OxfordAQA International GCSE Physics: Wave properties, sound and ultrasound (9203) – 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 follow the OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards, sections 3.3.1 (general properties of waves) and 3.3.3 (sound and ultrasound). Parts of questions 5, 9, 11 and 12, and all of question 10, draw on ultrasound content that is marked “P” in the specification, for full Physics.

Links: study guide, revision notes, course hub, printable checklist and the 10-minute diagnostics. Take sound as travelling at 340 m/s in air unless the question says otherwise.

Questions

1. Describe what is meant by a wave. [2]

2. State whether each wave is transverse or longitudinal: (a) ripples on a pond, (b) a musical note in air, (c) light. [3]

3. A tuning fork vibrates at 256 Hz. Calculate the wavelength of the sound in air. Give your answer to 2 significant figures. [3]

4. In a ripple tank the vibrating bar has a frequency of 8.0 Hz. The wavelength is 0.030 m in the deep region and 0.018 m in the shallow region.

(a) Calculate the wave speed in each region. [3] (b) State what happens to the frequency of the waves as they enter the shallow region. [1]

5. A doorway is 0.30 m wide at the end of a quiet corridor. A friend speaks in the corridor.

(a) Sound of wavelength 0.25 m passes through the gap. Explain what happens to the waves. [2] (b) A narrow beam of ultrasound with a wavelength of 4 mm meets the same gap. Explain why it behaves differently. [2]

6. An oscilloscope shows the signal from a microphone. The time base is 1.0 ms per division. One complete cycle of the trace is 2.5 divisions long. Calculate the frequency of the sound. [3]

7. Two traces, A and B, from the same microphone are shown on the same scales. Trace B is twice as tall as trace A and shows twice as many cycles across the screen. Compare sound B with sound A. [3]

8. A shout in a quiet valley is heard again 0.90 s later as an echo from a rock face. Calculate the distance to the rock face. [3]

9. A dog can hear sounds up to about 45 kHz.

(a) Calculate the wavelength in air of a sound of frequency 45 kHz. [2] (b) Explain why a person cannot hear a 45 kHz sound. [1]

10. Ultrasound is used in medicine.

(a) Describe how an ultrasound scan produces information about the inside of the body. [3] (b) Describe how ultrasound is used to treat kidney stones. [2]

11. A scan uses ultrasound pulses with a speed of 1560 m/s in the body. The oscilloscope shows the transmitted pulse at 0, then echoes at 2.0 divisions and 5.6 divisions. The time base is 50 µs per division.

(a) Explain why only half of each echo time is used to find a depth. [1] (b) Calculate the depth of the first boundary. [2] (c) Calculate the depth of the second boundary. [2] (d) Calculate the distance between the two boundaries. [1] (e) Explain how an echo from the second boundary can arrive at all. [1]

12. A survey ship sends 40 kHz ultrasound pulses downwards. The speed in sea water is 1500 m/s. An echo from the sea bed arrives 0.36 s after the pulse is sent.

(a) Calculate the depth of the sea bed. [3] (b) Calculate the wavelength of the ultrasound in the water. [2] (c) Suggest two reasons why ultrasound is used instead of audible sound. [2]

Answers

1. A disturbance from a vibrating source [1] that carries energy (and information) along its path, with no net movement of the material itself [1]. [2] Examiner insight: Both halves are needed, so say what is transferred and what is not. “Moves energy” alone leaves the matter point unanswered.

2. (a) Transverse [1]. (b) Longitudinal [1]. (c) Transverse [1]. [3] Examiner insight: Give a one-word answer for each part. Sound is the one that most often gets labelled wrongly, so link it to compressions and rarefactions if you want to check yourself.

3. λ = v/f [1]; λ = 340/256 [1] = 1.328 m, so 1.3 m [1]. [3] Examiner insight: Show the rearranged equation and the substitution. The significant figures instruction means a 1.328 answer with no rounding may lose the last mark.

4. (a) v = fλ [1]. Deep: 8.0 x 0.030 = 0.24 m/s [1]. Shallow: 8.0 x 0.018 = 0.144 m/s (0.14 m/s) [1]. (b) It stays the same [1]. [4] Examiner insight: Part (b) is a standard trap. The speed and wavelength change on refraction, and a clear statement that frequency is set by the source is what answers it.

