Module 3 of 9Lesson 1 of 24 · P20 Waves: terms, v = fλ and wave graphs
BGCSE Science Double Award · Physics · Form 5 · Section 3 Waves, light and sound
Lesson P20 · about 10 minutes

Waves: terms, v = fλ and wave graphs

Throw a stone into Gaborone Dam. Rings of ripples spread out across the water. A leaf floating nearby does not ride away on the ripples — it just bobs up and down. Something is travelling outwards, but it is not the water.

Prof KitsoWhat travels is energy. In 2024 most candidates did not notice that time was on the x-axis of a wave graph, and called the period a wavelength. Time on the axis gives the period; distance gives the wavelength.
leafbobs energy spreads out stone
Papers 1, 3 and 4Tested in 5 theory papers (2010–2022)🔬 Live ripple tank + wave graphs

Learn4 cards

What a wave does

A carries energy from one place to another without carrying matter with it.

The water particles only vibrate up and down about their rest position. The energy moves along the wave. That is why the leaf stays in the same place.

rest position leaf movesup and down only energy travels this way
Energy moves along; the water only moves up and down.

Crest, trough, wavelength, amplitude

A is a high point, a a low point.

The λ is the distance between two neighbouring crests (one complete wave). The A is the maximum of a particle from its rest position.

Both are read from a displacement–distance graph.

12345-4-2024distance / mdisplacement / cmwavelength λ A troughcrest
Here λ = 2.0 m and A = 2 cm.

Frequency and period

The f is the number of complete waves per unit time (per second). Unit: (Hz).

The T is the time for one complete wave. Read it from a displacement–time graph.

They are linked: f=1T{f=\dfrac{1}{T}}.

f=1Tf=\dfrac{1}{T}
0.10.20.30.40.5-4-2024time / sdisplacement / cmperiod T = 0.20 s
One complete wave takes 0.20 s, so f = 5 Hz.

The wave equation in a ripple tank

In a a vibrating dipper makes straight . Each wavefront joins points on the same crest.

The dipper makes f waves each second, each one λ long. So the waves travel f × λ metres each second: that is the wave speed v.

v=fλv=f\lambda
vibrating dipper λ v wavefronts
Gap between neighbouring wavefronts = λ.

See it live

Change the frequency of the dipper and the wavelength. Watch the speed v = fλ work out step by step. Then look at the two graphs side by side: on the distance graph measure λ and A; on the time graph measure the period T and check that f=1T{f=\dfrac{1}{T}}.

Ripple tank labChange f and λ, see v = fλ, and compare the displacement–distance and displacement–time graphs

In a real ripple tank: find f by counting the waves the dipper makes in 10 s and dividing by 10 (or freeze the waves with a stroboscope). Find λ by measuring across several wavefronts of the shadow on the paper with a ruler, then dividing by the number of gaps.

Worked example3 examples

Example 1Grade C
The graph shows how the displacement of a floating cork changes with time as water waves pass it.
0.20.40.60.81-6-4-20246time / sdisplacement / cm
The x-axis is time, so this graph gives the period.

(a) State the amplitude of the wave.

readhighest displacement from the rest position (0) = 6 cm
answeramplitude = 6 cm

(b) State the period of the wave.

readone complete wave runs from 0 s to 0.4 s
answerT=0.4 sT=0.4\text{ s}

(c) Calculate the frequency.

formulaf=1T{f=\dfrac{1}{T}}
substitutef=10.4{f=\dfrac{1}{0.4}}
answerf=2.5 Hzf=2.5\text{ Hz}

(d) Sketch, on the same axes, a wave with the same amplitude but double the frequency.

perioddouble the frequency → half the period: T=0.2 sT=0.2\text{ s}
plansame height (6 cm); one complete wave every 0.2 s, so 5 waves in 1.0 s
answertwice as many waves, same height:
0.20.40.60.81-6-4-20246time / sdisplacement / cm
Grey: the original wave. Orange: double the frequency.
Not 12 cmThe amplitude is measured from the rest position to a crest, not from a crest to a trough. Crest to trough here is 12 cm — twice the amplitude.
Example 2Grade B
In a ripple tank the dipper vibrates 10 times every second. A student measures 15 cm from the first wavefront to the sixth wavefront. Calculate the speed of the waves in m/s.
count gapsfirst to sixth wavefront = 5 gaps = 5 wavelengths
formulaλ=distancenumber of gaps{\lambda=\dfrac{\text{distance}}{\text{number of gaps}}}
substituteλ=15 cm5{\lambda=\dfrac{15\text{ cm}}{5}}
work outλ=3.0 cm\lambda=3.0\text{ cm}
in metresλ=0.030 m\lambda=0.030\text{ m}
formulav=fλv=f\lambda
substitutev=10×0.030v=10\times0.030
answerv=0.30 m/sv=0.30\text{ m/s}
Count the gapsSix wavefronts have only five gaps between them. Dividing by 6 gives λ = 2.5 cm and a wrong speed of 0.25 m/s.
Example 3Grade A
Graph 1 and Graph 2 describe the same wave on a long rope.
0.40.81.21.62-505distance / mdispl. / cmGraph 10.020.040.060.080.1-505time / sdispl. / cmGraph 2
Graph 1 has distance on the x-axis; Graph 2 has time.

