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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover statements P5.1f to P5.1i of the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026): sound at boundaries, interfaces, conversions between sound and vibrations, and the limits of hearing. Every question is Higher Tier only. Use v = f λ throughout, and take the speed of sound in air as 340 m/s unless the question gives another value. Read the study guide and revision notes first if you need to. Course pages: the OCR GCSE Physics hub and the checklist.

Questions

1. (Higher Tier only) A sound wave travels from air into steel.

(a) State which quantity of the wave does not change. [1] (b) State what happens to the wavelength, and give the reason. [2]

2. (Higher Tier only) A note of frequency 275 Hz travels in air. It then passes into a liquid where the speed of sound is 1400 m/s. Calculate the wavelength of the note in the air and in the liquid. Give each answer to 2 significant figures. [4]

3. (Higher Tier only) A note of frequency 1750 Hz has a wavelength of 3.2 m as it travels along a metal bar. Calculate the speed of sound in the metal, then calculate the wavelength of the same note when it leaves the bar and travels in air. [4]

4. (Higher Tier only) A pulse of ultrasound meets the boundary between two different tissues. Describe what can happen to the pulse at the boundary. [3]

5. (Higher Tier only) Before an ultrasound scan, a gel is spread on the skin. Explain why. [3]

6. (Higher Tier only) A survey vessel fires a downward pulse of sound into the sea, where sound travels at 1500 m/s.

(a) The echo from the seabed is detected 0.52 s after firing. Find the depth of the seabed. [2] (b) An earlier echo, from a shoal of fish, is detected 0.30 s after firing. Find the depth of the shoal. [2] (c) Calculate how far the shoal is above the seabed. [1]

7. (Higher Tier only) Describe the path of a sound wave through the ear, from the air outside to the signal in the nerve. Say what form the disturbance takes at each stage. [4]

8. (Higher Tier only) A student strikes a tuning fork and holds the base against a wooden table.

(a) Give two examples of devices or objects, other than the table, where a sound wave and a vibration of a solid are converted into one another. For each, say which way the conversion goes. [2] (b) Explain why the sound is louder when the fork touches the table. [2]

9. (Higher Tier only) Explain why the ear can only respond to sound over a limited range of frequencies. [4]

10. (Higher Tier only) A hammer strikes a steel rail. A listener stands 850 m away with an ear pressed to the rail and hears the sound twice: once through the steel and once through the air. The speed of sound is 5100 m/s in steel and 340 m/s in air.

(a) Calculate the time for the sound to reach the listener through the air. [2] (b) Calculate the time for the sound to reach the listener through the steel. [2] (c) Calculate the time interval between the two sounds. [1] (d) The sound in the steel has frequency 1700 Hz. Calculate its wavelength in the steel. [1]

11. (Higher Tier only) A hearing survey records the highest frequency each person can hear.

Person	Age (years)	Highest frequency heard (kHz)
A	25	19.5
B	45	16.0
C	65	11.0

(a) Calculate the wavelength in air of the highest frequency heard by person C. Give your answer in metres to 2 significant figures. [2] (b) Describe the pattern in the data. [1] (c) An alarm emits a steady tone of 15 kHz. State which people could hear it and explain your answer. [2] (d) Suggest two reasons why the highest audible frequency falls with age. [2] (e) A bat produces ultrasound of frequency 85 kHz. Calculate its wavelength in air in millimetres, and explain why none of the three people can hear it. [2]

Answers

1. (a) The frequency [1]. (b) The wavelength increases [1], because the speed is greater in steel and the frequency is fixed, so λ = v / f is larger [1]. [3] Examiner insight: Say that the frequency of the wave is unchanged, not the speed or wavelength. Give a reason with the wavelength, not just the direction of change.

2. Air: λ = v / f = 340 / 275 [1] = 1.236 m, so 1.2 m [1]. Liquid: the frequency is still 275 Hz [1], so λ = 1400 / 275 = 5.09 m, so 5.1 m [1]. [4] Examiner insight: Show the substitution before the answer. If the frequency were wrongly changed in the liquid, the working would show where the error began, which is why a written line of working matters.

3. Speed: v = f λ = 1750 × 3.2 [1] = 5600 m/s [1]. In air the frequency is still 1750 Hz [1], so λ = 340 / 1750 = 0.19 m [1]. [4] Examiner insight: This is the same equation rearranged in two directions. State that the frequency is unchanged, because it shows the idea behind the method as well as the arithmetic.

4. Some of the pulse is reflected back from the boundary [1]. Some is transmitted into the second tissue, where its speed and wavelength change [1]. Some is absorbed, transferring energy to the thermal store of the tissue [1]. [3] Examiner insight: Name all three effects and say what happens in each. Three separate points are needed, so “it bounces and goes through” alone leaves out one idea.

