Calculate wave speed with consistent wavelength units
Use v = fλ and distinguish wave speed from a medium particle's motion.
Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.
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.
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.
Sources for this lesson
Pause and explain
A wave has 8.0 Hz frequency and 0.25 m wavelength. What is its speed?
Worked example
A supplied wave has frequency 8.0 Hz and wavelength 25 cm. Find wave speed.
Show the worked solution
- 25 cm = 0.25 m.
- Use v = fλ.
- v = 8.0 × 0.25 = 2.0 m/s.
Answer 2.0 m/s.
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Test this topic →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.
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
- λ = 0.25 m.
- v = fλ.
- 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.
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
- f = 2000 Hz.
- λ = 0.170 m.
- 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.
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
- v = 4.0 × 0.15 = 0.60 m/s.
- Pattern distance = 0.60 × 2.0 = 1.20 m.
- 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.
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.
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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.