IGCSE Physics Sound Questions and Worked Solutions (3.4) | 0625

By /
IGCSE Physics 0625 · Topic 3.4

IGCSE Physics Sound Questions and Worked Solutions

50 topic-wise practice questions on Sound (Unit 3: Waves), written to the Cambridge IGCSE Physics 0625 syllabus for 2026–2028, each with a step-by-step worked solution.

50 questions25 Core15 Extended10 Challenge124 marksTextbook §3.4 · pp. 167–174
Questions tagged Core are for all candidates (Papers 1 & 3). Questions tagged Extended are Supplement content (Papers 2 & 4). Questions 41–50 are harder Challenge questions. Try each question first, then tap Show solution to check your working.

Topic guide: Sound

Sound is a longitudinal wave produced by vibrating sources. It needs a medium to travel through. Loudness depends on amplitude and pitch on frequency. Echoes and ultrasound are used to measure distances.

Key formulas

v = s ÷ tspeed of sound from distance and timeCore
echo: distance = v × t ÷ 2sound travels there and backCore
v = f λwave equationCore

Prerequisite knowledge

  • Waves and v = fλ (3.1)
  • Speed = distance ÷ time (1.2)

Exam tips & common mistakes

  • Halve the time (or double the distance) for echoes
  • 20 kHz = 20 000 Hz
  • Do not say sound waves are transverse

Key facts from the syllabus (3.4)

  • Sound is produced by vibrations and is longitudinal
  • Human hearing range: 20 Hz to 20 000 Hz
  • Sound needs a medium; it cannot travel through a vacuum
  • Speed of sound in air: about 330–350 m/s
  • Method: measure distance and time (e.g. echo from a wall or two observers far apart)
  • Larger amplitude → louder; higher frequency → higher pitch
  • An echo is a reflection of sound; ultrasound has frequency above 20 kHz
  • Extended compressions and rarefactions; sound is generally faster in solids than liquids, and in liquids than gases
  • Extended ultrasound uses — non-destructive testing, medical scanning of soft tissue, sonar (depth = speed × time ÷ 2)

Textbook: §3.4 Sound, pp. 167–174

1
Core[1 mark]

Describe how sound is produced.

Show solutionHide solution

Method: Vibrations.

By vibrating sources
2
Core[1 mark]

State whether sound waves are transverse or longitudinal.

Show solutionHide solution

Method: Longitudinal.

Longitudinal
3
Core[1 mark]

State the range of frequencies audible to humans.

Show solutionHide solution

Method: Syllabus.

20 Hz to 20 000 Hz
4
Core[2 marks]

Explain why sound cannot travel through a vacuum.

Show solutionHide solution

Method: Needs a medium.

Sound needs a medium (particles) to vibrate; a vacuum has no particles
5
Core[1 mark]

State the approximate speed of sound in air.

Show solutionHide solution

Method: Syllabus.

About 330–350 m/s
6
Core[4 marks]

Describe a method using a measurement of distance and time to determine the speed of sound in air.

Show solutionHide solution

Method: Clap–echo or two observers.

Stand a measured distance (e.g. 100 m) from a wall; clap and time the echo with a stopwatch (repeat over many claps); speed = 2 × distance ÷ time
7
Core[2 marks]

State how the loudness and pitch of a sound depend on the wave.

Show solutionHide solution

Method: Amplitude and frequency.

Larger amplitude → louder; higher frequency → higher pitch
8
Core[1 mark]

State what an echo is.

Show solutionHide solution

Method: Reflection.

A reflection of sound waves
9
Core[1 mark]

Define ultrasound.

Show solutionHide solution

Method: > 20 kHz.

Sound with a frequency higher than 20 kHz
10
Core[3 marks]

A student stands 85 m from a wall and hears an echo 0.50 s after clapping. Calculate the speed of sound.

Show solutionHide solution

Method: Distance travelled = 2 × 85.

v = 170 ÷ 0.50 = 340 m/s
11
Core[2 marks]

A person sees lightning and hears thunder 3.0 s later. Calculate the distance to the lightning (speed of sound 340 m/s).

Show solutionHide solution

Method: s = vt.

s = 340 × 3.0 = 1020 m
12
Core[2 marks]

Calculate the wavelength of a 512 Hz note in air (v = 340 m/s).

Show solutionHide solution

Method: λ = v ÷ f.

λ = 340 ÷ 512 = 0.664 m
13
Core[2 marks]

Describe how a loudspeaker produces sound.

Show solutionHide solution

Method: Vibrating cone.

Its cone vibrates, pushing and pulling the air to create sound waves
14
Core[2 marks]

Explain why astronauts on the Moon must use radios to talk.

Show solutionHide solution

Method: No air.

Sound cannot travel through a vacuum; radio waves can
15
Core[2 marks]

State the effect on the pitch of a note when its frequency increases.

Show solutionHide solution

Method: Higher pitch.

The pitch becomes higher
16
Core[2 marks]

State the effect on the loudness of a sound when its amplitude decreases.

Show solutionHide solution

Method: Quieter.

It becomes quieter
17
Core[2 marks]

State whether a sound of frequency 25 kHz can be heard by humans and explain.

Show solutionHide solution

Method: Range 20 Hz–20 kHz.

No, it is above 20 kHz (ultrasound)
18
Core[2 marks]

State whether a sound of frequency 10 Hz can be heard by humans.

Show solutionHide solution

Method: Below range.

No, it is below 20 Hz
19
Core[3 marks]

Two students stand 330 m apart. One fires a starting pistol and the other starts a stopwatch on seeing the flash and stops it on hearing the bang, recording 1.0 s. Calculate the speed of sound.

