Why is frequency the reciprocal of the period?

The number of waves in one second is one divided by the period.

Why is an ultrasound wavelength so short?

With a fixed speed a higher frequency gives a shorter wavelength.

Why does the wavelength grow in water?

Sound is over four times faster in water, so at the same frequency the wavelength grows too.

S5 Physics: Waves and Sound Practice Paper 3

2026-11-01

Tutor Prep Guide · Secondary Physics

Paper three warms up with v = f lambda, a frequency from 340 over 0.5 m and a period reciprocal, then times thunder at 3 s, halves a sonar echo and sizes a 40 kHz wave, before a string fundamental and a wave crossing into water, ending with the missing halving.

Quick answer　320 m/s, 680 Hz, 40 Hz, 20 Hz, 1 020 m, 375 m, 0.0085 m, 2 m with 1 m, 0.5 m with 2.21 m, and divide by two.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10 questions
Suggested time	25 minutes
Difficulty mix	Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1
Chapter	Chapter: S5 Physics: Waves and Sound – Tutor Lesson Plan
Level	S5

02　Questions (10 items)

FoundationA wave has frequency 80 Hz and wavelength 4 m. Find its speed.
FoundationSound travels at 340 m/s and a note has wavelength 0.5 m. Find its frequency.
FoundationA wave has a period of 0.025 s. Find its frequency.
FoundationA source makes 60 vibrations in 3 s. Find the frequency.
IntermediateThunder is heard 3 s after the lightning is seen (sound speed 340 m/s). How far away did it strike?
IntermediateA sonar receives the seabed echo 0.5 s after sending (sound speed in water 1 500 m/s). Find the depth.
IntermediateAn ultrasound has frequency 40 kHz and the speed of sound in air is 340 m/s. Find the wavelength.
ChallengeA string has a fundamental of 120 Hz and the wave speed on it is 240 m/s. Find the fundamental wavelength and the string length.
ChallengeA 680 Hz sound passes from air (340 m/s) into water (1 500 m/s) with its frequency unchanged. Find the wavelength in air and in water.
Error redoA student answers a sonar depth with 1 500 x 0.5 = 750 m. What is wrong?

03　Answers and one-line notes

#	Answer	One-line note
1	320 m/s (80 x 4)	Wave speed is v = f lambda.
2	680 Hz (340 / 0.5)	Rearranging v = f lambda gives f = v / lambda.
3	40 Hz (1 / 0.025)	Frequency is the reciprocal of the period.
4	20 Hz (60 / 3)	Frequency counts the vibrations per second.
5	1 020 m (340 x 3)	Light arrives almost at once, so the delay measures the sound.
6	375 m (1 500 x 0.5 / 2)	The echo covers the depth twice, so divide by two.
7	0.0085 m, which is 8.5 mm	40 kHz is 40 000 Hz, so lambda is 340 / 40 000.
8	2 m and a length of 1 m	lambda is v / f = 2; with both ends fixed the fundamental is half a wave, so the length is half the wavelength.
9	0.5 m in air and 2.21 m in water	lambda is v / f: 340 / 680 = 0.5 and 1 500 / 680 is about 2.21.
10	The echo travels out and back, so the depth needs another division by two: 375 m	The time covers the whole return trip, so one way is only half of it.

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.

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