answer 7 questions. If accepted, this result can contribute to your course progress.

Recent attempts

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Check again

Check again: General properties of waves

A text-first General Waves assessment with labelled controls and explicit motion, graph and equation data.

About 10 minutes

Check again

Answer 6 questions. If accepted, this result can contribute to your course progress.

Recent attempts

History is stored only in this browser.

No completed attempts are saved yet.

Review

Review: General properties of waves

A text-first General Waves assessment with labelled controls and explicit motion, graph and equation data.

About 10 minutes

Review

Answer 7 questions. A scheduled review can contribute to your course progress only when it is due and the result is accepted.

Recent attempts

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Course and syllabus information
Course
SEC G2 Science Physics component
Edition
SEC G2 Science Physics component 2027