What a wave is: energy on the move, transverse and longitudinal, and the words that describe every wave
By the end of this lesson you will be able to
- Define a wave as a transfer of energy without a transfer of matter, and describe what the particles of the medium actually do.
- Distinguish transverse from longitudinal waves by the direction of vibration, and classify light, sound, water, seismic and string waves.
- Label crest, trough, amplitude, wavelength, period, wavefront, ray and normal on wave diagrams, and state the unit of each.
- Calculate frequency from period with f = 1 ÷ T, and period from frequency, with units and significant figures.
- Interpret displacement–distance and displacement–time graphs, and never confuse the wavelength with the period.
- Process ripple-tank data to find how depth affects wave speed, and evaluate the method.
You met this before
- Sound is made by vibrating objects and is passed on from particle to particle, so it needs a medium; it cannot cross a vacuum (Year 3 Lesson 4).
- Light travels in straight lines at 3.0 × 10⁸ m/s in a vacuum and needs no medium, which is how sunlight reaches us across space (Year 3 Lesson 1).
- Speed = distance ÷ time, in m/s; sound in air travels at about 340 m/s.
- Frequency is the number of vibrations each second, measured in hertz (Hz); a higher frequency is heard as a higher pitch and a larger vibration as a louder sound (Year 3 Lesson 4).
- Energy is transferred from one store to another; a wave is one way to carry it from place to place (Year 1 Energy).
You see a lightning flash and hear the thunder 3.0 s later. How far away was the strike, and why can you ignore the time the light took?
Reveal
distance = speed × time
distance = 340 m/s × 3.0 s
distance = 1 020 m ≈ 1.0 km
Light time = 1 020 m ÷ 3.0 × 10⁸ m/s
Light time = 3.4 × 10⁻⁶ s
Three millionths of a second is far too short to notice, so the whole 3.0 s is the sound’s journey. That is the “three seconds per kilometre” rule.
Revise it in full: Year 3 · Physics · Unit 12 · Lesson 4 — Sound: how it travels · Year 3 · Physics · Unit 12 · Lesson 1 — Light: sources, rays and shadows
The puzzle
In a full football stadium, a “Mexican wave” races round the stands. From across the pitch it looks like a solid thing travelling at about 12 m/s — faster than a sprinter.
Yet not one spectator leaves their seat. Each person stands up, raises their arms and sits down again, a moment after the person beside them.
So what exactly is moving round the stadium? Not people. Not seats. Something is travelling, and it arrives on the far side — but it is a pattern, and the effort that makes it.
Every wave in this unit works like that: light from the Sun, sound from a speaker, a ripple on a pond, an earthquake crossing a continent. This lesson gives you the words and the pictures to describe all of them.
A cork floats on a still pond. You drop a stone in 2 m away and ripples spread out past the cork. Predict what the cork does as the ripples pass, and whether it ends up nearer the edge of the pond.
Compare your answer
The cork bobs up and down (with a small circling motion) as each ripple passes, then settles back where it was. It does not ride along with the ripples to the edge.
The ripples carry energy outwards — they could make a leaf bob on the far side — but the water, and the cork on it, only oscillates about the same place. That single idea separates a wave from a current.
The science — structured notes
1.1 A wave transfers energy, not matter
A wave is a disturbance that transfers energy from one place to another without transferring matter. Something must be disturbed: the particles of a material (the medium) in a mechanical wave, or electric and magnetic fields in light.
In a mechanical wave — sound, a ripple, a wave on a rope, an earthquake — each particle vibrates (oscillates) about a fixed rest position and passes the disturbance to its neighbour. The pattern moves on; the particles do not.
Watch the red point in the diagram. It goes up and down, and it is back where it started after every cycle. Meanwhile the shape slides one wavelength to the right. The energy put in by the vibrating source on the left is carried along the rope to whatever is at the other end.
| Family | What vibrates | Needs a medium? | Examples |
|---|---|---|---|
| Mechanical | particles of a solid, liquid or gas | yes — cannot cross a vacuum | sound, water waves, waves on strings and springs, seismic waves |
| Electromagnetic | electric and magnetic fields | no — travels through a vacuum at 3.00 × 10⁸ m/s | radio, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays |
“Ocean waves carry water from the middle of the ocean to the beach.”