(a) Use the correct graph to find the wavelength.

chooseGraph 1 (distance on the x-axis)
readone complete wave from 0 m to 0.8 m
answerλ=0.8 m\lambda=0.8\text{ m}

(b) Use the other graph to find the frequency.

readGraph 2: one complete wave from 0 s to 0.04 s, so T=0.04 sT=0.04\text{ s}
formulaf=1T{f=\dfrac{1}{T}}
substitutef=10.04{f=\dfrac{1}{0.04}}
answerf=25 Hzf=25\text{ Hz}

(c) Calculate the speed of the wave.

formulav=fλv=f\lambda
substitutev=25×0.8v=25\times0.8
answerv=20 m/sv=20\text{ m/s}

(d) A radio station broadcasts at 100 MHz. Radio waves travel at 3.0×108 m/s3.0\times10^{8}\text{ m/s}. Calculate the wavelength.

convert100 MHz=100×106 Hz100\text{ MHz}=100\times10^{6}\text{ Hz}
standard formf=1.0×108 Hzf=1.0\times10^{8}\text{ Hz}
formulaλ=vf{\lambda=\dfrac{v}{f}}
substituteλ=3.0×1081.0×108{\lambda=\dfrac{3.0\times10^{8}}{1.0\times10^{8}}}
answerλ=3.0 m\lambda=3.0\text{ m}
Prefixes firstM (mega) means ×106\times10^{6} (a million). In 2024 most candidates did not convert MHz to Hz before using v = fλ.

Misconception busterfrom the BEC examiner reports

On a displacement–time graph, the length of one complete wave is “the wavelength”.
Read the x-axis first. Time on the axis: one complete wave is the period T. Distance on the axis: it is the wavelength λ.

Examiner report: 2024 Paper 3, Q3(b)

Finding the frequency by measuring a wave diagram drawn against distance.
A distance diagram gives λ. Frequency needs time: use f=1T{f=\dfrac{1}{T}} from a time graph, or f=vλ{f=\dfrac{v}{\lambda}}.

Examiner report: 2024 Paper 1, Item 10

“Frequency is the number of waves per time taken.”
“Frequency is the number of complete waves per unit time (per second).”

Examiner report: General Science (1430) 2025 Paper 1, Q3(a)

Using 100 MHz as 100 Hz in v = fλ, or writing a wrong symbol for wavelength.
Convert prefixes first: 100 MHz = 1.0 × 10⁸ Hz. Use the standard symbols v, f and λ (lambda).

Examiner report: 2024 Paper 3, Q3(d)

Prof KitsoBefore you read any wave graph, put your finger on the x-axis label and say it out loud: “time — so this is the period” or “distance — so this is the wavelength”.

Quick check5 questions · Paper 1 style

1What is the wavelength of a wave?

  1. ✗ Not this one. Crest to the next trough is only half a wavelength.
  2. ✗ Not this one. Trough to crest is a height: twice the amplitude.
  3. ✓ Right. One complete wave: crest to the next crest.
  4. ✗ Not this one. That is the period, a time, not a distance.

2On a displacement–time graph, one complete wave takes 0.5 s. What is the frequency?

  1. ✓ Right. f=1T{f=\dfrac{1}{T}} gives 10.5=2 Hz{\dfrac{1}{0.5}=2\text{ Hz}}.
  2. ✗ Not this one. 0.5 is the period in seconds, not the frequency.
  3. ✗ Not this one. The decimal point has moved: 10.5\dfrac{1}{0.5} is 2, not 5.
  4. ✗ Not this one. This treats 0.5 s as 0.02 s; read the time axis carefully.

3A wave has a frequency of 50 Hz and a wavelength of 4.0 m. What is its speed?

  1. ✗ Not this one. This is fλ\dfrac{f}{\lambda}. The equation multiplies.
  2. ✗ Not this one. This is λf\dfrac{\lambda}{f}, upside down and still a division.
  3. ✗ Not this one. The two numbers were added.
  4. ✓ Right. v=fλv=f\lambda gives 50 × 4.0 = 200 m/s.

4In a ripple tank, the distance from the first wavefront to the fifth wavefront is 8.0 cm. What is the wavelength?

