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S4 Physics Waves Practice Paper 2 | Rearranging v = f x lambda and Ultrasound Depth Sounding: 10 Questions with Answers
S4 Physics Waves Practice Paper 2 | Rearranging v = f x lambda and Ultrasound Depth Sounding: 10 Questions with Answers
The second set of ten wave questions, practising rearrangement of v = f x lambda for wavelength and frequency, ultrasound depth sounding and the time lag of thunder, each with an answer and a one-line explanation.
Direct answers This paper has 10 questions, all using v = f x lambda or distance = speed x time; about 20 minutes is suggested.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 20 minutes |
| Difficulty mix | Foundation 3 · Intermediate 4 · Challenge 2 · Error redo 1 |
| Chapter | S4 Physics: Waves |
| Level | S4 (DSE) |
02 Questions (10 items)
- FoundationA wave has a frequency of 250 Hz and a wavelength of 1.4 m. Find its speed.
- FoundationThe speed of sound is 340 m/s and its frequency is 170 Hz. What is the wavelength?
- FoundationA wave travels at 12 m/s with a wavelength of 3 m. What is its frequency?
- IntermediateA ship sends ultrasound to the seabed and receives the echo 0.6 s later. How deep is the sea? (speed of sound in water 1 500 m/s)
- IntermediateHow do the frequency, the speed and the wavelength change when a wave passes from deep water into shallow water?
- IntermediateA wave travels 0.5 m in 0.25 s and its frequency is 4 Hz. Find its speed and its wavelength.
- IntermediateWhat is the period of a wave of frequency 250 Hz?
- ChallengeThunder is heard 3 s after the lightning is seen. How far away was the lightning? (speed of sound 340 m/s)
- ChallengeWhat are the wavelengths of a 500 Hz sound wave in air at 340 m/s and in water at 1 500 m/s?
- Error redoA student writes ‘wavelength = speed x frequency’. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | v = f x lambda = 250 x 1.4 = 350 m/s. | v = f x lambda is the core formula of this chapter. |
| 2 | lambda = v / f = 340 / 170 = 2 m. | To find the wavelength, divide v by f. |
| 3 | f = v / lambda = 12 / 3 = 4 Hz. | To find the frequency, divide v by lambda. |
| 4 | s = 1 500 x 0.6 / 2 = 450 m. | The echo covers the distance twice, so divide by 2. |
| 5 | The frequency is unchanged, the speed decreases and the wavelength decreases as well. | Frequency is set by the source; the medium only changes the speed. |
| 6 | v = 0.5 / 0.25 = 2 m/s; lambda = v / f = 2 / 4 = 0.5 m. | First find the speed from the distance and the time. |
| 7 | T = 1 / f = 1 / 250 = 0.004 s. | Period and frequency are reciprocals. |
| 8 | s = 340 x 3 = 1 020 m, about 1.02 km. | Light travels so fast that seeing the lightning can be treated as instant. |
| 9 | In air, lambda = 340 / 500 = 0.68 m; in water, lambda = 1 500 / 500 = 3 m. | The frequency does not change, so a greater speed gives a longer wavelength. |
| 10 | The formula should be lambda = v / f. Multiplying by the frequency would mean that a higher frequency gives a longer wavelength, which is the opposite of what happens. | Remember v = f x lambda; the other two forms are rearrangements. |
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.
Work through all ten questions under timed conditions first, then check each answer against its one-line explanation and redo the ones you got wrong.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.