Reveal answer
False. In deep water each bit of water moves in a small circle and returns to where it was; the wave carries energy, which is released when it breaks on the beach. (Surf does push water up the beach as the wave breaks and slows — but that is the end of the wave, not how it travels.)
1.2 Transverse and longitudinal
Waves come in two kinds, decided by one comparison: the direction the particles (or fields) vibrate compared with the direction the energy travels.
- Transverse: the vibrations are perpendicular (at 90°) to the direction of energy transfer. Examples: all electromagnetic waves including light; waves on a rope or guitar string; seismic S-waves; ripples on water (treated as transverse at MYP).
- Longitudinal: the vibrations are parallel to the direction of energy transfer. Examples: sound in any medium; seismic P-waves; a slinky pushed and pulled along its length.
A longitudinal wave has no crests and troughs. Instead it has compressions, where the particles are squeezed closer than normal, and rarefactions, where they are spread further apart. One wavelength is the distance from one compression to the next.
Classify each as transverse or longitudinal: (1) the vibration of a drum skin sending sound into the room; (2) the sound itself; (3) radio waves from a tower; (4) the S-wave of an earthquake; (5) ultrasound in the body.
Reveal answer
(1) The drum skin moves up and down while waves spread across it — transverse on the skin. (2) The sound it pushes into the air is longitudinal. (3) Radio: electromagnetic, so transverse. (4) S-wave (secondary, shear): transverse. (5) Ultrasound is sound: longitudinal.
The drum is a good trap: the source can vibrate one way while the wave it launches in another medium is of a different type.
“Light must be longitudinal, because it travels in straight lines, the same direction as the beam.”
Reveal answer
The direction of travel is the same for every wave; the test is the direction of vibration. In light, the electric and magnetic fields oscillate at 90° to the direction the light travels, so light is a transverse electromagnetic wave — the exact words the November 2021 physics paper wanted.
Evidence: polarising sunglasses cut glare by blocking one direction of vibration, which is only possible for a transverse wave.
1.3 The words that describe every wave
Two different graphs describe a wave, and the exam uses both. A displacement–distance graph is a snapshot of the whole wave at one instant. A displacement–time graph follows one particle as time passes. They look alike; they tell you different things.
| Term | Meaning | Symbol and unit |
|---|---|---|
| Rest position | where a particle sits when there is no wave; the centre line of the graph | — |
| Crest / trough | highest / lowest point of a transverse wave | — |
| Amplitude | maximum displacement from the rest position — rest line to crest, not crest to trough | A, in m (or cm) |
| Wavelength | distance between two neighbouring points in the same state of vibration, e.g. crest to crest, or compression to compression | λ (lambda), in m |
| Period | time for one complete oscillation, or for one wavelength to pass a point | T, in s |
| Frequency | number of complete oscillations (or waves passing a point) per second | f, in hertz, Hz (1 Hz = 1 s⁻¹) |
| Wave speed (velocity) | distance a crest (or any point of the pattern) travels per second, in a stated direction | v, in m/s |
Frequency and period are opposites of the same measurement: if 5 waves pass each second, each one takes a fifth of a second.
f = 1 ÷ T
T = 1 ÷ f
(1) A displacement–time graph for a point on a wave shows 3 complete cycles in 1.20 s. Calculate the period and the frequency. (2) A buoy has a frequency of 0.25 Hz. Calculate its period.
(1) T = total time ÷ number of cycles
T = 1.20 s ÷ 3
T = 0.40 s
f = 1 ÷ T
f = 1 ÷ 0.40 s
f = 2.5 Hz
(2) T = 1 ÷ f
T = 1 ÷ 0.25 Hz
T = 4.0 s
Always measure across several cycles and divide: reading one cycle off a graph (or timing one bob of a buoy) multiplies the reading error.
An ECG trace scrolls across a screen at 25 mm/s. The distance between neighbouring tall peaks is 20 mm. Calculate the period and frequency of the heartbeat, and the heart rate in beats per minute.
T = distance on screen ÷ scroll speed
T = 20 mm ÷ 25 mm/s
T = 0.80 s
f = 1 ÷ 0.80 s
f = 1.25 Hz
heart rate = 1.25 beats/s × 60 s/min
heart rate = 75 beats per minute
The distance on the screen stands for time, because the paper (or trace) moves at a steady speed. Converting a horizontal length into time is exactly how you read the period off an oscilloscope in Lesson 2.
