Point a sensor at one spot in a wave and record what happens there second by second, and you get a displacement–time graph. It looks like a wave, but its horizontal axis is time, not distance. Reading it correctly hands you the period and amplitude straight off the page, for transverse and longitudinal waves alike. The trap is mistaking it for a "snapshot" of the wave's shape, which it is not.
A motion sensor watches a single floating cork and plots its height against time as waves pass.
Before reading: on this graph, what does the distance between two peaks tell you, the wavelength or the period of the wave? Write your prediction and why.
The horizontal axis is time, so this is not a photo of the wave
A displacement–time graph is the story of just one particle: it shows where that single particle is, above or below rest, at each moment as the wave sweeps past it.
Imagine fixing your eyes on one cork as waves roll by, and plotting its height every instant. The result oscillates up and down, but the horizontal axis is time, so it is not a snapshot of the wave's shape across space. It tells you the history of one point. Because a longitudinal particle also moves back and forth in time, its displacement–time graph has the very same smooth oscillating shape.
A displacement–time graph tracks one particle over time, not the wave's shape in space. The same oscillating shape applies to both transverse and longitudinal waves.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? True or false: a displacement–time graph is a "photo" showing the wave's shape spread out across space.
Two numbers straight off the axes, then frequency for free
We just saw that the d–t graph is the history of one particle. That raises a question: exactly which wave quantities can we read from it? This card answers it → the period (from the time axis) and the amplitude (from the displacement axis).
From a displacement–time graph you read the period as the time for one full repeat and the amplitude as the biggest displacement from zero, and once you have the period, frequency comes free from $f = 1/T$.
The period $T$ is the time between two matching points one cycle apart (peak to next peak). The amplitude $A$ is the maximum height of the graph above (or below) the zero line. You cannot read wavelength from this graph, there is no distance axis. But frequency follows immediately: $f = 1/T$.
A displacement–time graph shows a particle reaching peaks of $+4\ \text{cm}$, with successive peaks at $t = 0.10\ \text{s}$ and $t = 0.30\ \text{s}$. Find the amplitude, period and frequency.
From a displacement–time graph: read the period $T$ as the time for one full cycle (peak to next peak), the amplitude $A$ as the maximum displacement, then find $f = 1/T$. Wavelength cannot be read from a d–t graph.
Add the highlighted method to your notes before the check below.
Quick check: A particle's displacement–time graph repeats every $0.25\ \text{s}$. What is the frequency of the wave?
Pattern A, Read then calculate
A displacement–time graph shows a particle oscillating with peaks of $+2.0\ \text{cm}$ and a full cycle taking $0.040\ \text{s}$. In your book: (1) state the amplitude in metres; (2) state the period; (3) calculate the frequency; (4) explain in one sentence why you cannot find the wavelength from this graph alone.
Two truths, one lie: Two truths and a lie about displacement–time graphs. Which statement is the lie?
Dot point: Analyse displacement–time graphs for longitudinal and transverse waves
Why a longitudinal particle gives a transverse-looking graph
We just read period and amplitude from a graph. That raises a question: does a longitudinal wave's d–t graph look different? This card answers it → no, because we plot displacement against time either way.
Even though a sound particle moves back and forth rather than up and down, plotting its displacement from rest against time still gives the same smooth oscillation, so a longitudinal d–t graph is read exactly like a transverse one.
A particle in a sound wave oscillates along the direction of travel. If we call "displacement to the right" positive and "to the left" negative, the displacement–time graph is the same smooth oscillating curve as for a transverse particle. The peaks correspond to moments of maximum displacement one way; the troughs to maximum displacement the other way. So period and amplitude are read in exactly the same manner for both wave types, the graph hides which type it is.
A longitudinal wave's displacement–time graph has the same oscillating shape as a transverse wave's, because both plot a particle's displacement against time. Period and amplitude are read identically for both.
Pause, copy the highlighted point into your book before the check below.
Last check: Decide if the statement is true or false.
