2.1 Progressive waves
⤓ Presentation⤓ PowerPointPresentation: opens offline in any browser, simulations live. PowerPoint: editable.
Unit 2 Waves and Electricity, AS content, including the speed-of-sound core practical. Superposition and stationary waves are in 2.2; intensity, refraction and polarisation are in 2.3.
What this page covers
7 points
- What a progressive wave is: energy transferred, no net transfer of material
- Transverse and longitudinal waves, with examples
- Amplitude, wavelength, frequency, period and wave speed
- Using and justifying \(v = f\lambda\)
- Longitudinal waves in terms of molecule displacement and pressure
- Displacement–distance and displacement–time graphs, and phase along a wave
- Core practical: the speed of sound in air with a two-beam oscilloscope
In one minute
6 points
- A progressive wave transfers energy through a medium (or through space) without any net transfer of the medium itself.
- Transverse: particles oscillate at right angles to the direction of energy transfer. Longitudinal: particles oscillate parallel to it, forming compressions and rarefactions.
- Amplitude \(A\) is the maximum displacement from equilibrium; wavelength \(\lambda\) is the distance between neighbouring points oscillating in phase; \(f = 1/T\).
- In one period the wave moves one wavelength, so \(v = f\lambda\). The source sets \(f\); the medium sets \(v\).
- A displacement–distance graph is a snapshot and shows \(\lambda\); a displacement–time graph follows one point and shows \(T\).
- Points a distance \(\Delta x\) apart differ in phase by \(360^\circ\times\Delta x/\lambda\); points one wavelength apart are in phase.
Contents
12 sections
- What is a progressive wave, and how do transverse and longitudinal waves differ?
- What do amplitude, wavelength, frequency and period mean?
- Where does v = fλ come from, and how do you use it?
- How do you describe a longitudinal wave using displacement and pressure?
- How do you read displacement–distance and displacement–time graphs?
- How do you measure the speed of sound with an oscilloscope?
- Exam tips for 2.1
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is a progressive wave, and how do transverse and longitudinal waves differ?
Drop a stone into a pond and ripples spread out, yet a cork floating on the surface only bobs up and down as they pass; it is not carried to the bank. Something travels outwards, but it is not the water. That "something" is energy, and the travelling disturbance that carries it is a wave.
Each particle of the medium oscillates about a fixed equilibrium position and passes the disturbance on to its neighbour a little later in time. Electromagnetic waves need no medium at all: what oscillates there is an electric and a magnetic field.
- Transverse examples: waves on a stretched string or rope, all electromagnetic waves (light, radio, X-rays), the S-waves of an earthquake.
- Longitudinal examples: sound and ultrasound in any medium, a push–pull pulse along a slinky spring, the P-waves of an earthquake.
- Surface water waves are a mixture (water particles move in loops), but are usually modelled as transverse.
Defining a transverse wave as one where "the wave moves up and down". The particles (or fields) oscillate; the wave travels forwards. A full definition names both the direction of oscillation and the direction of energy transfer.
What do amplitude, wavelength, frequency and period mean?
| Quantity | Meaning | Unit |
|---|---|---|
| displacement \(x\) or \(d\) | distance and direction of a particle from its equilibrium position | m |
| amplitude \(A\) | the maximum displacement from the equilibrium position | m |
| wavelength \(\lambda\) | the distance between two neighbouring points oscillating in phase, e.g. crest to crest or compression to compression | m |
| period \(T\) | the time for one complete oscillation of a particle | s |
| frequency \(f\) | the number of complete oscillations per unit time | Hz = s⁻¹ |
| wave speed \(v\) | the distance travelled by the wave (by a crest, say) per unit time | m s⁻¹ |
A student shakes one end of a long rope. A photograph shows that the vertical distance from a crest to a trough is 0.36 m and that exactly four complete wavelengths fit along 6.0 m of rope. A marked point on the rope makes 15 complete oscillations in 10.0 s. Determine the amplitude, the wavelength, the frequency and the period of the wave.