A student counts 30 ripples passing a stick in 12.0 s. Calculate the frequency and the period, with units.
Reveal answer
f = number of waves ÷ time
f = 30 ÷ 12.0 s
f = 2.5 Hz
T = 1 ÷ f
T = 1 ÷ 2.5 Hz
T = 0.40 s
Check: 30 waves × 0.40 s gives 12.0 s.
A student marks the amplitude of a wave from the top of a crest to the bottom of a trough, 6.0 cm, and writes “A = 6.0 cm”. What is wrong, and what is the amplitude?
Reveal answer
Amplitude is measured from the rest position to a crest (or to a trough). Crest to trough is twice the amplitude, so A = 6.0 cm ÷ 2 = 3.0 cm. This is one of the commonest lost marks on a wave diagram.
1.4 Wavefronts, rays and the normal
Seen from above, water waves in a ripple tank form lines. A wavefront is a line joining points that are in step — for example, a line along a crest. Neighbouring wavefronts are one wavelength apart.
A ray is an arrow showing the direction the wave (and its energy) travels. Rays are always at 90° to the wavefronts.
Where a ray meets a surface, you draw the normal: a line at 90° to the surface at that point. Every angle in Lessons 3 and 4 — incidence, reflection, refraction — is measured from the normal, never from the surface.
A lamp above the tank makes the waves visible: each crest acts like a weak lens and focuses light into a bright line on the paper below, each trough spreads it into a dark one. So the pattern on the paper is a picture of the wavefronts, and a ruler on the paper measures the wavelength.
A ray meets a flat barrier making an angle of 70° with the surface of the barrier. State the angle of incidence and explain why.
Reveal answer
The angle of incidence is measured between the ray and the normal, which is at 90° to the surface. So:
i = 90° − 70°
i = 20°
Measuring from the surface is the classic mistake; the laws of reflection and refraction only work with angles from the normal.
1.5 Energy, amplitude and speed
The energy a wave carries grows with its amplitude: a larger vibration means more energy at the source and more delivered at the far end. For mechanical waves, doubling the amplitude roughly quadruples the energy carried. A loud sound has a larger amplitude than a quiet one of the same pitch; a storm swell has a larger amplitude than a ripple.
The speed of a wave is set by the medium, not by the source. Sound travels at the same speed whether it is loud or soft, high or low. Water waves depend on depth. In water that is shallow compared with the wavelength:
v = √(g × d)
where g is 9.8 m/s² and d the depth in m. You will test this in the ripple tank below.
On 26 December 2004 an earthquake off Sumatra launched a tsunami across the Indian Ocean. Taking the average depth of the ocean along its path as 4 000 m, calculate the wave speed, and estimate the travel time to the coast of Somalia, about 5 000 km away.
v = √(g × d)
v = √(9.8 m/s² × 4 000 m)
v = √(39 200 m²/s²)
v = 198 m/s (about 710 km/h, the speed of an airliner)
t = distance ÷ speed
t = 5.0 × 10⁶ m ÷ 198 m/s
t = 2.5 × 10⁴ s
t = 2.5 × 10⁴ s ÷ 3 600 s/h
t = 7.0 hours
The waves reached Somalia roughly seven hours after the earthquake and still killed people there. With a warning system and seven hours, almost everyone could have reached high ground. The Indian Ocean had no such system in 2004; it has one now.
Near the coast the tsunami moves into water 10 m deep. Calculate its new speed, and explain why its height grows so much.
Reveal answer
v = √(9.8 m/s² × 10 m)
v = 9.9 m/s
The front slows to a twentieth of its ocean speed while the water behind is still arriving fast, so the wave bunches up: the wavelength shortens (v = fλ with f unchanged, Lesson 2) and the same energy is squeezed into a shorter stretch of water — so the amplitude rises, from under a metre in mid-ocean to several metres or more at the shore.
An earthquake sends out a P-wave (longitudinal, 6.0 km/s) and an S-wave (transverse, 3.5 km/s). A seismometer records the S-wave 60 s after the P-wave. How far away was the earthquake?
