Freeze a whole wave in an instant, a single photograph of every particle at one moment, and you get a displacement–distance graph. Now the horizontal axis is distance, so this graph finally shows you the wavelength. Pair it with the period from a displacement–time graph and you can find the wave's speed. The catch: these two graphs look near-identical, so reading the axis label is the whole game.
A camera takes one flash photo of a long wave on a rope, capturing every point at the same instant.
Before reading: on this photo, what does the distance between two peaks tell you, the wavelength or the period? And could this single photo ever tell you the wave's speed on its own? Write your prediction.
The horizontal axis is distance, so this one does show wavelength
A displacement–distance graph is a single frozen photo of the entire wave at one instant, showing where every particle sits along the wave's path right now.
Instead of watching one particle over time, a displacement–distance graph captures all the particles at one frozen instant, plotting each one's displacement against its position. Because the horizontal axis is distance, the gap between two matching points (peak to next peak) is the wavelength $\lambda$. The vertical axis still gives the amplitude. What it cannot show is timing, a single snapshot has no period.
A displacement–distance graph is a snapshot of the whole wave at one instant. The distance for one cycle (peak to next peak) is the wavelength $\lambda$; the maximum displacement is the amplitude $A$. It cannot give the period.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? On a displacement–distance graph, what does the distance between two successive peaks represent?
Wavelength from one graph, period from the other → speed
We just saw that a distance graph gives wavelength but not period. That raises a question: how do we get the wave's speed if no single graph shows it? This card answers it → combine the two graphs with $v = f\lambda$.
Neither graph alone gives speed, but the distance graph hands you the wavelength and the time graph hands you the period, and $v = \lambda/T$ ties them together into a speed.
Read $\lambda$ from the displacement–distance graph and $T$ (or $f$) from the displacement–time graph of the same wave. Then the wave equation gives the speed:
$$v = f\lambda = \frac{\lambda}{T}$$
A displacement–distance graph of a wave shows successive peaks $0.60\ \text{m}$ apart. A displacement–time graph of the same wave shows a full cycle taking $0.40\ \text{s}$. Find the wave's speed.
To find wave speed from graphs: read $\lambda$ from a displacement–distance graph and $T$ (or $f$) from a displacement–time graph of the same wave, then use $v = \lambda/T = f\lambda$.
Add the highlighted method to your notes before the check below.
Quick check: True or false: a single displacement–distance graph, on its own, lets you calculate the wave's speed.
A displacement–distance graph is a snapshot of the whole medium; a displacement–time graph follows one particle
Use the simulation. Keep $v = 2.0\ \text{m s}^{-1}$ and raise the frequency from $1.0\ \text{Hz}$ to $2.0\ \text{Hz}$. What happens to the wavelength on the displacement–distance graph?
Use the simulation. True or false: moving the tracked particle from $x = 1\ \text{m}$ to $x = 2\ \text{m}$ changes the period on its displacement–time graph.
Pattern A, Compare then calculate
In your book make a two-column table: "displacement–time" vs "displacement–distance". For each, write (a) what the horizontal axis is, (b) which quantity you can read for the repeat (period or wavelength), and (c) what you cannot read. Then: a wave's distance graph gives $\lambda = 0.80\ \text{m}$ and its time graph gives $T = 0.20\ \text{s}$, calculate the speed and the frequency.
Odd one out: Three of these can be read directly from a displacement–distance graph. Which one is the odd one out?
Dot point: Analyse displacement–distance graphs for longitudinal and transverse waves
How a sound wave's snapshot still reads as a smooth curve
We just used a distance graph to find wavelength. That raises a question: how does a longitudinal wave, with no up-down shape, appear on a displacement–distance graph? This card answers it → the displacement curve is smooth; compression centres occur at zero crossings where its spatial gradient is negative.
A longitudinal displacement–distance graph records each particle's forward or backward shift from rest; its extrema are not compression centres.
For a sound wave we plot each particle's displacement along the direction of travel (forward positive, backward negative) against its equilibrium position. One wavelength is the distance between successive points in the same phase, such as successive displacement peaks. Compression occurs where neighboring particles converge: for small displacements this is where the spatial displacement gradient is negative, including a zero crossing with negative slope. Rarefaction occurs where the gradient is positive, including a zero crossing with positive slope. Displacement maxima and minima have zero gradient, so they are neither compression nor rarefaction centres.