Amplitude is half the crest-to-trough height: \(A = 0.36\div2\) → A = 0.18 m.
Wavelength = length ÷ number of wavelengths: \(\lambda = 6.0\div4\) → λ = 1.5 m.
Frequency = oscillations ÷ time: \(f = 15\div10.0\) → f = 1.5 Hz; period \(T = 1\div1.5\) → T = 0.667 s.
Where does v = fλ come from, and how do you use it?
\(v\) = wave speed (m s⁻¹), \(f\) = frequency (Hz), \(\lambda\) = wavelength (m).
Why it is true: in one period \(T\) the source makes one oscillation and the wave moves forward one wavelength, so \(v = \lambda/T = f\lambda\). The argument uses nothing about the kind of wave, so it holds for every kind of wave.
- The frequency is set by the source and does not change when the wave moves into a new medium.
- The speed is set by the medium (its stiffness, density, tension, temperature …). In one medium, all frequencies of sound travel at the same speed.
- So in a given medium, doubling \(f\) halves \(\lambda\); moving into a medium where \(v\) is larger makes \(\lambda\) larger in the same ratio.
- Watch the prefixes: kHz = 10³ Hz, MHz = 10⁶ Hz, GHz = 10⁹ Hz; mm, μm and nm for wavelengths.
A loudspeaker at the side of a swimming pool plays a steady note of frequency 1.20 kHz. The speed of sound is 340 m s⁻¹ in air and 1480 m s⁻¹ in water. Calculate the wavelength of the sound in the air and in the water.
Rearrange: \(\lambda = v/f\), with \(f = 1200\ \text{Hz}\).
In air: \(\lambda = 340\div1200\) → λ = 0.283 m. In water (same frequency, larger speed): \(\lambda = 1480\div1200\) → λ = 1.23 m.
The wavelength rises by the same factor as the speed, 4.35, because \(f\) is fixed by the loudspeaker.
A vibrating bar dips into a ripple tank at 12 Hz. On a photograph, eight successive bright lines (wave crests) span a distance of 0.105 m. (a) Calculate the speed of the ripples. (b) Calculate the time for a crest to cross the 0.45 m width of the tank.
(a) Eight crests enclose seven gaps: \(\lambda = 0.105\div7 = 0.0150\ \text{m} = 15.0\ \text{mm}\). Then \(v = f\lambda = 12\times0.0150\) → v = 0.180 m s⁻¹.
(b) The wave moves at constant speed: \(t = 0.45\div0.180\) → t = 2.5 s.
Dividing by 8 instead of 7 gives \(\lambda = 13.1\ \text{mm}\) and \(v = 0.158\ \text{m s}^{-1}\), which is 12.5% too small.
When you are told how many crests, compressions or bright lines fit in a distance, count the gaps: \(n\) crests in a row span \((n - 1)\) wavelengths. Sketching three or four crests settles it in seconds.
How do you describe a longitudinal wave using displacement and pressure?
In a sound wave each air molecule is displaced along the direction of travel. Take displacement to the right (the direction of travel) as positive.
- A compression is where molecules are closest together: the molecules just behind it are displaced forwards and those just in front are displaced backwards, so they crowd in. Pressure and density are above the undisturbed value.
- A rarefaction is where molecules are furthest apart; pressure and density are below the undisturbed value.
- At the centre of a compression or a rarefaction the displacement itself is zero; the molecules with the largest displacement sit halfway between them, where the pressure is normal.
- So the pressure variation is a quarter of a wavelength out of step with the displacement graph.
- The wavelength is the distance from one compression to the next (or rarefaction to rarefaction); a compression and the nearest rarefaction are \(\lambda/2\) apart.
A tuning fork produces a sound of frequency 680 Hz in air, where the speed of sound is 340 m s⁻¹. (a) Calculate the distance between neighbouring compressions. (b) State the distance from a compression to the nearest rarefaction. (c) Calculate the time between successive compressions reaching a listener's ear. (d) State how far an air molecule travels along with the wave in one second.
(a) Compression to compression is one wavelength: \(\lambda = 340\div680\) → 0.500 m.
