OCR GCSE Physics A: Sound waves at boundaries and hearing (J249)

This guide covers four statements from topic P5.1, Wave behaviour, in the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026): P5.1f, P5.1g, P5.1h and P5.1i. All four are shown in bold in the specification, so they are tested in the Higher Tier only. Foundation Tier candidates do not need them. The statements on wave speed, frequency and wavelength in general (P5.1a to P5.1e) are taught elsewhere. Course pages: the OCR GCSE Physics hub and the checklist. If you are unsure where you stand, try the free diagnostics first.

Companion pages: the revision notes and the practice questions.

What you must be able to do

Statement	What you must do	Tier
P5.1f	Show how changes in velocity, frequency and wavelength are inter-related when sound passes from one medium to another	Higher Tier only
P5.1g	Describe reflection, transmission and absorption of waves at a material interface (examples such as ultrasound and sonar)	Higher Tier only
P5.1h	Describe, with examples, processes that convert wave disturbances between sound waves and vibrations in solids (a simple structure of the parts of the ear is expected)	Higher Tier only
P5.1i	Explain why such processes only work over a limited frequency range, and the relevance of this to human hearing (including why hearing changes with ageing)	Higher Tier only

You need one equation, which you should already know: wave speed (m/s) = frequency (Hz) × wavelength (m), or v = f λ. Everything in P5.1f is that equation used carefully.

Sound crossing from one medium to another (P5.1f)

A sound wave is a series of compressions and rarefactions passed from particle to particle. The speed depends on the material. As a general pattern, sound travels slowest in gases, faster in liquids and fastest in solids, because the particles are closer together and pass the disturbance on more quickly. Questions give you the speeds you need. Typical values are about 340 m/s in air, about 1480 m/s in water and about 5100 m/s in steel, but always use the numbers in the question.

The key idea: the source sets the frequency

Think about what happens at the boundary. Each compression arriving at the boundary pushes on the next material and starts a compression there. So the number of compressions leaving per second must equal the number arriving per second. If it did not, compressions would pile up or vanish at the boundary.

• Frequency does not change when sound moves between media. It is fixed by the source.
• Speed changes, because the new material has different properties.
• Wavelength must therefore change, since v = f λ with f fixed. Rearranged, λ = v / f. When the speed goes up, the wavelength goes up in the same proportion.

This gives a very useful shortcut: λ₂ / λ₁ = v₂ / v₁.

Worked example 1: air to water

A loudspeaker in air produces a note of frequency 425 Hz. The speed of sound in air is 340 m/s. The same note then travels into water, where the speed is 1480 m/s.

1. Wavelength in air: λ = v / f = 340 / 425 = 0.80 m.
2. The frequency stays at 425 Hz in the water.
3. Wavelength in water: λ = 1480 / 425 = 3.48 m, so 3.5 m to 2 significant figures.
4. Check with the ratio: 1480 / 340 = 4.35, and 3.48 / 0.80 = 4.35.

So the wavelength is about 4.4 times longer in water, the frequency is unchanged, and the speed is 4.4 times greater.

Worked example 2: working out an unknown speed

A tone of frequency 2000 Hz has wavelength 0.75 m in water. It then enters air, where its wavelength is measured as 0.17 m.

1. Speed in water: v = f λ = 2000 × 0.75 = 1500 m/s.
2. Frequency in air is still 2000 Hz.
3. Speed in air: v = 2000 × 0.17 = 340 m/s.

Worked example 3: steel rail

A hammer strikes a steel rail and produces a sound of frequency 1.2 kHz. Take the speed in steel as 5100 m/s and in air as 340 m/s.

1. Convert: 1.2 kHz = 1200 Hz.
2. Wavelength in steel: 5100 / 1200 = 4.25 m.
3. Wavelength in air: 340 / 1200 = 0.283 m, so 0.28 m.

A common slip is to say the frequency in the air is different. It is not. The rail and the air next to it vibrate at the same rate.

Reflection, transmission and absorption at an interface (P5.1g)

An interface (or boundary) is where two different materials meet, for example air and skin, or water and a rock layer. When a wave hits one, three things can happen, and usually all three happen to some degree.

Effect	What happens to the wave	Energy
Reflection	Part of the wave bounces back into the first material, giving an echo	Stays as wave energy in the first material
Transmission	Part of the wave passes into the second material, with a changed speed and wavelength	Carried on as wave energy in the second material
Absorption	Part of the wave is taken in by the material	Transferred to the thermal store of the material, which warms slightly

The fractions depend on the two materials. A large difference between the materials, such as air against a solid wall, gives strong reflection and little transmission. Soft, fibrous or porous materials absorb more sound, which is why curtains and foam panels reduce echoes in a room.

