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Wave quantities, graphs and speed

Key idea: Wave speed equals frequency multiplied by wavelength. Wavelength is read from a graph of displacement against distance, and the period from a graph of displacement against time.

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  • SEC G3 Physics 2027
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Five quantities describe any wave: how big its vibrations are, how long each repeat is, how often it repeats, how long one repeat takes and how fast it travels. You can read them from two kinds of wave graph, and three of them are linked by the most useful equation in the topic, v = fλ.

Five wave quantities

Amplitude and wavelength in rope and particle modelsA spatial rope model has an amplitude arrow from its equilibrium line to a crest and a wavelength arrow between neighbouring crests. A longitudinal particle model shows alternating crowded and spread-out regions, a wavelength between compression centres, and a separate double arrow for local particle vibration parallel to rightward wave propagation.Transverse rope: one instantEquilibriumAmplitude AOne wavelength λ: neighbouring crestsWave travels rightLongitudinal particle model: one instantCompressionRarefactionOne wavelength λWave travels rightParticles vibrate to and fro
Amplitude and wavelength in rope and particle modelsA spatial rope model has an amplitude arrow from its equilibrium line to a crest and a wavelength arrow between neighbouring crests. A longitudinal particle model shows alternating crowded and spread-out regions, a wavelength between compression centres, and a separate double arrow for local particle vibration parallel to rightward wave propagation.Transverse rope: one instantEquilibriumAmplitude AOne wavelength λ: neighbouring crestsWave travels rightLongitudinal particle model: one instantCompressionRarefactionOne wavelength λWave travels rightParticles vibrate to and fro
Amplitude measures displacement from equilibrium. Wavelength is the spacing of neighbouring crests or compressions. The lower arrow shows vibration direction, not amplitude or a particle path to the next compression. Illustrative spatial models; not graphs of measured data.
QuantityMeaningSI unit
Amplitude, Athe maximum displacement of a point from its rest (equilibrium) positionmetre (m)
Wavelength, λthe shortest distance between two points at the same stage of vibration, such as two neighbouring crests or two neighbouring compressionsmetre (m)
Frequency, fthe number of complete waves produced each second, equal to the frequency of the sourcehertz (Hz)
Period, Tthe time taken to produce one complete wavesecond (s)
Wave speed, vthe distance travelled by a wavefront each secondmetre per second (m/s)

Two quantities are often confused:

  • Amplitude is measured from the rest position to a crest, not from a crest to a trough. The crest-to-trough height is twice the amplitude.
  • Wave speed is how fast the crests move along. It is not how fast a point of the medium moves as it vibrates.

Frequency counts waves per second, while period is seconds per wave, so each is the reciprocal of the other:

T = 1/f f = 1/T

Worked example 1

Waves per second and seconds per wave

Problem

A source makes 20 complete vibrations in 5.0 s at a steady rate. Find its frequency and period.

Show full solution

  1. Count vibrations per second

    Method

    Divide the number of vibrations by the time.

    Reason

    Frequency is the number of vibrations each second.

    Working

    Answer

    f = 20/(5.0 s) = 4.0 Hz

  2. Find the time for one vibration

    Method

    Divide the time by the number of vibrations.

    Reason

    The rate is steady, so each vibration takes the same time.

    Working

    Answer

    T = (5.0 s)/20 = 0.25 s

  3. State both quantities

    Working

    Answer

    The frequency is 4.0 Hz and the period is 0.25 s. Check: 1/4.0 = 0.25.

Spot the mistake 2

Is the period the same number as the frequency?

About 4 min

Learner response

A source vibrates at 50 Hz. A student says its period is 50 s. Calculate the correct period.

Diagnose, then calculate

Unit: s
Show full solution

  1. Use the reciprocal

    Method

    Divide 1 by the frequency.

    Reason

    50 waves each second means each wave takes one fiftieth of a second.

    Working

    Answer

    T = 1/f = 1/(50 Hz) = 0.020 s

Reading wave graphs

Two kinds of graph look alike but answer different questions. Always read the horizontal axis first.

  • A displacement–distance graph is a snapshot of the whole wave at one instant, like a photograph. Its horizontal axis is distance, so the repeat along it is the wavelength.
  • A displacement–time graph follows one point of the medium as time passes. Its horizontal axis is time, so the repeat along it is the period.