(b) Compression to rarefaction is half a wavelength → 0.250 m.
(c) Successive compressions arrive one period apart: \(T = 1\div680\) → 1.47 ms.
(d) Molecules only oscillate about their equilibrium positions → no net distance; each vibrates 680 times a second about a fixed position.
Saying that at a compression the molecules have their maximum displacement. A compression is where neighbouring molecules have moved towards it from both sides; the molecule at its centre has zero displacement at that instant.
How do you read displacement–distance and displacement–time graphs?
| Displacement–distance graph | Displacement–time graph | |
|---|---|---|
| What it shows | a snapshot of every point along the wave at one instant | the history of one point over time |
| Horizontal repeat length | the wavelength \(\lambda\) | the period \(T\) |
| Peak height | the amplitude \(A\) | the amplitude \(A\) |
| For a longitudinal wave | positive displacement means moved forwards along the direction of travel, not "up" | the same convention, for one molecule |
- To find the wave speed you usually need both: \(\lambda\) from the distance graph and \(T\) (so \(f\)) from the time graph, then \(v = \lambda/T\).
- A snapshot taken a little later has the same shape, shifted a distance \(v\Delta t\) in the direction of travel: a quarter of a wavelength after a quarter of a period.
- To find which way a particle is moving, imagine the shifted snapshot and see whether that point's displacement has gone up or down.
Phase along a wave
Phase describes where a point is in its cycle of oscillation. One complete cycle is 360° or \(2\pi\) rad. Moving one wavelength along a progressive wave takes you through one whole cycle, so two points a distance \(\Delta x\) apart have a phase difference
- Points \(\lambda, 2\lambda, 3\lambda\) … apart are in phase; points \(\lambda/2, 3\lambda/2\) … apart are in antiphase (180°, \(\pi\) rad).
- In a progressive wave every point within one wavelength has a different phase, and all have the same amplitude.
- The link between phase difference and path difference for two separate sources is in 2.2.
The graphs above show a transverse wave on a long stretched spring travelling in the +x direction. The displacement–distance graph has crests 0.80 m apart; the displacement–time graph for the point at x = 0 has crests 25 ms apart. (a) Calculate the speed of the wave. (b) Calculate the phase difference between points P (x = 0.10 m) and Q (x = 0.40 m), in degrees and in radians. (c) State the direction in which the point at x = 0 is moving at t = 0. (d) Calculate the time for a crest to travel 2.0 m along the spring.
(a) \(\lambda\) from the distance graph, \(T\) from the time graph: \(f = 1\div0.025 = 40\ \text{Hz}\), \(v = 40\times0.80\) → v = 32 m s⁻¹.
(b) \(\Delta x = 0.30\ \text{m}\): \(\Delta\phi = 360^\circ\times0.30\div0.80\) → 135° = 3π/4 rad = 2.36 rad.
(c) Just after \(t = 0\) the time graph becomes positive → moving upwards (in the +d direction). Check with the snapshot: shifting the solid curve a little to the right (towards the dashed curve) raises the point at x = 0.
(d) A crest moves at the wave speed: \(t = 2.0\div32\) → t = 0.0625 s (62.5 ms).
Before reading any value off a wave graph, read the horizontal axis label. A repeat length in metres is a wavelength; a repeat length in seconds is a period. Reading 0.80 as a period, or 25 ms as a wavelength, gives a speed with no meaning.
How do you measure the speed of sound with an oscilloscope?
This is a core practical. The oscilloscope compares the phase of the sound arriving at a microphone with the phase of the signal driving the loudspeaker. Moving the microphone one wavelength further away brings the two traces back into step, so the distance moved between "in-phase" positions is the wavelength. With the frequency known, \(v = f\lambda\).
Method in outline
- Connect a signal generator to a loudspeaker and also to channel 1 of a two-beam oscilloscope; connect a microphone to channel 2.
- Set a frequency of a few kilohertz, so that the wavelength (around 10 cm) is short enough for several to fit along a metre rule; read \(f\) from the generator (or from the period on the screen).