Reveal answer
Let the distance be d (in km). Both waves travel d, the S-wave taking 60 s longer:
d ÷ 3.5 − d ÷ 6.0 = 60
d × (0.2857 − 0.1667) = 60
d × 0.1190 = 60
d = 504 km ≈ 500 km
Three stations each give a distance; where the three circles cross is the epicentre. The same timing idea drives Lesson 2.
Try it — a full investigation: water depth and wave speed in a ripple tank
The lab is a school ripple tank. A straight bar dips once and sends a single pulse across the water; a phone films its shadow in slow motion as it crosses two lines 0.40 m apart.
You set the depth (the independent variable, 2–20 mm). You can also change the liquid (tap water or dense salt solution) and use a depth gauge that has a hidden fault. The lab calculates the speed from the time; your job is the science.
Common mistakes to avoid
- “The wave carries the water (or air) along.” Particles oscillate about fixed positions; energy moves on.
- Amplitude measured crest to trough. Measure from the rest position to a crest: half of crest-to-trough.
- Reading a wavelength off a time axis. On a displacement–time graph crest to crest is the period T.
- Confusing period and frequency. T in seconds, f in hertz, and f = 1 ÷ T; a bigger period means a smaller frequency.
- “Sound is transverse because the speaker cone moves back and forth.” The cone moves along the direction the sound travels — that is what makes sound longitudinal.
- Angles measured from the surface. Draw the normal first; measure from it.
- “A louder sound travels faster.” Speed depends on the medium; amplitude sets the energy, not the speed.
- Hz written as “hz” or “HZ”. The unit is Hz, named after Heinrich Hertz; s⁻¹ is also accepted.
Study smart: how this is examined
Study routine. Today: draw both wave graphs from memory and label every term with its unit. In two days: explain aloud, without notes, the difference between transverse and longitudinal waves using a rope and a slinky, then do the ECG EXAMPLE. In a week: run the ripple tank as a full investigation, plot v² against d, and take the quiz.
How we found out — listening to earthquakes to see inside the Earth
No one has drilled more than about 12 km into the Earth, yet we know it has a solid iron inner core, a liquid outer core and a solid rocky mantle. That knowledge came almost entirely from waves.
In 1906 the Irish geologist Richard Dixon Oldham studied records of large earthquakes from seismometers around the world. He noticed that P-waves arriving at stations on the far side of the Earth came late, as if slowed and bent by a distinct central region. He argued for a core.
In 1914 the German-American seismologist Beno Gutenberg calculated the depth of the boundary between mantle and core at about 2 900 km — a figure still used today.
In 1926 the British scientist Harold Jeffreys concluded that the outer core must be liquid. The decisive evidence is that transverse S-waves are never detected passing straight through the core: a liquid cannot support the sideways shearing of a transverse wave, so it stops them, leaving an “S-wave shadow” on the far side of the planet.
Then, in 1936, the Danish seismologist Inge Lehmann noticed faint P-waves arriving in the shadow zone where none were expected. She explained them in a paper with one of the shortest titles in science, “P′”, by proposing a solid inner core that reflected them. Later measurements, including those after underground nuclear tests in the 1960s, confirmed it.
The idea that changed minds was simple: longitudinal waves pass through solids and liquids, transverse waves only through solids — and the Earth was quietly showing which was which.
Explain why S-waves cannot travel through the liquid outer core, while P-waves can.
Reveal answer
S-waves are transverse: the rock is displaced sideways, at 90° to the direction of travel, and a neighbouring layer is dragged sideways with it. A solid resists this shearing, so the disturbance is passed on. A liquid’s particles simply slide past each other, so there is nothing to pull the next layer sideways and the wave dies out.
P-waves are longitudinal: they squeeze and stretch the material along the direction of travel, and liquids resist compression, so P-waves pass through (slowed and bent at the boundary).
Where it is used now
- Earthquakes in Botswana. On 3 April 2017 a magnitude 6.5 earthquake struck the Central District near Moiyabana, one of the largest ever recorded in the country; its waves were felt as far away as Johannesburg. Seismometers in the region timed its P- and S-waves to locate it.
- Tsunami warnings. Pressure sensors on the deep-ocean floor detect the tiny rise of a passing tsunami and radio it to warning centres; the √(g d) speed gives the arrival time along every coast.
