Waves in Air, Fluids and Solids

A wave is a periodic disturbance that travels through space (or a material) carrying energy from a source to elsewhere. The particles of the medium oscillate about a fixed position; the wave passes through them but they do not travel with the wave.

The classic British GCSE demonstration is a floating cork (or rubber duck) on a pond. Throw a stone in; ripples spread out; the cork bobs up and down on the spot. The ripples carry energy outwards, but the water that holds the cork stays where it is.

Two big ideas every AQA examiner wants to see:

1. The wave transfers energy, not the medium.
2. The wave can also transfer information (this matters for the EM spectrum: radio carries radio programmes, microwaves carry phone calls, light carries images, etc.).

Why this matters in the exam. AQA Paper 2 (2023) examiner report: "Many candidates wrote that water 'moves with the wave' across the pond. This was not credited." Always say the medium oscillates — it does not flow with the wave.

In a transverse wave, the particles oscillate perpendicular (at right angles) to the direction of energy transfer.

Everyday examples in the AQA spec:

• Ripples on the surface of water (the water surface goes up and down; the wave moves horizontally).
• A wave on a stretched rope (flick one end up and down — a hump travels along the rope).
• A wave on a slinky spring (when you wiggle one end side-to-side).
• All electromagnetic waves — radio, microwaves, infrared, visible light, UV, X-rays, gamma rays.
• S-waves (secondary seismic waves) inside the Earth.

The shape you draw on the page is a sine curve. The high points are called peaks (or crests); the low points are called troughs.

British exam tip. Light is the AQA spec's most-asked example. State that "light is a transverse electromagnetic wave" — that single phrase wins marks in the EM spectrum questions (4.6.2.1).

In a longitudinal wave, the particles oscillate parallel to the direction the wave travels. The wave pattern looks like compressions (squashed regions, where particles are bunched up) and rarefactions (stretched regions, where particles are spread out).

Examples in the AQA spec:

• Sound waves in air, water and solids (the most-asked example).
• A slinky spring pushed and pulled along its length.
• P-waves (primary seismic waves) inside the Earth.

Sound cannot travel through a vacuum — there are no particles to compress and rarefy. EM waves can cross a vacuum because their oscillation is in electric and magnetic fields, not in matter. This is a favourite 6-mark long-answer point of comparison.

Why a slinky helps. Holding the end of a slinky and pushing it forward creates a visible compression that travels along the spring while the coils themselves only move a few centimetres back and forth. This is exactly what air molecules do when sound passes through them.

AQA loves direct comparison questions. The mark scheme typically wants two distinguishing features and at least one example of each.

Feature	Transverse	Longitudinal
Particle motion	Perpendicular to wave direction	Parallel to wave direction
Visual shape	Sine curve with peaks & troughs	Compressions & rarefactions
Needs a medium?	EM waves: NO. Others: yes.	Yes — always
Examples (AQA spec)	EM waves, ripples on water, rope waves, S-waves	Sound in air, slinky push-pull, P-waves

Both types share these features:

• They transfer energy without transferring matter.
• They can be reflected (4.6.1.3), refracted (4.6.2.2) and absorbed.
• They obey $v = f\lambda$ (4.6.1.2).

A quick mental check. Imagine you are the medium. If the wave passes and you wobble up and down, it's transverse. If you slide forwards and backwards, it's longitudinal.

AQA past-paper warning. In Paper 2 (2024), candidates were asked to describe sound as longitudinal. A common loss-of-mark answer said "sound makes air go up and down" — sound makes the air move along the direction the sound travels, not vertically.

Every wave can be described with the same five quantities. Get comfortable identifying them on a sine-curve diagram.

• Amplitude $A$ — the maximum displacement of a particle from its rest (equilibrium) position. Measured from the centre line up to a peak (NOT from peak to trough). SI unit: metre (m).
• Wavelength $\lambda$ — the distance between two equivalent points on adjacent waves (peak-to-peak, trough-to-trough, or any same-phase points). SI unit: metre (m).
• Frequency $f$ — the number of complete waves passing a fixed point per second. SI unit: hertz (Hz). 1 Hz = 1 wave per second.
• Period $T$ — the time for one complete wave to pass a point. SI unit: second (s).
• Wave speed $v$ — the speed at which the wave (energy) moves. SI unit: metre per second (m/s).

