Wavelength, frequency and speed are not three unrelated numbers, they are locked together by one short equation, $v = f\lambda$. It works for ripples, sound, light and earthquakes alike. Master it and you can find any one of the three quantities from the other two, which is the single most-used skill in the whole Waves topic.
In one second, $3$ complete waves pass a post, and each wave is $2\ \text{m}$ long.
Before reading: how far does the front of the wave travel in that one second, and so what is the wave's speed? Write your reasoning.
Speed is just "length of each wave" times "how many per second"
The wave equation is plain common sense in symbols: if each wave is $\lambda$ metres long and $f$ of them pass every second, then $f\lambda$ metres of wave sweep past you each second, and that is the speed.
Speed is distance ÷ time. In one second, $f$ complete waves pass a point, and each is $\lambda$ long, so the total length passing per second is $f \times \lambda$. That total length per second is the wave's speed:
$$v = f\lambda$$
$v$ in m s⁻¹ · $f$ in Hz · $\lambda$ in m
Since $f = 1/T$, an equivalent form is $v = \lambda / T$. Rearranging gives $f = v/\lambda$ and $\lambda = v/f$, so any one quantity can be found from the other two.
The wave equation is $v = f\lambda$: wave speed = frequency × wavelength. With $f = 1/T$ it can also be written $v = \lambda/T$. Rearrange for any unknown: $f = v/\lambda$, $\lambda = v/f$.
Pause, copy the highlighted equation into your book before moving on.
Did you get this? True or false: in a fixed medium, doubling a wave's frequency doubles its speed.
Pick the rearrangement, convert units, substitute
We just built the equation $v = f\lambda$. That raises a question: how do we use it when we are given two quantities and asked for the third? This card answers it → rearrange, convert to SI units, then substitute.
Every wave-equation problem is the same three moves: decide which quantity is missing, rearrange $v = f\lambda$ for it, and substitute, being careful to put everything in metres, hertz and seconds first.
The only trap is units. A frequency in MHz must become Hz ($\times 10^6$); a wavelength in cm must become metres ($\div 100$). Do the conversion before substituting.
A radio station broadcasts at $f = 100\ \text{MHz}$. Radio waves travel at the speed of light, $v = 3.0 \times 10^8\ \text{m s}^{-1}$. Find the wavelength of these radio waves.
To solve $v = f\lambda$ problems: (1) identify the unknown; (2) convert all quantities to SI units (m, Hz, s); (3) rearrange and substitute. Use $\lambda = v/f$, $f = v/\lambda$ or $v = f\lambda$.
Add the highlighted method to your notes before the check below.
Quick check: A wave has frequency $5.0\ \text{Hz}$ and wavelength $0.40\ \text{m}$. What is its speed?
Pattern A, Calculate then check units
In your book solve all three, showing the rearrangement each time: (1) A sound wave of frequency $440\ \text{Hz}$ travels at $343\ \text{m s}^{-1}$ in air, find its wavelength. (2) A water wave of wavelength $0.5\ \text{m}$ travels at $1.5\ \text{m s}^{-1}$, find its frequency, then its period. (3) A wave of period $0.004\ \text{s}$ and wavelength $0.68\ \text{m}$, find its frequency, then its speed.
Odd one out: Three of these are correct rearrangements of the wave equation. Which one is the odd one out?
Dot point: Solve problems using v = fλ and T = 1/f
Why higher frequency means shorter wavelength in one medium
We just solved for individual quantities. That raises a question: what happens to the wavelength if we change the frequency without changing the medium? This card answers it → wavelength must shrink, because $v$ is fixed.
In one medium the wave speed is set by the medium, so $v$ stays constant, which means frequency and wavelength must trade off: push the frequency up and the wavelength comes down by the same factor.
Because $v = f\lambda$ and $v$ is fixed for a given medium, $f$ and $\lambda$ are inversely proportional. Double the frequency and the wavelength halves; quarter the frequency and the wavelength quadruples. This is why a high-pitched (high-frequency) sound has a shorter wavelength than a low-pitched one in the same air, and why high-energy gamma rays have far shorter wavelengths than radio waves even though both are light travelling at $c$.