- Wave energy. Machines at sea convert the up-and-down motion of swell into electricity — a direct use of energy carried by waves without the water travelling.
- Medicine. An ECG trace is a displacement–time graph of the heart’s electrical activity; reading its period and frequency is a daily hospital skill.
- Music. A guitar string carries transverse waves; the sound it makes in the air is longitudinal. Tuning is adjusting the frequency, in hertz, to match a standard such as 440 Hz for the note A.
Science now — and the scientist you need to be
What is happening in 2026
The phone in your pocket can feel an earthquake. Android phones contain accelerometers that detect shaking, and Google’s Android Earthquake Alerts System, running since 2020, combines signals from many phones to spot the faster P-wave and warn people a few seconds before the more damaging S-waves arrive. A 2025 study in Science reported how it performed across many countries.
Meanwhile, the fibre-optic cables that carry internet traffic (Lesson 4) are being turned into seismometers. A technique called distributed acoustic sensing fires laser pulses along a fibre and times the tiny reflections, so a single cable can act like thousands of sensors along its length. Researchers use it to record earthquakes, traffic and even whale calls on the seabed.
Careers here range from seismology and ocean engineering to acoustics, telecommunications and medical physics. All of them start with the vocabulary of this lesson and the discipline of reading a graph’s axes before reading its shape.
The mindshift
What you need to be able to do
Say what is moving and what is not. Classify a wave by comparing two directions. Label a graph after reading its axes. Measure many cycles and divide. Linearise a square-root law by squaring. And before trusting a line through the origin, ask whether your zero was really zero.
See an example
Five cycles took 312 frames at 240 frames per second.
time for five cycles = 312 frames ÷ 240 frames/s
time for five cycles = 1.30 s
T = 1.30 s ÷ 5
T = 0.26 s
f = 1 ÷ T
f = 1 ÷ 0.26 s
f = 3.8 Hz
A careful assistant says sound in air is longitudinal but may add that sound in solids can also travel as transverse (shear) waves — true, and a good reason to check the medium in any question.
Key terms
| Wave | A disturbance that transfers energy from place to place without transferring matter. |
| Medium | The material a mechanical wave travels through; electromagnetic waves need none. |
| Transverse wave | Vibrations perpendicular to the direction of energy transfer (light, waves on strings, S-waves). |
| Longitudinal wave | Vibrations parallel to the direction of energy transfer (sound, P-waves). |
| Compression / rarefaction | Regions of a longitudinal wave where particles are closer together / further apart than normal. |
| Amplitude, A | Maximum displacement from the rest position, in m. |
| Wavelength, λ | Distance between neighbouring points in step, such as crest to crest, in m. |
| Period, T | Time for one complete oscillation, in s. |
| Frequency, f | Oscillations per second, in Hz; f = 1 ÷ T. |
| Wavefront / ray / normal | Line joining points in step / arrow showing direction of travel at 90° to wavefronts / line at 90° to a surface where a ray meets it. |
Think like a scientist
Criterion C A student investigated how the depth of water in a ripple tank affects the speed of a straight pulse. She timed the pulse between two lines 0.40 m apart using slow-motion video. Her results are below.
| Depth / mm | Time trial 1 / s | Time trial 2 / s | Time trial 3 / s |
|---|---|---|---|
| 4 | 2.04 | 1.99 | 2.01 |
| 8 | 1.44 | 1.42 | 1.45 |
| 12 | 1.17 | 1.15 | 1.18 |
| 16 | 1.01 | 1.12 | 1.01 |
| 20 | 0.90 | 0.91 | 0.89 |
(a) Identify the anomalous result, suggest a specific cause, and calculate the mean time at each depth, excluding it. [3]
(b) Calculate the speed and the speed squared at each depth, and present them in a processed-data table with units in the headings. [4]
(c) Describe the graph you would draw to test whether v² is directly proportional to depth: axes, units, scale and line. [3]
(d) Calculate the gradient of the line, with its unit, and compare it with a known constant. [3]
(e) State whether the data support the hypothesis “doubling the depth multiplies the speed by √2”, quoting values. [2]
(f) Explain one random error and one systematic error in this method, and suggest one improvement. [3]
Model answer
(a) 1.12 s at 16 mm (trial 2) is anomalous: 0.11 s above the other two, which agree [1]. A likely cause: the dipper was pushed in more slowly, making a broad pulse whose crossing frame was judged late; or the video frames were counted from the wrong line [1].