A displacement–time graph for a particle in a longitudinal sound wave has the same smooth oscillating shape as one for a transverse wave.
Match them up: Match each feature of a displacement–time graph to what it tells you.
Dot point: Analyse displacement–time 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–time graph for a wave shows peaks of $\pm 3.0\ \text{cm}$ and a complete cycle every $0.050\ \text{s}$. State the amplitude, determine the period and frequency, and explain why the wavelength cannot be read from this graph.
1 mark: amplitude $= 3.0\ \text{cm} = 0.030\ \text{m}$ · 1 mark: period $T = 0.050\ \text{s}$ · 1 mark: $f = 1/T = 20\ \text{Hz}$ · 1 mark: wavelength needs a distance axis, which a d–t graph does not have
ApplyBand 4(3 marks) 2. A displacement–time graph for one particle in a wave shows crests at $t = 0.2\ \text{s}$ and $t = 0.8\ \text{s}$, and the particle reaches a maximum displacement of $4.0\ \text{cm}$ above the zero line. State the period, the frequency and the amplitude, explaining how each is read from the graph.
1 mark: period $= 0.6\ \text{s}$ from crest-to-crest on the time axis · 1 mark: frequency $= 1/0.6 \approx 1.7\ \text{Hz}$ · 1 mark: amplitude $= 4.0\ \text{cm}$, read from the zero line to the crest
Dot point: Analyse displacement–time graphs for longitudinal and transverse waves
AnalyseBand 5(3 marks) 3. A student is given a displacement–time graph for a longitudinal sound wave and says: "This can't be right, sound waves don't look like a sine curve." Explain why a displacement–time graph of a longitudinal wave can legitimately be a smooth oscillating curve.
1 mark: the graph plots one particle's displacement, not the shape of the medium · 1 mark: in a longitudinal wave the particle oscillates back and forth about a rest position · 1 mark: that back-and-forth displacement varies smoothly with time, giving the curve
Dot point: Analyse displacement–time 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 amplitude is the maximum displacement, $3.0\ \text{cm} = 0.030\ \text{m}$. The period is the time for one complete cycle, read off the time axis as $T = 0.050\ \text{s}$. The frequency is the reciprocal of the period: $f = 1/T = 1/0.050 = 20\ \text{Hz}$. The wavelength cannot be read from this graph because its horizontal axis is time, not distance, wavelength is a distance and requires a displacement–distance graph.
Q2 (3 marks): On a displacement–time graph the horizontal axis is time, so the gap between successive crests is the time for one complete oscillation, the period: $T = 0.8 - 0.2 = 0.6\ \text{s}$. The frequency is its reciprocal, $f = 1/T = 1/0.6 = 1.7\ \text{Hz}$ (2 s.f.). The amplitude is the maximum displacement of the particle from its rest position, read vertically from the zero line up to a crest, so $A = 4.0\ \text{cm}$. Note that no wavelength can be obtained from this graph, because the horizontal axis carries time rather than distance.
Q3 (3 marks): The student is confusing a graph with a picture. A displacement–time graph does not show the shape of the medium; it shows how the displacement of a single particle varies with time. In a longitudinal wave that particle oscillates back and forth along the direction of travel, and its displacement from the rest position is positive in one direction and negative in the other. Because the oscillation is smooth and repeating, plotting that displacement against time produces a smooth oscillating curve, exactly as it does for a transverse wave. The curve is not a picture of the compressions and rarefactions; it is a record of one particle's motion, and it correctly yields the period and amplitude of the sound wave.
You can now read amplitude and period straight off a displacement–time graph and turn the period into frequency, for transverse and longitudinal waves alike. Next lesson reads wavelength from the other kind of wave graph, displacement–distance.
↩ Back to focus-area hubAt the start you predicted whether the gap between two peaks gave wavelength or period. Because the horizontal axis is time, that gap is the period, not the wavelength. If you guessed wavelength, you were reading the graph as a snapshot in space, but a d–t graph follows one particle through time.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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