On both, the amplitude is read on the vertical axis, from zero to a crest.

Rope displacement at one instant

Displacement in cm against distance along the rope in m, from 0 m to 8 m, at one instant. The rope has a smooth repeating shape between +3 cm and −3 cm: crests at 1 m and 5 m, troughs at 3 m and 7 m, and zero displacement at 0, 2, 4, 6 and 8 m. The wave travels towards increasing distance.

Scroll across the figure to read all labels.

Displacement in cm against distance along the rope in m, from 0 m to 8 m, at one instant. The rope has a smooth repeating shape between +3 cm and −3 cm: crests at 1 m and 5 m, troughs at 3 m and 7 m, and zero displacement at 0, 2, 4, 6 and 8 m. The wave travels towards increasing distance.Displacement in cm against distance along the rope in m, from 0 m to 8 m, at one instant. The rope has a smooth repeating shape between +3 cm and −3 cm: crests at 1 m and 5 m, troughs at 3 m and 7 m, and zero displacement at 0, 2, 4, 6 and 8 m. The wave travels towards increasing distance.
A snapshot of one rope at t = 0 s. The wave travels towards increasing distance.
Open full-size graph
What to noticeConstructed sinusoidal model, not measured readings. Zero displacement is equilibrium.
View figure data
Figure data: Rope displacement at one instant
Distance along rope (m)Rope
0.0000.000000
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0.2501.148050
0.3751.666711
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1.0003.000000
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3.000-3.000000
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4.0000.000000
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4.7502.771639
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5.7501.148050
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6.0000.000000
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6.750-2.771639
6.875-2.942356
7.000-3.000000
7.125-2.942356
7.250-2.771639
7.375-2.494409
7.500-2.121320
7.625-1.666711
7.750-1.148050
7.875-0.585271
8.0000.000000

Worked example 3

Reading a snapshot of a rope

Problem

The displacement–distance graph above shows a rope at one instant. Its highest displacement is + 3 cm, and neighbouring crests are at 1 m and 5 m. Find the amplitude and the wavelength.

Show full solution

  1. Read the amplitude

    Method

    Measure from the rest position, zero, up to a crest.

    Reason

    Amplitude is the maximum displacement from rest, not the crest-to-trough height.

    Working

    Answer

    A = 3 cm

  2. Read the wavelength

    Method

    Subtract the positions of two neighbouring crests.

    Reason

    Neighbouring crests are one wavelength apart. A crest and the next trough are only half a wavelength apart.

    Working

    Answer

    λ = 5 m-1 m = 4 m

  3. State the readings

    Working

    Answer

    The amplitude is 3 cm and the wavelength is 4 m.

Displacement of the rope point 2.0 m along

Displacement in cm against time in s, from 0 s to 0.8 s, for the point 2.0 m along the same rope. The point starts at zero displacement moving up, reaches +3 cm at 0.10 s and 0.50 s, −3 cm at 0.30 s and 0.70 s, and passes through zero at 0, 0.20, 0.40, 0.60 and 0.80 s.

Scroll across the figure to read all labels.

Displacement in cm against time in s, from 0 s to 0.8 s, for the point 2.0 m along the same rope. The point starts at zero displacement moving up, reaches +3 cm at 0.10 s and 0.50 s, −3 cm at 0.30 s and 0.70 s, and passes through zero at 0, 0.20, 0.40, 0.60 and 0.80 s.Displacement in cm against time in s, from 0 s to 0.8 s, for the point 2.0 m along the same rope. The point starts at zero displacement moving up, reaches +3 cm at 0.10 s and 0.50 s, −3 cm at 0.30 s and 0.70 s, and passes through zero at 0, 0.20, 0.40, 0.60 and 0.80 s.
The same wave, recorded at one point of the rope 2.0 m from its end.
Open full-size graph
What to noticeConstructed sinusoidal model, not measured readings. Zero displacement is equilibrium.
View figure data
Figure data: Displacement of the rope point 2.0 m along
Time (s)Marked point
0.00000.000000
0.01250.585271
0.02501.148050
0.03751.666711
0.05002.121320
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0.07502.771639
0.08752.942356
0.10003.000000
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0.7500-2.121320
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0.80000.000000

Worked example 4

Following one point through time

Problem

The displacement–time graph above follows one point 2.0 m along the same rope. Its displacement is greatest at 0.10 s and again at 0.50 s. Find the period and the frequency.