- Place the microphone on the axis of the loudspeaker, next to a metre rule fixed to the bench. Adjust the time-base and channel gains so that both traces show a few cycles.
- Move the microphone until the two traces are in phase (peaks line up) and record its position on the rule.
- Move it slowly away, counting each time the traces come back into phase, through several (say 6–10) wavelengths. Record the final position.
- Wavelength = distance moved ÷ number of times the traces came back into phase (the number of wavelengths, not counting the starting position). Repeat the run, take a mean, then \(v = f\lambda\).
Getting a good result
- Many wavelengths: judging exactly where the traces line up is uncertain by several millimetres; spreading that over 6 wavelengths makes the percentage uncertainty about six times smaller.
- Antiphase is sharper to spot on some screens: in-phase and antiphase positions alternate every \(\lambda/2\), so you can count either, as long as you know which.
- Reduce reflections from the bench and walls (soft material on the bench, other objects moved away), which can distort the microphone trace.
- Alternatively, record each in-phase position, plot position against the number of wavelengths and take the gradient as \(\lambda\); an unknown zero (where inside the loudspeaker the sound "starts") then only shifts the intercept.
- Safety: keep the volume low, especially at higher frequencies.
The signal generator is set to 2.50 kHz. In three runs, a student starts with the traces in phase and moves the microphone away until they have come back into phase six times (six wavelengths). The distances moved are 82.3 cm, 81.8 cm and 82.9 cm. Determine the speed of sound and its percentage uncertainty, assuming the frequency is exact.
Mean distance \(= (82.3 + 81.8 + 82.9)\div3 = 82.3\ \text{cm}\); uncertainty (half the range) \(= (82.9 - 81.8)\div2 = \pm0.55\ \text{cm}\).
Six wavelengths in the mean distance: \(\lambda = 82.33\div6 = 13.72\ \text{cm} = 0.1372\ \text{m}\).
\(v = f\lambda = 2500\times0.1372\) → v = 343 m s⁻¹.
The percentage uncertainty in the distance carries straight through to \(v\): \(0.55\div82.3\times100 = 0.67\%\) → v = (343 ± 2) m s⁻¹.
Judging a single wavelength of 13.7 cm to ± 0.5 cm would give 3.6%, which is the reason for measuring over many wavelengths.
Taking the distance between an in-phase position and the next antiphase position as the wavelength. That distance is only \(\lambda/2\), and the speed comes out half its true value.
Exam tips for 2.1
For transverse and longitudinal waves, name what oscillates and compare its direction with the direction of energy transfer (or of wave travel): "perpendicular to" or "parallel to".
Convert kHz, MHz and GHz to Hz, and mm, cm or nm to m, before using \(v = f\lambda\). Then sense-check: sound in air is a few hundred m s⁻¹; light in a vacuum is 3.00 × 10⁸ m s⁻¹.
\(n\) crests (or compressions, or in-phase positions) in a row span \(n - 1\) wavelengths. "The microphone was moved through six wavelengths" is different from "six in-phase positions were seen".
Distance on the horizontal axis gives \(\lambda\); time gives \(T\). Write "from the graph, λ = …" or "T = …" as a line of working, so the reading is clear even if a later step slips.
Find \(\Delta x/\lambda\) first, then multiply by 360° or \(2\pi\). Give the unit (degrees or rad) every time; "135" alone is ambiguous.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| A wave carries the medium along with it. | A floating cork bobs in place as ripples pass under it. | A progressive wave transfers energy; each particle only oscillates about a fixed position. |
| Amplitude is the height from crest to trough. | Crest to trough is two amplitudes. | Amplitude is the maximum displacement from the equilibrium position. |
| A higher frequency sound travels faster. | In one medium all frequencies of sound arrive together, or music would be scrambled at a distance. | The medium sets \(v\); a higher \(f\) just means a shorter \(\lambda\). |
| The frequency changes when a wave enters a new medium. | The oscillations at the boundary must match on both sides, one for one. | Frequency is set by the source; \(v\) and \(\lambda\) change together. |
| At a compression the molecules have their largest displacement. | Molecules on either side are displaced towards the compression; the one at its centre is at its equilibrium position. | At a compression the displacement is zero and the pressure is greatest. |
| A displacement–time graph shows the wavelength. | Its horizontal axis is time, so its repeat length is a time. | A displacement–time graph gives the period; a displacement–distance graph gives the wavelength. |
How this topic connects
2.2 Superposition and stationary waves
Phase difference is the key to interference; two progressive waves travelling in opposite directions build a stationary wave, and the speed of a wave on a string sets its harmonics.