Means (2 d.p., matching the data): 2.01, 1.44, 1.17, 1.01, 0.90 s [1].
(b) Speed = 0.40 m ÷ mean time, for example at 4 mm:
v = 0.40 m ÷ 2.01 s
v = 0.199 m/s
v² = 0.0396 m²/s²
| depth / m | mean time / s | speed / m s⁻¹ | speed² / m² s⁻² |
|---|---|---|---|
| 0.004 | 2.01 | 0.199 | 0.0396 |
| 0.008 | 1.44 | 0.278 | 0.0772 |
| 0.012 | 1.17 | 0.342 | 0.117 |
| 0.016 | 1.01 | 0.396 | 0.157 |
| 0.020 | 0.90 | 0.444 | 0.198 |
Speeds: 0.199, 0.278, 0.342, 0.396, 0.444 m/s [1]; v²: 0.0396, 0.0772, 0.117, 0.157, 0.198 m²/s² [1]. Depth converted to m: 0.004, 0.008, 0.012, 0.016, 0.020 m [1]. Table with headings depth / m, mean time / s, speed / m s⁻¹, speed² / m² s⁻², units only in headings, consistent 3 s.f. [1].
(c) v² (m²/s²) on the vertical axis against depth (m) on the horizontal [1]; even scales from zero, for example 0.004 m and 0.04 m²/s² per large square, using more than half the grid [1]; points as crosses and a single ruled line of best fit, extended to test whether it passes through the origin [1].
(d) Using the origin and the point (0.020 m, 0.198 m²/s²) on the line:
gradient = 0.198 m²/s² ÷ 0.020 m
gradient = 9.9 m/s² [1] (accept 9.6–10.0)
Unit: (m²/s²) ÷ m gives m/s² [1]. This matches g, 9.8 m/s², within about 1 %, as v² = g d predicts [1].
(e) Supported [1]: from 8 mm to 16 mm the speed rose from 0.278 to 0.396 m/s, a factor of 1.42 — within 1 % of √2 (1.41); and v² ÷ d stays between 9.6 and 9.9 m/s² at every depth, so v² is directly proportional to d [1].
(f) Random error: the blurred shadow of the pulse makes the exact frame on which it crosses a line uncertain by one or two frames (about ±0.01 s), scattering the times — reduced by the mean of three and by the line of best fit [1].
Systematic error: if the tank is not level, the depth at the timing lines differs from the depth read at the centre, shifting every speed the same way; or a gauge with a zero error adds the same amount to every depth, giving the line an intercept on the depth axis [1].
Improvement: level the tank with a spirit level and zero the gauge on the glass; time over a longer distance, such as 0.60 m, to reduce the percentage timing error [1]. (Vague answers such as “mistakes” or “more trials” score 0.)
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth.
Select the two longitudinal waves from this list: light from a lamp · sound from a drum · ripples on a pond · ultrasound in the body · microwaves in an oven · S-wave from an earthquake.
Reveal answer
Sound from a drum [1]; ultrasound in the body [1]. (Both are sound, a longitudinal mechanical wave; light and microwaves are transverse electromagnetic; ripples and S-waves are transverse.)
A displacement–distance graph of a wave on a rope shows crests at 0.30 m, 1.10 m and 1.90 m. The highest point is 0.12 m above the rest position. State the amplitude and wavelength, and state what else you would need to find the frequency.
Reveal answer
A = 0.12 m [1]
λ = 1.10 m − 0.30 m
λ = 0.80 m [1]
The frequency needs a time measurement: the period from a displacement–time graph (or the wave speed, using v = fλ from Lesson 2) [1].
A loudspeaker cone is filmed in slow motion. It completes 20 oscillations in 0.080 s. Calculate its period and frequency, giving the frequency in standard form.