Show full solution

  1. Measure one complete vibration

    Method

    Subtract the times of two neighbouring peaks.

    Reason

    Between them, the point has made exactly one complete vibration. The horizontal axis is now time.

    Working

    Answer

    T = 0.50 s-0.10 s = 0.40 s

  2. Convert to frequency

    Method

    Take the reciprocal of the period.

    Reason

    Frequency is the number of vibrations each second.

    Working

    Answer

    f = 1/(0.40 s) = 2.5 Hz

  3. State the readings

    Working

    Answer

    The period is 0.40 s and the frequency is 2.5 Hz.

Check your understanding 1 Read the two wave graphs

Use the two rope graphs above. Read their wavelength and period, then calculate frequency and wave speed.

Unit: m
Unit: s
Show the steps Optional working
Unit: s
Unit: Hz
Show the steps Optional working
Unit: m
Unit: s
Unit: Hz
Unit: m/s
Show answer

Both graphs show displacement in cm vertically. The snapshot has distance in m horizontally; the trace has time in s. Amplitude is 3 cm on both, from equilibrium to a crest.

Neighbouring spatial crests at 1 and 5 m give wavelength 4 m. Time peaks at 0.10 and 0.50 s give period 0.40 s, so frequency is 2.5 Hz and wave speed is 2.5 × 4 = 10 m/s. Show these intervals on their own axes. A snapshot has no elapsed time, while one point’s time trace gives no spatial spacing; the speed calculation uses both graphs’ stated same-wave relationship.

A spatial gradient is local slope of the rope, with displacement per distance, not distance travelled per time. Even a time-graph gradient describes local vibration velocity rather than wave propagation speed. At a crest, local vertical velocity can be zero while the travelling wave still propagates.

Check your understanding 2 Two zero crossings

On the displacement–time graph above, the point passes through zero displacement at 0 s and again at 0.20 s. Is the time between them one period?

Show answer

No. At 0 s the point is moving up through zero; at 0.20 s it is moving down through zero. That is only half a vibration, so 0.20 s is half a period. One full period runs between two crossings in the same direction, for example from 0 s to 0.40 s.

Wave speed: v = fλ

In one period, a source sends out one complete wave, and the front of the wave moves forward by one wavelength. So the wave travels a distance λ in a time T:

v = λ/T

Since 1/T = f, this becomes the wave equation:

v = fλ λ = v/f f = v/λ

Use metres, seconds and hertz: convert centimetres to metres and kilohertz to hertz before substituting. For waves travelling at a fixed speed, a higher frequency gives a shorter wavelength: more waves are squeezed into each second of travel.

Worked example 5

Finding a wavelength

Problem

A wave on a rope travels at 3.0 m s⁻¹ and has a frequency of 2.0 Hz. Find its wavelength.
Show full solution

  1. Rearrange the wave equation

    Method

    Make the wavelength the subject of v = fλ.

    Reason

    Divide both sides by f.

    Working

    Answer

    λ = v/f

  2. Substitute

    Method

    Divide the speed by the frequency.

    Reason

    Metres per second divided by waves per second leaves metres per wave.

    Working

    Answer

    λ = (3.0 m s⁻¹)/(2.0 Hz) = 1.5 m

Guided practice 6

Finding the speed of a sound wave

About 4 min

Problem

A sound wave has a frequency of 250 Hz and a wavelength of 1.4 m. Find its speed.

Choose the equation, then calculate

Unit: m/s

Hints

Hint 1

choose the equation

You know f and λ and want v.

Hint 2

check the units

Hertz times metres gives metres per second, so no conversion is needed.

Show full solution

  1. Apply the wave equation

    Method

    Multiply the frequency by the wavelength.

    Reason

    250 waves pass each second, and each is 1.4 m long.