2.3 Intensity, refraction and polarisation
Refraction is a change of speed at constant frequency, exactly the \(v = f\lambda\) reasoning above; only transverse waves can be polarised, which is evidence that light is transverse.
2.5 Photons, and oscillations (Unit 5)
For electromagnetic waves \(c = f\lambda\), and the frequency fixes each photon's energy through \(E = hf\). Each particle of a wave carrying a pure tone performs simple harmonic motion; the sine-shaped graphs on this page are the same functions.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(v = f\lambda\) | wave speed (m s⁻¹) from frequency (Hz) and wavelength (m) | yes |
| \(f = 1/T\) | frequency from period (e.g. from a displacement–time graph or an oscilloscope) | no — learn it |
| \(v = \lambda/T\) | wave speed straight from the two graphs | no — derive it from \(v = f\lambda\) and \(f = 1/T\) |
| \(\Delta\phi = 360^\circ\times\Delta x/\lambda = 2\pi\Delta x/\lambda\) | phase difference between two points on a wave | no — learn it |
| \(\lambda\) = distance ÷ number of wavelengths | wavelength from several crests or in-phase positions | no — learn it (count the gaps) |
Check your understanding
1. A displacement–time graph for one point on a wave repeats every 4.0 ms. Calculate the frequency.
\(f = 1/T = 1\div4.0\times10^{-3} = 250\ \text{Hz}\).
2. A radio station broadcasts at 95.0 MHz. Calculate the wavelength of its radio waves.
\(\lambda = c/f = 3.00\times10^{8}\div95.0\times10^{6} = 3.16\ \text{m}\).
3. Describe the motion of the air molecules as a sound wave passes, and say where the pressure is greatest.
Each molecule oscillates back and forth about a fixed position, parallel to the direction the sound travels. Where neighbouring molecules have moved towards each other there is a compression, with pressure (and density) above normal; the pressure is greatest at the centre of a compression.
4. Two points on a progressive wave of wavelength 0.48 m are 0.12 m apart. Calculate their phase difference.
\(\Delta\phi = 360^\circ\times0.12\div0.48 = 90^\circ\) (\(\pi/2\) rad).
5. In the speed-of-sound practical, why is the microphone moved through several in-phase positions rather than one?
The position at which the traces line up can only be judged to within a few millimetres. Measuring the distance for many wavelengths and dividing makes this uncertainty a much smaller percentage of the measured distance.
6. Ultrasound of frequency 2.0 MHz travels through soft tissue at 1540 m s⁻¹. Calculate its wavelength.
\(\lambda = v/f = 1540\div2.0\times10^{6} = 7.7\times10^{-4}\ \text{m} = 0.77\ \text{mm}\).
Specification checklist
- (33) I can explain what amplitude, frequency, period, wavelength and wave speed mean, and use \(f = 1/T\).
- (34) I can use \(v = f\lambda\) and explain why it is true, including when a wave changes medium.
- (35) I can describe a longitudinal wave in terms of how the molecules are displaced and where the pressure is higher (compressions) or lower (rarefactions) than normal.
- (36) I can describe a transverse wave and give examples of transverse and longitudinal waves.
- (37) I can draw and read displacement–distance and displacement–time graphs for transverse and longitudinal waves, and work out the phase difference between two points on a wave.
- (38) I can describe how to measure the speed of sound in air with a signal generator, loudspeaker, microphone and two-beam oscilloscope, and calculate the result and its uncertainty.