中文（繁體）
中文（简体）
English
S3 Physics: Sound Waves, Frequency, Speed and Echo (Paper 3: challenge)
S3 Physics: Sound Waves, Frequency, Speed and Echo (Paper 3: challenge)
The third paper carries the formula into real instruments, from deriving d = v times t divided by 2 to sonar depth sounding and ultrasound scanning.
Direct answers Label every question as one way or two way, and that single habit carries this paper.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S3 Physics: Sound Waves, Frequency, Speed and Echo |
| Level | S3 Physics |
02 Questions (10 items)
- FoundationWhy is the echo ranging formula d = v times t divided by 2?
- FoundationWhy do two sounds less than about 0.1 s apart merge into one?
- FoundationIf the speed rises when sound enters another medium at the same frequency, what happens to the wavelength?
- IntermediateAn ultrasound scan at 1,500 m/s gets a return after 0.0002 s. What is the depth?
- IntermediateWith 5,000 m/s in steel and a return after 0.004 s, how deep is the flaw?
- IntermediateAt 1,500 m/s in water, what is the wavelength of 30 kHz ultrasound?
- ChallengeA wave has wavelength 0.85 m at 400 Hz. What is the speed?
- ChallengeFor the same frequency, how many times longer is the wavelength in water at 1,500 m/s than in air at 340 m/s?
- ChallengeA research vessel gets a seabed echo after 0.4 s in water at 1,500 m/s. How deep is the sea?
- Error redoA student uses d = v times t for an echo and gets 204 m from 0.6 s. What is the right answer and the cause?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | The sound travels to the target and back, covering a distance of 2d | The total path is speed times time and equals twice the one-way distance, so it is halved. |
| 2 | The ear retains a sound briefly, so closely spaced copies cannot be separated | This is why rooms need absorbing panels, otherwise reflections smear the sound together. |
| 3 | It grows, because wavelength is speed divided by frequency | The source fixes the frequency, so a larger speed must give a longer wavelength. |
| 4 | 0.15 m, which is 15 cm | Taking 1,500 times 0.0002 and halving gives 0.15 m, a short time for a short depth as scanning needs. |
| 5 | 10 m, which is 5,000 times 0.004 divided by 2 | A high speed in a solid means a very short time still probes ten metres. |
| 6 | 0.05 m | Dividing 1,500 by 30,000 gives 0.05 m, much longer than the same frequency would give in air. |
| 7 | 340 m/s | Multiplying 400 by 0.85 gives 340 m/s, which matches the speed of sound in air. |
| 8 | About 4.41 times | At a fixed frequency wavelength is proportional to speed, and 1,500 over 340 is about 4.41. |
| 9 | 300 m, which is 1,500 times 0.4 divided by 2 | The round trip is 600 m so the depth is 300 m, the basis of any echo sounder. |
| 10 | 102 m; the student forgot to halve the two-way path | Writing 2d = v times t first and then solving for d keeps the factor of two in place. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
The challenge paper uses real measurements: sonar, ultrasound scanning and thunder timing all rest on one echo model, so write the two-way path first and then substitute the numbers.Tutor Registration →
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.