On a longitudinal displacement–distance graph, read wavelength from one full spatial cycle. A negative displacement gradient marks compression; a positive gradient marks rarefaction. Thus their centres occur at zero crossings of opposite slope, not at displacement peaks or troughs.
Pause, copy the highlighted point into your book before the check below.
Last check: Decide if the statement is true or false.
On a displacement–distance graph, the wavelength is the distance between two successive peaks, for both transverse and longitudinal waves.
Fill the gaps: Complete the comparison of the two wave graphs, then press Check.
A displacement–distance graph is a of the whole wave at one instant, and its horizontal axis carries . The repeating length read from it is the . To find the wave speed you must combine it with the taken from a displacement–time graph.
Dot point: Analyse displacement–distance graphs for longitudinal and transverse waves
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.
AnalyseBand 5(4 marks) 1. A displacement–distance graph of a wave shows a wavelength of $1.2\ \text{m}$, and a displacement–time graph of the same wave shows a period of $0.30\ \text{s}$. Explain what each graph shows, and calculate the wave's speed.
1 mark: d–distance graph is a snapshot giving wavelength · 1 mark: d–time graph follows one particle giving period · 1 mark: correct use of $v = \lambda/T$ · 1 mark: $v = 4.0\ \text{m s}^{-1}$
ApplyBand 5(3 marks) 2. A displacement–distance graph of a wave shows crests at $x = 1.0\ \text{m}$ and $x = 3.5\ \text{m}$, with no crest between them. A separate displacement–time graph for the same wave gives a period of $0.50\ \text{s}$. Determine the wavelength and the wave speed, and state why one graph alone is not enough.
1 mark: wavelength $= 2.5\ \text{m}$ · 1 mark: $v = \lambda/T = 5.0\ \text{m s}^{-1}$ · 1 mark: explains that the distance graph gives $\lambda$ but not $T$, so speed needs both
Dot point: Analyse displacement–distance graphs for longitudinal and transverse waves
AnalyseBand 5(3 marks) 3. Two graphs of the same wave are drawn side by side and look almost identical in shape. Explain how you would decide which is the displacement–time graph and which is the displacement–distance graph, and state what each repeating interval means.
1 mark: read the horizontal axis label and units · 1 mark: seconds indicates time, so the repeat is the period · 1 mark: metres indicates distance, so the repeat is the wavelength
Dot point: Analyse displacement–distance graphs for longitudinal and transverse waves
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): The displacement–distance graph is a snapshot of the whole wave at one instant; its horizontal axis is distance, so the spacing between successive peaks is the wavelength, $\lambda = 1.2\ \text{m}$. The displacement–time graph follows a single particle over time; its horizontal axis is time, so the repeat interval is the period, $T = 0.30\ \text{s}$. Combining them, the wave speed is $v = \lambda/T = 1.2 / 0.30 = 4.0\ \text{m s}^{-1}$.
Q2 (3 marks): On a displacement–distance graph the horizontal axis is position, so the gap between adjacent crests is one wavelength: $\lambda = 3.5 - 1.0 = 2.5\ \text{m}$. The wave speed follows from $v = f\lambda$ with $f = 1/T$, which is $v = \lambda/T = 2.5/0.50 = 5.0\ \text{m s}^{-1}$. One graph alone is not enough because each carries only half the information: the displacement–distance graph gives the wavelength and amplitude but no timing, while the displacement–time graph gives the period and amplitude but no length along the wave. Speed is a distance divided by a time, so it needs one quantity from each.
Q3 (3 marks): The two graphs cannot be told apart by their shape, because both are smooth repeating curves for the same wave. The only reliable way to distinguish them is to read the horizontal axis label and its units. If the horizontal axis is time, measured in seconds, the graph tracks one particle as time passes and the repeating interval along that axis is the period $T$. If the horizontal axis is position, measured in metres, the graph is a snapshot of the whole medium at one instant and the repeating interval is the wavelength $\lambda$. The amplitude is the same on both, since it is read from the vertical axis in each case.
You can now read wavelength from a displacement–distance graph, tell it apart from a displacement–time graph by the axis label, and combine both to find wave speed. Next lesson applies wave speed and wavelength to a real-world First Nations technique for reading water depth.
↩ Back to focus-area hubAt the start you predicted whether a single photo could give the wave's speed. The distance between peaks on the photo is the wavelength, but a single snapshot has no timing, so it cannot give speed alone, you also need the period from a displacement–time graph. Two still pictures, combined through $v = f\lambda$, reveal the speed.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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