    Working

    Answer

    v = fλ = 250 Hz × 1.4 m = 350 m s⁻¹

Exam-style question 1

Frequency of a water wave

4 marks

Examination question

A water wave travels at 0.80 m s⁻¹ and has a wavelength of 4.0 cm. Calculate its frequency. [4 marks]

Write your answer before viewing the mark scheme

Show the steps Optional working
Unit: m
Unit: Hz
Show solution step by step

  1. Convert the wavelength

    1 mark

    Method

    Convert centimetres to metres.

    Reason

    The speed is in metres per second.

    Working

    Answer

    4.0 cm = 0.040 m

  2. Calculate the frequency

    3 marks

    Method

    Rearrange v = fλ and divide the speed by the wavelength.

    Reason

    That gives the number of waves passing each second.

    Working

    Answer

    f = v/λ = (0.80 m s⁻¹)/(0.040 m) = 20 Hz

Another rope at one instant

Displacement in cm against distance along the rope in m, from 0 m to 1.2 m, at one instant. The rope has a smooth repeating shape between +2 cm and −2 cm: crests at 0.15 m and 0.75 m, troughs at 0.45 m and 1.05 m, and zero displacement at 0, 0.30, 0.60, 0.90 and 1.20 m.

Scroll across the figure to read all labels.

Displacement in cm against distance along the rope in m, from 0 m to 1.2 m, at one instant. The rope has a smooth repeating shape between +2 cm and −2 cm: crests at 0.15 m and 0.75 m, troughs at 0.45 m and 1.05 m, and zero displacement at 0, 0.30, 0.60, 0.90 and 1.20 m.Displacement in cm against distance along the rope in m, from 0 m to 1.2 m, at one instant. The rope has a smooth repeating shape between +2 cm and −2 cm: crests at 0.15 m and 0.75 m, troughs at 0.45 m and 1.05 m, and zero displacement at 0, 0.30, 0.60, 0.90 and 1.20 m.
A snapshot of a second rope at t = 0 s. The wave travels right.
Open full-size graph
What to noticeConstructed sinusoidal model, not measured readings. Zero displacement is equilibrium.
View figure data
Figure data: Another rope at one instant
Distance along rope (m)Rope
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Check your understanding 3 Speed from a snapshot

The snapshot above shows a second rope. Its source vibrates at 5.0 Hz. Read the wavelength from the graph and find the wave speed.

Show the steps Optional working
Unit: m
Unit: m/s
Show answer

Neighbouring crests are at 0.15 m and 0.75 m, so λ = 0.60 m. Then

v = fλ = 5.0 Hz × 0.60 m = 3.0 m s⁻¹

Use the horizontal spacing of the crests, not the amplitude, and take the frequency from the question: a snapshot cannot give it.

Check your understanding 4 Doubling the frequency

The frequency of a ripple-tank dipper is doubled, and the ripples are found to travel at the same speed as before. What happens to the wavelength? Explain.

Show answer

It halves. With v unchanged, λ = v/f, so doubling f halves λ. Twice as many waves are sent out each second over the same distance, so each one must be half as long.

Check your understanding 5 Change frequency at fixed speed

A rope wave has neighbouring crests at 0.20 and 0.80 m. At one fixed point, neighbouring peaks occur at 0.050 and 0.250 s. Calculate its speed. The frequency doubles while speed stays unchanged. Find the new period and wavelength.

Show the steps Optional working
Unit: m
Unit: s
Unit: m/s
Unit: s
Unit: m
Show answer

λ = 0.80-0.20 = 0.60 m and T = 0.250-0.050 = 0.200 s. Thus f = 1/T = 5.0 Hz and v = fλ = 3.0 m/s.

The new frequency is 10 Hz, so the new period is 0.100 s and the new wavelength is 3.0/10 = 0.30 m. Frequency does not determine amplitude; we would need new displacement information to find it.

Common mistakes

  • Measuring amplitude from crest to trough instead of from the rest position.
  • Measuring a wavelength from a crest to the next trough. That is half a wavelength.
  • Reading a “wavelength” from a displacement–time graph. Its repeat is the period.
  • Writing the period as the same number as the frequency. Use T = 1/f.
  • Forgetting to convert centimetres to metres, or kilohertz to hertz, before using v = fλ.
  • Saying a higher frequency makes a wave travel faster. At a fixed speed, it makes the wavelength shorter.

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