Ultrasound

Ultrasound is sound with a frequency above the upper limit of human hearing. Pulses are sent into the body from a probe. At each boundary between tissues, some of the pulse is reflected and the rest is transmitted onward. The probe detects the echoes. Echoes that arrive later come from deeper boundaries, and a computer builds up an image from the timings. A gel is spread on the skin because an air gap between the probe and the body would reflect almost all of the ultrasound at the first interface, so very little would get in.

Sonar

Sonar works the same way in water. A ship sends a pulse of sound downwards and times the echo from the seabed or from a shoal of fish. The sound has travelled to the object and back, so the distance to the object is half of speed × time.

Worked example 4: sonar depth

A pulse is sent from a ship and the echo returns after 0.36 s. The speed of sound in sea water is 1500 m/s.

1. Total distance travelled = speed × time = 1500 × 0.36 = 540 m.
2. This is there and back, so the depth = 540 / 2 = 270 m.

Worked example 5: ultrasound in tissue

Take the speed of ultrasound in soft tissue as 1540 m/s. A probe sends a pulse of frequency 2.5 MHz into the body. An echo from one boundary returns 78 microseconds later.

1. Wavelength in the tissue: λ = v / f = 1540 / 2 500 000 = 6.16 × 10⁻⁴ m, which is 0.62 mm.
2. Total path = 1540 × 78 × 10⁻⁶ = 0.12 m.
3. Depth of the boundary = 0.12 / 2 = 0.060 m, which is 6.0 cm.

The very short wavelength is useful. Waves reflect from detail that is about as large as the wavelength or larger, so a short wavelength lets small structures show up.

Converting between sound waves and vibrations in solids (P5.1h)

A sound wave in air is a pressure disturbance. When it meets a solid surface, the changing pressure pushes the surface to and fro, so the solid vibrates at the same frequency as the sound. The process also works in reverse: a vibrating solid pushes the air next to it and makes a sound wave. Examples:

• A loudspeaker cone vibrates and pushes on the air to make sound. A signal generator connected to a loudspeaker lets you choose the frequency and listen.
• A microphone diaphragm is a thin sheet that vibrates when sound waves hit it.
• A tuning fork pressed onto a table makes the table vibrate, and the large surface passes the vibration to the air, so the sound is louder.
• A window pane can buzz when a lorry with a strong bass note passes.

The ear

A simple structure of the ear is expected. Follow the chain of conversions:

1. The outer ear (the pinna and ear canal) collects sound waves in air and guides them inward.
2. The sound makes the eardrum, a thin membrane, vibrate.
3. The eardrum’s vibration passes to three small bones, the ossicles, which carry it across the air-filled middle ear.
4. The last of the ossicles pushes on a membrane in the cochlea, a coiled, fluid-filled organ. The vibration becomes a pressure wave in the fluid.
5. Tiny hair cells in the cochlea respond and produce electrical signals, which travel along the auditory nerve to the brain.

The sound wave in air becomes a vibration of solid membrane and bones, then a wave in liquid, and finally a nerve signal.

Why the range is limited, and human hearing (P5.1i)

Every device that converts sound to vibration, and every part of the ear, has mass and stiffness. A mass on a spring cannot follow an oscillation that is too fast, because its inertia stops it changing direction quickly enough. Very slow, small pressure changes also produce only a weak response. So each conversion works properly only over a limited range of frequencies, and outside it the vibration is too small to be useful.

For a typical young person the range of hearing is roughly 20 Hz to 20 000 Hz (20 kHz). Sound above 20 kHz is ultrasound. Sound below 20 Hz is infrasound. Other species have different ranges because their ears have different structures. Dogs can hear well above 20 kHz, which is why a dog whistle works, and bats use much higher frequencies for echo location.

Worked example 6: wavelengths at the ends of the range

Using 340 m/s for air:

1. At 20 Hz: λ = 340 / 20 = 17 m.
2. At 20 000 Hz: λ = 340 / 20 000 = 0.017 m (1.7 cm).

Ageing

Hearing changes with age. The ability to hear the highest frequencies is usually lost first, so the upper limit falls. Causes include damage to the hair cells, which the body does not replace, and wear from loud noise over many years. Lower frequencies are usually affected later. Someone whose upper limit has fallen to 14 kHz can still hear speech but can no longer hear a 16 kHz tone.

Common errors

• Saying the frequency changes at a boundary. Only speed and wavelength change.
• Forgetting to halve the distance for an echo.
• Using kHz or MHz without converting to Hz.
• Writing “the sound is absorbed” when you mean transmitted. Absorbed energy ends up in the thermal store of the material.
• Describing the eardrum as converting sound to electricity. It converts a sound wave in air into a vibration of a solid.
• Answering “why is the range limited” with “because the ear is not good enough”. Mention mass, stiffness or the structures that must vibrate.

Next steps

Work through the revision notes, then the practice questions. Tick off progress on the checklist and see the hub.

Official syllabus

OCR, GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), first assessment 2018. Statements P5.1f to P5.1i.