Common Foundation Tier mistake: measuring amplitude from peak to trough. That's twice the amplitude — the AQA mark scheme rejects it.

For longitudinal waves the wavelength is the distance from one compression to the next compression (or rarefaction to rarefaction). Amplitude relates to how much the particles are displaced from equilibrium during the oscillation.

Period $T$ is the time for one complete oscillation. Frequency $f$ is how many oscillations happen per second.

$T = \frac{1}{f} \qquad \text{or} \qquad f = \frac{1}{T}$

• $T$ in seconds (s).
• $f$ in hertz (Hz).

Worked example. Mains UK electricity oscillates at 50 Hz. What is the period?

$T = \frac{1}{f} = \frac{1}{50} = 0.02\,\text{s} = 20\,\text{ms}$

Worked example (reverse). A wave has a period of 0.004 s. What is the frequency?

$f = \frac{1}{T} = \frac{1}{0.004} = 250\,\text{Hz}$

Quick sanity check. A small period (fraction of a second) means a high frequency. A long period (e.g. 2 s for a slow pendulum) means a low frequency (0.5 Hz here).

The single most important equation in this topic:

$v = f \, \lambda$

• $v$ in m/s.
• $f$ in Hz.
• $\lambda$ in m.

Why it works. If $f$ waves pass a point each second and each wave is $\lambda$ metres long, then $f\lambda$ metres of wave pass the point each second — that's the speed.

Worked example 1 — finding speed. A water wave has wavelength 0.50 m and frequency 4.0 Hz. What is its speed?

$v = f\lambda = 4.0 \times 0.50 = 2.0\,\text{m/s}$

Worked example 2 — finding wavelength. A radio wave has frequency 100 MHz and travels at $3 \times 10^8$ m/s. Find its wavelength.

$\lambda = \frac{v}{f} = \frac{3 \times 10^8}{100 \times 10^6} = 3\,\text{m}$

A typical FM radio wavelength — handy for matching to aerial sizes.

Worked example 3 — finding frequency. A sound wave in air travels at 340 m/s and has wavelength 0.68 m. Find the frequency.

$f = \frac{v}{\lambda} = \frac{340}{0.68} = 500\,\text{Hz}$

On the AQA equation sheet? Yes — $v = f\lambda$ is printed on the equation sheet (2024+ exams). You still need to be confident rearranging it for $f$ or $\lambda$ in the exam.

AQA's Required Practical 8 asks you to measure wave speed using two methods. Both come up in past papers.

Method A — wave on a stretched string.

Apparatus: signal generator + vibration generator + string + pulley + masses + metre ruler + strobe (optional).

Procedure (short version):

1. Clamp one end of the string to the vibration generator. Pass the other end over a pulley and hang masses to keep tension constant.
2. Switch on the signal generator and slowly vary the frequency until a clear standing-wave pattern appears (you see one or more 'loops').
3. Measure the wavelength by counting whole half-wavelengths in the standing wave and using a metre ruler.
4. Read the frequency from the signal generator.
5. Calculate wave speed: $v = f\lambda$.

Sources of error: difficulty pinpointing the exact resonance; parallax when reading the ruler; string tension changing if masses swing.

Method B — ripples in a water tank.

Apparatus: ripple tank with vibrating dipper, strobe lamp or stop-frame camera, metre ruler, stopwatch.

Procedure (short version):

1. Set the dipper to produce continuous ripples at a known frequency.
2. Use a strobe (or a snapshot photo) to 'freeze' the ripples on the white screen below.
3. Measure the wavelength using a ruler placed on the screen (measure 10 wavelengths and divide by 10 for accuracy).
4. Calculate wave speed: $v = f\lambda$.

Alternatively time how long a single wavefront takes to cross a known distance $d$ and use $v = d/t$.

Why measure 10 wavelengths. Random error in a single ruler reading is roughly the same whether you measure 1 wavelength or 10. Dividing the 10-wavelength reading by 10 effectively shrinks the relative error by a factor of 10.