In a fixed medium, $v$ is constant, so frequency and wavelength are inversely proportional ($f \propto 1/\lambda$): raising the frequency shortens the wavelength by the same factor, and vice versa.
Pause, copy the highlighted relationship into your book before the check below.
Last check: Decide if the statement is true or false.
In the same medium, a wave with double the frequency of another has half the wavelength.
Match them up: Match each quantity in the wave equation to its SI unit.
Dot point: Solve problems using v = fλ and T = 1/f
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. A sound wave in air has a frequency of $256\ \text{Hz}$ and a speed of $343\ \text{m s}^{-1}$. Calculate its wavelength and its period, showing your working and units.
1 mark: correct formula $\lambda = v/f$ · 1 mark: $\lambda \approx 1.34\ \text{m}$ · 1 mark: correct formula $T = 1/f$ · 1 mark: $T \approx 3.9 \times 10^{-3}\ \text{s}$
ApplyBand 4(3 marks) 2. A wave on a string has a period of $0.040\ \text{s}$ and a wavelength of $1.5\ \text{m}$. Calculate its frequency and its speed, showing your working and stating the units.
1 mark: $f = 1/T = 25\ \text{Hz}$ · 1 mark: correct substitution into $v = f\lambda$ · 1 mark: $v = 37.5\ \text{m s}^{-1}$ with units
Dot point: Solve problems using v = fλ and T = 1/f
AnalyseBand 5(3 marks) 3. A student is told a wave has a period of $0.025\ \text{s}$ and a wavelength of $0.80\ \text{m}$. They write $v = T\lambda = 0.025 \times 0.80 = 0.020\ \text{m s}^{-1}$. Identify the error, calculate the correct speed, and explain how the size of the answer alone should have warned them.
1 mark: identifies that the wave equation uses frequency, not period, so $T$ must first be converted with $f = 1/T$ · 1 mark: $f = 40\ \text{Hz}$ and $v = f\lambda = 32\ \text{m s}^{-1}$ · 1 mark: explains that $0.020\ \text{m s}^{-1}$ is implausibly slow for a wave of that wavelength and period, since one wavelength passes every $0.025\ \text{s}$
Dot point: Solve problems using v = fλ and T = 1/f
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): Rearranging the wave equation, $\lambda = v/f = 343 / 256 = 1.34\ \text{m}$ (3 s.f.). The period is the reciprocal of the frequency: $T = 1/f = 1/256 = 3.9 \times 10^{-3}\ \text{s}$. Both results use SI units (metres and seconds) and are quoted to 2–3 significant figures.
Q2 (3 marks): First convert the period to a frequency, since the wave equation is written in terms of frequency: $f = 1/T = 1/0.040 = 25\ \text{Hz}$. Then substitute into the wave equation: $v = f\lambda = 25 \times 1.5 = 37.5\ \text{m s}^{-1}$. The wave travels at $37.5\ \text{m s}^{-1}$, or $38\ \text{m s}^{-1}$ to two significant figures.
Q3 (3 marks): The error is that the wave equation is $v = f\lambda$, which uses the frequency, not the period. The student substituted $T$ where $f$ belongs. The period must first be converted: $f = 1/T = 1/0.025 = 40\ \text{Hz}$. Then $v = f\lambda = 40 \times 0.80 = \mathbf{32\ \text{m s}^{-1}}$. The size of the original answer should itself have been a warning. The wave covers one whole wavelength, $0.80\ \text{m}$, in one period, $0.025\ \text{s}$, so the speed must be roughly $0.80 \div 0.025$, which is tens of metres per second. A result of $0.020\ \text{m s}^{-1}$ is smaller than that by a factor of about $1600$, and it implies the wave would take forty seconds to advance a single wavelength. Checking whether an answer is the right order of magnitude catches substitution errors of this kind before they are carried into later work.
You can now use $v = f\lambda$ and $T = 1/f$ to find any wave quantity from the others, with units handled correctly. Next lesson reads these same quantities straight off displacement–time wave graphs.
↩ Back to focus-area hubAt the start you reasoned that $3$ waves of $2\ \text{m}$ passing each second means $6\ \text{m}$ travels per second, a speed of $6\ \text{m s}^{-1}$. That is exactly $v = f\lambda = 3 \times 2$. You discovered the wave equation yourself before it was even written down.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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