Reveal answer
T = 0.080 s ÷ 20
T = 0.0040 s (4.0 ms) [1]
f = 1 ÷ T
f = 1 ÷ 0.0040 s
f = 250 Hz [1]
f = 2.5 × 10² Hz [1 for standard form with unit]
Two astronauts on a spacewalk, 2 m apart, can see each other clearly but cannot hear each other shout without their radios. Explain.
Reveal answer
Sound is a mechanical wave that needs particles to pass on the vibration; the near-vacuum between the two astronauts has almost no particles, so the sound of a shout inside one helmet cannot cross to the other [1]. Light is an electromagnetic wave, which needs no medium and crosses the vacuum [1].
Describe the motion of the air particles in front of a loudspeaker as a sound wave passes, using the terms compression, rarefaction and direction of energy transfer.
Reveal answer
Each air particle oscillates backwards and forwards about a fixed position, parallel to the direction of energy transfer [1]. Where neighbouring particles are pushed together there is a compression (higher pressure); where they are pulled apart there is a rarefaction (lower pressure) [1]. The compressions and rarefactions travel outwards at the speed of sound, but the particles themselves do not travel with the wave — they return to where they started [1].
A buoy at sea rises and falls 12 times in one minute. The crests of the swell are 24 m apart. Calculate the period, the frequency and the speed of the waves.
Reveal answer
T = 60 s ÷ 12
T = 5.0 s [1]
f = 1 ÷ 5.0 s
f = 0.20 Hz [1]
In one period a crest moves one wavelength: v = λ ÷ T
v = 24 m ÷ 5.0 s
v = 4.8 m/s [1]
Lesson 2 writes this as v = fλ.
7. A seismometer 900 km from an earthquake records P-waves at 6.0 km/s and S-waves at 3.5 km/s. Calculate the time between their arrivals. A warning system needs 5 s to issue an alert after detecting the P-wave. How much warning do people 900 km away get before the S-wave arrives, and why do people 30 km from the epicentre get almost none?
Reveal answer
time = distance ÷ speed
P arrival = 900 km ÷ 6.0 km/s
P arrival = 150 s
S arrival = 900 km ÷ 3.5 km/s
S arrival = 257 s
gap = 257 s − 150 s
gap = 107 s [1]
warning = 107 s − 5 s
warning = 102 s [1]
At 30 km: P arrives after 30 km ÷ 6.0 km/s, which is 5.0 s
S arrives after 30 km ÷ 3.5 km/s, which is 8.6 s
gap = 8.6 s − 5.0 s
gap = 3.6 s, less than the 5 s needed [1]
Near the epicentre the two waves have not had time to separate, so the S-wave arrives before the alert — the “blind zone” of every early-warning system [1].
8. A tsunami crosses 1 500 km of ocean with an average depth of 3 000 m. Calculate its travel time in hours. Then explain, using the idea that the frequency of a wave does not change when it slows, why its wavelength must shorten as it approaches the coast.
Reveal answer
v = √(9.8 m/s² × 3 000 m)
v = 171 m/s [1]
t = 1.5 × 10⁶ m ÷ 171 m/s
t = 8.8 × 10³ s
t = 2.4 hours [1]
The crests arrive at the coast at the same rate at which they were produced — the frequency is fixed by the source [1]. Each crest travels a wavelength in one period, so λ = v × T: when v falls in shallow water with T unchanged, λ must fall in proportion — the crests crowd together [1].
Summary
- A wave transfers energy without transferring matter: particles (or fields) oscillate about fixed positions while the pattern moves on.
- Mechanical waves need a medium; electromagnetic waves, including light, cross a vacuum at 3.00 × 10⁸ m/s.
- Transverse: vibration perpendicular to energy transfer (light, strings, S-waves). Longitudinal: parallel, with compressions and rarefactions (sound, P-waves).
- Amplitude: rest position to crest. Wavelength: crest to crest (or compression to compression). Period: time for one cycle. Frequency: cycles per second.
- f = 1 ÷ T
- Displacement–distance graphs show λ; displacement–time graphs show T. Read the axis first.
- Rays are at 90° to wavefronts; the normal is at 90° to a surface, and angles are measured from it.
- The medium sets the wave speed; in shallow water v = √(g d), so v² against d is a straight line of gradient g.
Next time → The wave equation and sound: why v = f λ, the speed of sound and how to measure it well, and the timing of echoes in sonar, ultrasound and bats.