When a wave reaches a boundary between two materials, three things can happen — usually some of each:

• Absorbed. The wave's energy is transferred to the material (e.g. light hitting black paper warms it).
• Transmitted. The wave passes through into the new medium (and may also be refracted — covered in 4.6.2.2).
• Reflected. The wave bounces back into the first medium.

A perfect black surface absorbs all the light that hits it. A perfect mirror reflects all of it. Real surfaces are somewhere in between — a polished tabletop reflects partly and absorbs partly.

Why this matters for hearing in a room. Soft furnishings (carpet, curtains, soft chairs) absorb sound; smooth walls reflect it. Concert halls are designed so that the right amount of reflection reaches the audience without creating a confusing echo — too much absorption and the sound feels dead; too much reflection and it feels echoey.

For any reflection from a smooth surface:

$\text{angle of incidence (}i\text{)} \;=\; \text{angle of reflection (}r\text{)}$

Both angles are measured from the normal — an imaginary line at 90° to the surface, drawn from the point where the ray hits.

The four things mark schemes want on a reflection diagram:

1. The incident ray with an arrow pointing towards the surface.
2. A dashed normal line at the point of incidence.
3. Both angles labelled $i$ and $r$ measured from the normal.
4. The reflected ray with an arrow pointing away from the surface, on the other side of the normal.

Examiner reports (2023) flag candidates who measure angles from the mirror surface — this is wrong and loses both marks for the labelling.

Both kinds of reflection still obey $i = r$ at every microscopic point. The difference is the shape of the surface.

• Specular reflection happens on a smooth, polished surface (mirror, still water, polished metal). Parallel incoming rays reflect to give parallel outgoing rays — you see a clear image.
• Diffuse reflection happens on a rough surface (paper, brick, fabric). Each tiny part of the rough surface acts like a small mirror tilted in a different direction, so parallel incoming rays scatter in many directions — no clear image, but you can see the surface from many angles.

This is why a page of A4 paper looks bright from any angle (diffuse), but a mirror reveals an image only when you're looking at the right angle (specular).

Top-band exam tip. When asked "why can a piece of paper be seen from every angle but a mirror cannot?" the mark scheme wants both ideas: (1) paper is rough → diffuse reflection scatters light; (2) the mirror is smooth → all the reflected light goes in one direction. Only state the law of reflection if asked.

AQA's RP9 (Physics-only) for reflection asks you to measure the angle of reflection for several different angles of incidence and confirm $i = r$.

Apparatus. Ray box (with a single-slit attachment), plane mirror, A3 white paper, protractor, sharp pencil, ruler.

Method (short version):

1. Place the mirror flat on the paper. Draw a line along its back edge (so you know where the surface is when you remove the mirror).
2. Draw a normal at the centre of this line — a dashed line at 90°.
3. Aim the ray box so the single ray strikes the mirror exactly where the normal meets it, at a chosen angle of incidence (e.g. 20°).
4. Mark the incident ray with two crosses (one near the box, one near the mirror) and the reflected ray with two crosses (one near the mirror, one near the far edge).
5. Remove the mirror. Use the ruler to join each pair of crosses, giving clean straight rays.
6. Measure $i$ and $r$ from the normal with a protractor.
7. Repeat for at least five different values of $i$ (e.g. 10°, 20°, 30°, 40°, 50°).
8. Plot $r$ on the $y$-axis vs $i$ on the $x$-axis. The graph should be a straight line through the origin with gradient 1.

Sources of error. Thick ray (use a single slit, narrow); parallax when reading the protractor; the mirror's silvered surface is at the back of the glass so always draw along the back edge.

Graph analysis. Plot $r$ (y-axis) against $i$ (x-axis). A gradient of exactly 1.0 confirms the law. Any deviation suggests measurement error — for evaluation marks, comment on protractor resolution (typically ±0.5°) and ray-box beam thickness.

When a loudspeaker cone pushes forwards it squeezes the air molecules in front of it into a compression (a region of slightly higher pressure). When the cone moves backwards it leaves the molecules a little farther apart than usual — a rarefaction (a region of slightly lower pressure). The cone repeats this many times each second, sending a chain of compressions and rarefactions out through the air.