5. Without gel there is a layer of air between the probe and the skin [1]. Air and skin are very different materials, so almost all of the ultrasound would be reflected at this boundary [1]. The gel fills the gap, so more ultrasound is transmitted into the body [1]. [3] Examiner insight: Link the gel to the air gap and to reflection at the interface. “To make it work better” or “to keep the probe clean” does not address the physics.

6. (a) Distance there and back = 1500 × 0.52 [1] = 780 m, so depth = 780 / 2 = 390 m [1]. (b) 1500 × 0.30 = 450 m there and back [1], so depth = 225 m [1]. (c) 390 − 225 = 165 m [1]. [5] Examiner insight: Forgetting to halve is the most common slip, and it gives 780 m and 450 m. Writing “there and back” beside the first calculation shows you know why.

7. Sound waves in air are collected by the outer ear and travel along the ear canal to the eardrum [1]. The wave makes the eardrum vibrate, so the disturbance is now a vibration of a solid membrane [1]. The ossicles (three small bones) pass the vibration across the middle ear to the cochlea [1]. In the fluid-filled cochlea the vibration becomes a wave in liquid, and hair cells convert it into electrical signals that the auditory nerve carries to the brain [1]. [4] Examiner insight: A list of parts earns little unless each is paired with the change of form. Say what is vibrating at each stage.

8. (a) Loudspeaker cone: vibrating solid produces a sound wave in the air [1]. Microphone diaphragm (or eardrum): sound wave in the air makes a solid vibrate [1]. Any two correct examples with the correct direction are acceptable. (b) The fork vibrates the table, which has a large surface [1]. The large surface pushes on much more air, so a bigger sound wave is produced [1]. [4] Examiner insight: An example without a direction of conversion is incomplete, because the specification asks for the process, not only the object.

9. The parts of the ear that vibrate (the eardrum, ossicles and the structures in the cochlea) have mass and stiffness [1]. At very high frequencies the parts cannot change direction quickly enough to keep in step with the sound [1]. At very low frequencies the response is weak and the pressure changes are too slow to produce a useful vibration [1]. Outside the range the vibration is too small to stimulate the hair cells, so no signal reaches the brain [1]. [4] Examiner insight: Mention a physical property of the parts, such as mass, stiffness or inertia. Answers that only repeat “the ear cannot hear it” do not explain anything.

10. (a) t = distance / speed = 850 / 340 [1] = 2.5 s [1]. (b) t = 850 / 5100 [1] = 0.17 s (0.167 s) [1]. (c) 2.5 − 0.167 = 2.3 s [1]. (d) λ = 5100 / 1700 = 3.0 m [1]. [6] Examiner insight: Carry the unrounded steel time (0.1667 s) into the subtraction. Rounding it to 0.2 s first would give 2.3 s here by luck, but the same habit gives wrong answers for other numbers.

11. (a) 11.0 kHz = 11 000 Hz [1]. λ = 340 / 11 000 = 0.031 m [1]. (b) The older the person, the lower the highest frequency they can hear [1]. (c) Persons A and B [1], because their upper limits (19.5 kHz and 16.0 kHz) are above 15 kHz, but person C’s limit of 11.0 kHz is below it [1]. (d) Any two from: hair cells in the cochlea are damaged and are not replaced [1]; long-term exposure to loud noise wears the ear [1]. (e) 85 kHz = 85 000 Hz, λ = 340 / 85 000 = 0.004 m = 4.0 mm [1]. It is above the upper limit of all three people (the highest is 19.5 kHz), so it is ultrasound to them [1]. [9] Examiner insight: In part (c) you need both the comparison and the conclusion for each person. Quoting only the values from the table without comparing them to 15 kHz is a statement of the data, not an explanation.

Where marks are usually lost

• Changing the frequency when the sound crosses a boundary. It is the speed and wavelength that change.
• Leaving frequencies in kHz or MHz when substituting into v = f λ.
• Giving the round-trip distance as the depth in echo questions.
• Rounding too early in multi-step calculations, then subtracting two rounded values.
• Saying absorbed energy “disappears” instead of passing to the thermal store of the material.
• Writing “the sound goes through” for transmission without saying that speed and wavelength change.
• Listing the parts of the ear in order without describing the form of the disturbance at each stage.
• Giving no physical reason (mass, stiffness, inertia) for the limited hearing range.

Next steps

• Re-read the revision notes and the study guide.
• Tick off statements on the checklist and visit the OCR GCSE Physics hub.
• Take one of the free 10-minute diagnostics to find your next weak spot.
• Book a free trial class for help with Higher Tier waves.

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.