5. (a) The wavelength is comparable to the gap [1], so the waves diffract and spread out noticeably beyond the doorway (which is why the friend can be heard round the corner) [1]. (b) The wavelength of 4 mm is much smaller than the 0.30 m gap [1], so very little diffraction happens and the beam passes through almost straight [1]. [4] Examiner insight: Answers gain credit by comparing the wavelength with the gap size. Stating only that the gap is “big” or “small” is incomplete.

6. T = 2.5 x 1.0 ms [1] = 2.5 ms = 0.0025 s [1]. f = 1/T = 1/0.0025 = 400 Hz [1]. [3] Examiner insight: The period is read in ms, so change it to seconds before taking the reciprocal. Skipping that gives 0.4 Hz instead of 400 Hz.

7. B is louder, since its amplitude is greater [1]. B has a higher pitch, since it shows more cycles in the same time so its frequency is greater [1]. The frequency of B is twice that of A (the period is half) [1]. [3] Examiner insight: Link each observation to its physical quantity: height to amplitude to loudness, cycles per screen to frequency to pitch. Using “volume” with no reason does not show the link.

8. The sound travels to the rock and back, so the one-way time is 0.90/2 = 0.45 s [1]. s = vt = 340 x 0.45 [1] = 153 m, so 150 m (2 significant figures) [1]. [3] Examiner insight: Using 0.90 s directly gives 306 m, which is exactly double the correct answer. Say in words that the time is for a round trip.

9. (a) λ = v/f = 340/45 000 [1] = 0.00756 m, so 7.6 mm [1]. (b) 45 kHz is higher than the top of the human hearing range, about 20 kHz [1]. [3] Examiner insight: Convert kHz to Hz first. For part (b), compare the figure with the hearing range, rather than saying “it is too high”.

10. (a) Pulses of ultrasound are sent into the body [1]. Some of the ultrasound is partially reflected at each boundary between different tissues [1]. The time for the reflections to return to the detector is used to find the depth of the boundary, and the signals are processed into an image [1]. (b) A beam of high-intensity ultrasound is focused on the stone [1], which breaks it into small pieces that can pass out of the body in urine [1]. [5] Examiner insight: “Partially” matters in part (a), because some of the pulse must continue to deeper boundaries. In part (b), mention that the pieces leave the body without surgery.

11. (a) The echo time includes the journey out and the journey back, so it is twice the time to the boundary [1]. (b) Echo time = 2.0 x 50 = 100 µs [1], one way 50 µs; s = 1560 x 50 x 10^-6 = 0.078 m (7.8 cm) [1]. (c) Echo time = 5.6 x 50 = 280 µs, one way 140 µs [1]; s = 1560 x 140 x 10^-6 = 0.2184 m, so 0.22 m [1]. (d) 0.2184 - 0.078 = 0.1404 m, so 0.14 m [1]. (e) At the first boundary the ultrasound is only partially reflected, so some is transmitted and carries on to the second boundary [1]. [7] Examiner insight: Show the unit conversion from microseconds to seconds. Keep unrounded values until the end, then subtract: rounding both depths first can shift the final digit of the gap.

12. (a) One-way time = 0.36/2 = 0.18 s [1]; s = vt = 1500 x 0.18 [1] = 270 m [1]. (b) λ = v/f = 1500/40 000 [1] = 0.0375 m, so 3.8 cm (0.038 m) [1]. (c) It cannot be heard by people, so it causes no nuisance [1]; its short wavelength diffracts very little, so it forms a narrow beam that gives a precise location [1]. Other sensible reasons could also gain credit. [7] Examiner insight: Give two distinct reasons. Two wordings of the same idea, such as “narrow” and “does not spread”, tend to earn one point between them.

Where marks are usually lost

• Using the full echo time instead of half in s = vt.
• Forgetting to convert µs, ms, kHz or MHz before substituting.
• Writing that frequency changes on refraction.
• Describing pitch and loudness without naming frequency and amplitude.
• Answering a diffraction question without comparing wavelength and gap size.
• Calling ultrasound “very loud sound” instead of sound above the hearing range.
• Giving a final answer with no unit, or with the wrong number of significant figures when one is requested.
• Saying a pulse is “reflected” at a boundary without saying that only part of it is.

Next steps

• Revise with the revision notes.
• Re-read the study guide.
• Open the course hub and the printable checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class: /trial/.

Official syllabus

OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards, published by OxfordAQA. Sections 3.3.1 and 3.3.3.