That repeating pressure pattern is the sound wave. Each air molecule only wobbles a few micrometres back and forth — the wave is the pattern moving, not the air itself.

No medium, no sound. Because the wave is a vibration of particles, there must be particles to carry it. In a vacuum (e.g. interplanetary space) there is essentially nothing to compress, so sound cannot exist. That is why we never hear the Sun's huge solar flares — there is no continuous medium connecting us to them.

Speed of sound in different materials. Sound generally travels faster in denser, stiffer materials because the particles are closer together and pass on the vibration more quickly:

• Air (gas, ~340 m/s at 20 °C)
• Water (liquid, ~1 500 m/s)
• Steel (solid, ~5 000 m/s)

You can demonstrate this by tapping one end of a long metal railing — a friend with an ear pressed to the other end hears the tap before it arrives through the air.

When a sound wave reaches the outer ear, its compressions and rarefactions push and pull the eardrum (tympanic membrane). The eardrum is a thin, taut sheet — when the pressure outside rises, it is pushed inwards; when it falls, it springs back out. The membrane therefore vibrates at exactly the same frequency as the incoming sound.

These vibrations are then passed through three tiny bones in the middle ear (the ossicles: hammer, anvil and stirrup) into the cochlea, a fluid-filled spiral in the inner ear. Hairs in the cochlea convert the mechanical vibration into electrical nerve signals that the brain interprets as sound.

This sequence is essentially a chain of media conversions:

$\text{air (gas)} \;\to\; \text{eardrum (solid)} \;\to\; \text{ossicles (solid)} \;\to\; \text{cochlear fluid (liquid)}$

At each boundary the wave's speed and wavelength change (because each medium has a different stiffness and density), but its frequency stays the same — that is why a 440 Hz tuning fork still sounds like 440 Hz no matter which medium the sound has crossed.

Why this limits the range. Each part of this chain is a mechanical system with its own natural response. The eardrum can flex very rapidly but not infinitely fast; the cochlea can only sort frequencies up to a certain limit. Together, the system handles roughly 20 Hz to 20 kHz. Above that range the eardrum cannot vibrate fast enough; below it, the brain interprets the slow pressure changes as separate thuds rather than a tone.

AQA's specification states the human hearing range as approximately 20 Hz to 20 000 Hz (20 kHz). Sounds below 20 Hz are called infrasound and sounds above 20 kHz are called ultrasound — we cannot hear either of them.

Examples to remember:

• Lowest piano note (A0) ≈ 27 Hz — within our range.
• Standard concert pitch (A above middle C) = 440 Hz — comfortably mid-range.
• A teenager can often hear up to 18–20 kHz; many adults over 50 only reach about 12–14 kHz.
• Dogs can hear up to about 45 kHz; bats use ultrasound up to about 100 kHz for echolocation.

Why the range narrows with age. The hair cells inside the cochlea are easily damaged and do not regenerate. Loud music, industrial noise and ageing all gradually destroy hair cells, starting with those tuned to the highest frequencies. This is why hearing loss is usually noticed first as difficulty hearing high-pitched voices or a kettle whistle.

The link to spec 4.6.1.4. Examiners want you to connect the limit to the mechanical conversion in the ear. The bones, membranes and fluids in the ear simply cannot vibrate faster than about 20 000 times per second, and cannot pass on vibrations slower than about 20 per second as a coherent tone.

When a sound wave passes from one medium to another (e.g. air → glass → air, or air → eardrum → bone), three things happen:

1. Some energy is reflected at the boundary (this is why you hear an echo from a wall, and why some sound is reflected back from the eardrum without being heard).
2. Some energy is transmitted into the new medium.
3. The transmitted wave travels at a different speed in the new medium, so its wavelength changes. The relationship $v = f\lambda$ still holds; because $f$ stays the same and $v$ changes, $\lambda$ must change too.

For example, a 440 Hz note in air ($v ≈ 340$ m/s) has $\lambda = 340 / 440 ≈ 0.77$ m. The same note travelling in water ($v ≈ 1500$ m/s) has $\lambda = 1500 / 440 ≈ 3.4$ m. Same frequency; different wavelength.

