Edexcel IGCSE Physics · Spec 3.6-3.7
Wave Speed: Using the Equation
Worked examples using the wave speed equation.
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Wave Speed: Using the Equation
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Explained
The wave equation, and the conversions around it
Wave speed equals frequency multiplied by wavelength, v equals f times lambda. Speed in metres per second, frequency in hertz, wavelength in metres.
Frequency is the number of complete waves passing a point each second. Period is the time for one complete wave, so the two are reciprocals: period equals one over frequency, and frequency equals one over period.
Using it
A wave with a frequency of 15 Hz and a wavelength of 0.020 m has a speed of 15 times 0.020, which is 0.30 m/s.
To find a frequency, divide the speed by the wavelength. To find a wavelength, divide the speed by the frequency. As always, put the numbers into the equation as it stands and rearrange afterwards.
Units are where this goes wrong
Wavelengths are often given in centimetres, millimetres or nanometres, and every one of those must become metres before you multiply.
Divide by 100 for centimetres, by 1000 for millimetres, and by ten to the nine for nanometres. Write the conversion down as a separate line rather than doing it in your head, because it is the step examiners look for.
Frequency is sometimes given in kilohertz or megahertz, which need multiplying by a thousand or a million respectively.
What the mark scheme accepts and rejects
An Edexcel International GCSE Physics mark scheme for a three mark wave speed calculation awards one of those marks for the conversion of centimetres to metres, allowing a division by 100 seen anywhere in the working. It then states that not converting scores two marks at most.
So the conversion is not a preliminary to the physics. It is one third of the question, and it is awarded on its own even if what follows goes wrong.
The same series of mark schemes adds two more provisions worth knowing. A power of ten error costs one mark rather than all of them. And if no other marks are scored, selecting the correct formula still earns one, so writing v equals f times lambda before you attempt anything is never wasted.
They are strict about rounding, though. One accepts a range of values around the exact answer and then rejects a particular incorrectly rounded one, so rounding at the end rather than partway through matters.
The Doppler question
A moving source produces a pattern this equation explains neatly. As a source approaches, the wavefronts ahead of it bunch closer together, so the wavelength is smaller.
The speed of the wave has not changed, because that depends on the medium rather than on the source. So if the wavelength falls and the speed is constant, the frequency must rise, and a higher frequency sounds like a higher pitch.
A mark scheme for exactly this question credits four things: that the frequency is greater, that the wavefronts are closer together, that the wavelength decreases, and that the wave speed does not change. The last of those is the one candidates leave out, and it is the reason the rest follows.
Checking the answer
Sound in air is about 340 m/s and light in a vacuum is 3.0 times ten to the eight m/s. If you have calculated a speed for either and it is nowhere near, something has gone wrong, and it is almost always a unit conversion.
The two quantities also trade off. For a fixed speed, doubling the frequency halves the wavelength, so an answer where both went up in the same calculation is worth checking.
Spec 3.6-3.7
What you need to know
- Use wave speed equals frequency times wavelength
- Rearrange it to find frequency or wavelength
- Use period equals one over frequency
Active recall
Quick check
Answer each question before opening the answer.
What is the wave speed equation?
Wave speed = frequency × wavelength (v = f × λ).
A wave has frequency 50 Hz and wavelength 2 m. What is its speed?
v = f × λ = 50 × 2 = 100 m/s.
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