  1. ✗ Not this one. This divides by 5 wavefronts; there are only 4 gaps.
  2. ✓ Right. Five wavefronts have 4 gaps: 8.04=2.0 cm{\dfrac{8.0}{4}=2.0\text{ cm}}.
  3. ✗ Not this one. This is four wavelengths, not one.
  4. ✗ Not this one. This multiplies instead of dividing.

5The frequency of a wave is doubled. Its speed stays the same. What happens to the wavelength?

  1. ✗ Not this one. If f and λ both doubled, v = fλ would be four times bigger.
  2. ✗ Not this one. Then fλ would double and the speed would change.
  3. ✓ Right. v = fλ is fixed, so when f doubles λ must halve.
  4. ✗ Not this one. Nothing here is squared.

Exam cornerPaper 3 style · 7 marks

Fig. 3.1 shows the displacement–time graph of a water wave.

0.040.080.120.160.2-4-2024time / sdisplacement / cm
Fig. 3.1
(a)
Define amplitude. [1]
(b)
State the amplitude of the wave. [1]
(c)
Determine the frequency of the wave. [2]
(d)
The speed of the wave is 2.0 m/s. Calculate its wavelength. [2]
(e)
A student says that the wavelength can be read from Fig. 3.1 as 0.08. Explain why the student is wrong. [1]
Model answer

(a)

The maximum displacement of a particle from its rest position.

(b)

4 cm

(c)

readone complete wave from 0 s to 0.08 s, so T=0.08 sT=0.08\text{ s}
formulaf=1T{f=\dfrac{1}{T}}
substitutef=10.08{f=\dfrac{1}{0.08}}
answerf=12.5 Hzf=12.5\text{ Hz}

(d)

formulaλ=vf{\lambda=\dfrac{v}{f}}
substituteλ=2.012.5{\lambda=\dfrac{2.0}{12.5}}
answerλ=0.16 m\lambda=0.16\text{ m}

(e)

The x-axis of Fig. 3.1 is time, so 0.08 s is the period. Wavelength is a distance; it is read from a displacement–distance graph.
Marking key — how the examiner awards the marks
Mark pointMark
(a)the maximum displacement (of a particle) from the rest / equilibrium positionB1
(b)4 cm (unit needed)B1
(c)period T = 0.08 s seen, or 10.08\dfrac{1}{0.08}C1
12.5 Hz (accept 13 Hz)A1
(d)λ=2.012.5{\lambda=\dfrac{2.0}{12.5}} (their f)C1
0.16 m (16 cm)A1
(e)the x-axis shows time, so 0.08 s is the period (a time); wavelength is a distance, read from a displacement–distance graphB1

Total: 7 marks

Examiner tip. Write the formula, then the substitution, then the answer with its unit. If your frequency is wrong, the examiner can still give the C1 mark in the next part for using your value correctly.

Rememberone-page card to download

  • A wave transfers energy without transferring matter; particles vibrate about their rest position.
  • Wavelength λ: distance between neighbouring crests. Amplitude A: maximum displacement from the rest position.
  • Frequency f: number of complete waves per unit time (Hz). Period T: time for one complete wave (s).
  • Displacement–distance graph → λ and A. Displacement–time graph → T and A.
  • n wavefronts have (n − 1) gaps. Convert kHz and MHz to Hz first.

Formula card

v=fλv=f\lambdawave speed = frequency × wavelength
f=1T{f=\dfrac{1}{T}}frequency=1period{\text{frequency}=\dfrac{1}{\text{period}}}
λ=vf{\lambda=\dfrac{v}{f}}the same equation, rearranged
Download this card (PDF)

Science that earnsa skill you can sell

Low-cost ripple tanks for school labs

Many schools have no working ripple tank, so students only ever see wavefronts in a textbook. You can build a simple one: a clear tray of shallow water on a stand, an LED torch above it, white card below, and a small battery motor with an off-centre weight on a bar as the dipper.

Show a teacher it works: straight wavefronts on the card, closer together at higher frequency. Add a one-page worksheet for v = fλ.

Example costs
Clear plastic trayP60
LED torchP40
Small 3 V motor, switch and battery holderP45
Wooden stand, ruler, white card, glueP40
Costs for one tankP185
Selling price to a schoolP380
Profit per tankP195

Use batteries only — never mains electricity near water. Keep the motor and wires above the water, and use a plastic tray, not glass, so there are no sharp edges if it is dropped.

Example figures only — check local prices, and test every tank before you sell it.

Syllabus checklist Tap when you can do it

← P19 Heat transfer in everyday life: flask, car cooling, greenhouse effect, desert nightsNext: P21 Transverse and longitudinal waves →

Stadira Training Institute · Gaborone, Botswana · BGCSE Science Double Award (0569) Complete Tutor

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