This is exactly what happens inside the ear: the sound wave enters a denser medium each time, the wavelength stretches, and at the cochlea the frequency information is finally turned into nerve impulses. The mechanical complexity of the chain is one reason hearing is limited to a finite range — every conversion absorbs energy and filters extreme frequencies.

An everyday example. Stand by a swimming pool. You can hear children shouting in the air but, as soon as you put your head under the water, their voices sound muffled and strange. The sound that travelled through the air was partly reflected at the water's surface, partly transmitted (but at a different speed/wavelength), and the wave that reached your eardrum had a very different waveform from before.

Ultrasound is simply sound at a frequency above the upper limit of human hearing — above 20 000 Hz (20 kHz). Medical ultrasound transducers typically use frequencies between 1 MHz and 15 MHz. Higher frequency means shorter wavelength, which means smaller features can be resolved.

A transducer (a small handheld probe) acts as both a loudspeaker and a microphone. It sends a short pulse of ultrasound into the body. Whenever the pulse meets a boundary between two materials of different density — for example, the boundary between soft tissue and a fluid-filled cavity — some of the ultrasound is reflected back. The transducer detects the echo and a computer uses the time delay to build up an image.

Medical uses (Physics-only spec content):

• Pre-natal scanning — sending ultrasound through the mother's abdomen to image the foetus. Safe because, unlike X-rays, ultrasound is not ionising.
• Imaging the heart (echocardiography), the gall bladder, blood vessels and tendons.
• Breaking up kidney stones by focusing high-intensity ultrasound on them.

Industrial uses:

• Flaw detection in metal castings, weld joints and pipelines. A pulse is sent into the metal; if it meets a crack inside it reflects from the crack before reaching the back wall, giving an extra peak on the receiver.
• Measuring the thickness of pipes (corrosion makes the inside wall move closer to the outside).

Both ultrasound imaging and sonar use the same idea: send a pulse, measure how long the echo takes to come back, and convert that time into a distance using the wave equation for distance:

$s = v\,t$

The pulse travels out to the boundary and back again, so the time $t$ recorded is the round-trip time. The distance to the reflector is half the total path:

$\text{distance to reflector} = \dfrac{v\,t}{2}$

Worked example. An ultrasound transducer emits a pulse that returns 80 µs after transmission. The speed of ultrasound in soft tissue is 1500 m/s. How far below the transducer is the boundary?

$s = \dfrac{v\,t}{2} = \dfrac{1500 \times 80\times10^{-6}}{2} = 0.06 \text{ m} = 6 \text{ cm}$

Sonar and bats. The same principle is used by ships to find the depth of the sea bed (sonar) and by bats to locate insects (echolocation). Both send a pulse, time the echo and halve the path.

Earthquakes release energy as both transverse and longitudinal waves through the body of the Earth. Seismometers around the world record when each type arrives, and from the arrival times scientists work out where the earthquake's epicentre was and what the Earth must be made of inside.

P-waves (primary):

• Longitudinal (compressions and rarefactions).
• The fastest type, so they arrive first at a seismometer (P for "primary").
• Can travel through both solid and liquid layers of the Earth.

S-waves (secondary):

• Transverse (particles oscillate perpendicular to wave direction).
• Slower than P-waves, so they arrive second (S for "secondary").
• Can travel only through solids — a liquid has no rigidity to support the sideways shear motion.

The S-wave shadow zone. When the Earth is hit by a strong earthquake, P-waves are detected almost everywhere on the surface, but S-waves are only detected up to about 105° of arc around the epicentre — beyond that no S-waves arrive. The simplest explanation is that some part of the Earth's interior is liquid and absorbs the S-waves. This is one of the strongest pieces of evidence that the outer core is liquid.

P-waves are also refracted (bent) as they cross the core boundary, which is why a smaller "P-wave gap" appears too, and detailed analysis of these patterns has given us the layered Earth model: solid crust, solid mantle, liquid outer core, solid inner core.

Why this matters for the exam. AQA examiners want a clear cause-and-effect chain: "S-waves cannot travel through liquid → they are not detected on the opposite side of the Earth → therefore at least one inner layer must be liquid." That logic is the key marking point.