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IGCSE · Cambridge (CIE) · 0972

Calculate wave speed with consistent wavelength units

Use v = fλ and distinguish wave speed from a medium particle's motion.

General wave properties pathway · 0972 (9–1) · 3.1
Exam scope and assessment notes

Core and Extended · Physics 0972, 2026–2028, section 3.1. Core covers energy transfer, rope/spring/water-wave observations, wave features, speed, transverse and longitudinal models, reflection, refraction and ripple-tank diffraction. Extended compares wavelength effects at gaps and edges, including gap size. Written work is self-checked; human subject-teacher review, supervised practical competence and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.

Before you begin

  • Describe wavelength and frequency.
  • Convert centimetres, millimetres and kilohertz.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

What is 25 cm in metres?

What you will learn

  • Recall and use v = fλ.
  • Convert wavelengths and frequencies before calculating a propagation rate.
Original supplied regular-wave account uses f = 4.0 Hz and wavelength 0.15 m, so propagation speed is 0.60 m/s and pattern distance in 2.0 s is 1.20 m. The route diagram is qualitative, not a particle path, measured apparatus record or universal water-wave speed. Values refer to the same unchanged wave conditions.Same regular wave · supplied valuesFrequency: 4.0 cycles/sWavelength: 0.15 m per cyclev = fλ = 4.0 × 0.15Wave speed = 0.60 m/sPattern distance in 2.0 s: 1.20 mThe arrow is not a particle journeyAmplitude is a separate quantity
Original supplied regular-wave account uses f = 4.0 Hz and wavelength 0.15 m, so propagation speed is 0.60 m/s and pattern distance in 2.0 s is 1.20 m. The route diagram is qualitative, not a particle path, measured apparatus record or universal water-wave speed. Values refer to the same unchanged wave conditions.

Each cycle advances one complete repeat

For a regular travelling wave in the stated medium, one complete cycle advances the pattern by one wavelength. Frequency cycles each second therefore give wave speed v = fλ. With frequency in hertz and wavelength in metres, the resulting unit is m/s. This is a pattern propagation rate, not an individual particle's speed or the amplitude of its vibration.

A supplied water wave with f = 4.0 Hz and λ = 0.15 m has v = 0.60 m/s. In 2.0 s its pattern travels 1.20 m if the stated conditions keep the rate constant. Increasing amplitude is not a substitute for increasing wavelength in this equation; amplitude is a different wave feature.

Make units consistent before multiplying

Convert 25 cm to 0.25 m before multiplying by 8.0 Hz: v = 2.0 m/s. Convert 2.0 kHz to 2000 Hz and 170 mm to 0.170 m to get 340 m/s. A factor-of-one-hundred or one-thousand error is a unit mistake, not a new physical effect.

The equation connects the properties of the same wave under the same conditions. Do not multiply the frequency of one source by the wavelength of another unrelated pattern. In a real medium, propagation conditions may affect speed; use the supplied values and avoid assuming that every water or sound wave has a universal speed.

Pause and explain

A wave has 8.0 Hz frequency and 0.25 m wavelength. What is its speed?

Put the idea to work

Worked example

A supplied wave has frequency 8.0 Hz and wavelength 25 cm. Find wave speed.

Show the worked solution
  1. 25 cm = 0.25 m.
  2. Use v = fλ.
  3. v = 8.0 × 0.25 = 2.0 m/s.

Answer 2.0 m/s.

Try a topic test

Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Find v for f = 8.0 Hz and wavelength 25 cm.

Give me a hint

Convert the length to metres before applying the product.

Compare my reasoning
  1. λ = 0.25 m.
  2. v = fλ.
  3. v = 8.0 × 0.25 = 2.0 m/s.

2.0 m/s.

Look for these in your work

  • Use consistent units.
  • State a speed unit.
2 · Independent

Find speed for f = 2.0 kHz and λ = 170 mm.

Give me a hint

Convert both prefixes rather than only one.

Compare my reasoning
  1. f = 2000 Hz.
  2. λ = 0.170 m.
  3. v = 2000 × 0.170 = 340 m/s.

340 m/s.

Look for these in your work

  • Convert both values.
  • Multiply properties of the same wave.
3 · Transfer

A constant-speed pattern has f = 4.0 Hz and λ = 0.15 m. Find the pattern's distance in 2.0 s and explain what that does not measure.

Give me a hint

Find wave speed before using time; keep pattern and particle motion distinct.

Compare my reasoning
  1. v = 4.0 × 0.15 = 0.60 m/s.
  2. Pattern distance = 0.60 × 2.0 = 1.20 m.
  3. That distance is not travel of the same medium particle with the wave.

Pattern travels 1.20 m; this is not a particle's along-wave journey.

Look for these in your work

  • Use the supplied constant rate.
  • Identify the moving pattern.

Common mistakes

  • Convert wavelength before multiplying.
  • Use the same wave's frequency and wavelength.
  • Wave speed is not amplitude or particle vibration speed.
Recall without your notes

What can you explain now?

A wave has f = 2.0 kHz and λ = 170 mm. What is v?

Compare with the explanation

340 m/s.

2000 Hz × 0.170 m = 340 m/s.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Calculate v = fλ with prefixes converted and explain which motion the answer describes.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Physics (9-1)

Course scope and assessment

491 sequenced Physics lessons provide 1473 guided, independent and application problems. 58 of 58 syllabus sections have full linked teaching sequences; 0 sections remain pending. Three earlier starter overviews remain available but are not counted as full sequences. The map covers all six subject areas for 2026–2028 and keeps review and assessment limits visible. This is available content, not full programme completion, an exam grade or certified mastery. Written practice is self-checked. Supervised practical experience, cumulative assessment and human subject-teacher review remain pending. Paper method exercises do not demonstrate practical competence or complete either practical-assessment route. This is the separate 9–1 qualification 0972; Cambridge states that its 2026–2028 subject content is otherwise the same as 0625. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. The 2029 syllabus is separately published at https://www.cambridgeinternational.org/Images/764368-2029-syllabus.pdf and is not yet mapped here. This course supports exams in 2026, 2027 and 2028 only.

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