Show solutionHide solution

Method: v = d/t.

v = 330 m/s
20
Core[2 marks]

State one source of error in the previous experiment and how to reduce it.

Show solutionHide solution

Method: Reaction time.

Reaction time; use a longer distance or repeat and average
21
Core[3 marks]

A boy shouts towards a cliff 165 m away and hears the echo after 1.0 s. Calculate the speed of sound.

Show solutionHide solution

Method: Distance 330 m.

v = 330 ÷ 1.0 = 330 m/s
22
Core[3 marks]

An echo is heard 1.2 s after a clap. Speed of sound = 340 m/s. Calculate the distance to the wall.

Show solutionHide solution

Method: Halve.

d = 340 × 1.2 ÷ 2 = 204 m
23
Core[2 marks]

Calculate the frequency of a sound of wavelength 0.50 m in air (v = 340 m/s).

Show solutionHide solution

Method: f = v/λ.

f = 680 Hz
24
Core[2 marks]

State the range of the speed of sound in air given in the syllabus.

Show solutionHide solution

Method: Approx.

About 330–350 m/s
25
Core[2 marks]

Explain why thunder is heard after the lightning is seen.

Show solutionHide solution

Method: Light faster.

Light travels much faster than sound
26
Extended[2 marks]

Describe compression and rarefaction in a sound wave.

Show solutionHide solution

Method: Regions of high and low pressure.

Compression: particles close together (high pressure); rarefaction: particles spread apart (low pressure)
27
Extended[1 mark]

State the order of the speed of sound in solids, liquids and gases.

Show solutionHide solution

Method: Solids fastest.

Fastest in solids, then liquids, slowest in gases
28
Extended[3 marks]

A ship's sonar pulse returns after 0.80 s. Sound travels at 1500 m/s in sea water. Calculate the depth of the sea.

Show solutionHide solution

Method: Depth = v × t ÷ 2.

d = 1500 × 0.80 ÷ 2 = 600 m
29
Extended[3 marks]

An ultrasound pulse travels at 1540 m/s in soft tissue and returns from an organ boundary after 6.5 × 10⁻⁵ s. Calculate the depth of the boundary.

Show solutionHide solution

Method: d = vt ÷ 2.

d = 1540 × 6.5×10⁻⁵ ÷ 2 = 0.0500 m
30
Extended[2 marks]

Describe the use of ultrasound in non-destructive testing.

Show solutionHide solution

Method: Reflections from flaws.

Ultrasound pulses are sent into a material; reflections from cracks or flaws return earlier than from the far side, revealing the flaw
31
Extended[2 marks]

State two medical advantages of using ultrasound to scan soft tissue.

Show solutionHide solution

Method: Non-ionising.

It is non-ionising so it is safe (e.g. for foetuses); it gives real-time images of soft tissue
32
Extended[3 marks]

An ultrasound wave has a frequency of 2.0 MHz and travels at 1500 m/s in water. Calculate its wavelength.

Show solutionHide solution

Method: λ = v ÷ f.

λ = 1500 ÷ 2.0×10⁶ = 7.5 × 10⁻⁴ m
33
Extended[2 marks]

Sound travels at 5000 m/s in steel and 340 m/s in air. Explain why someone with an ear to a rail hears a train before hearing it through the air.

Show solutionHide solution

Method: Faster in solids.

Sound travels much faster in steel than in air, so it arrives sooner
34
Extended[3 marks]

Describe the motion of air particles as a sound wave passes.

Show solutionHide solution

Method: Oscillate along direction.

They vibrate back and forth along the direction of travel, forming compressions and rarefactions
35
Extended[2 marks]

State the distance between two neighbouring compressions in a sound wave.

Show solutionHide solution

Method: Wavelength.

One wavelength
36
Extended[3 marks]

Sound travels at 1500 m/s in water. A sound of frequency 3000 Hz passes from air (340 m/s) into water. Calculate its wavelength in each.

Show solutionHide solution

Method: λ = v/f.

Air: 0.113 m; water: 0.500 m
37
Extended[3 marks]

A sonar pulse returns from a shoal of fish after 0.12 s. Speed of sound in water = 1500 m/s. Calculate the depth of the shoal.

Show solutionHide solution

Method: Halve.

d = 1500 × 0.12 ÷ 2 = 90.0 m
38
Extended[2 marks]

Explain how ultrasound is used to detect cracks in a metal pipe.

Show solutionHide solution

Method: Early echo.

A crack reflects some ultrasound back earlier than the far wall, showing its position
39
Extended[3 marks]

An ultrasound pulse in a body takes 1.3 × 10⁻⁴ s to return from an organ. v = 1540 m/s. Calculate the depth of the organ in cm.

Show solutionHide solution

Method: Halve.

d = 1540 × 1.3×10⁻⁴ ÷ 2 = 0.100 m = 10.0 cm
40
Extended[2 marks]

Explain why sound travels faster in solids than in gases.

Show solutionHide solution

Method: Particles closer.

The particles are closer and more strongly linked, so vibrations are passed on more quickly

Challenge questions

Harder, multi-step questions at the top end of the exam. Attempt these once you are confident with the questions above.

41
ChallengeCore[4 marks]

Describe in detail an experiment to measure the speed of sound in air using two microphones and a timer.

Show solutionHide solution

Method: Two microphones.

Place two microphones a measured distance apart (e.g. 1.0 m) connected to a fast timer; make a sharp sound in line; the timer measures the delay; v = distance ÷ time; repeat and average
42
ChallengeCore[4 marks]

A person stands between two cliffs and claps. She hears echoes after 0.60 s and 1.4 s. Speed of sound = 340 m/s. Calculate the distance between the cliffs.

Show solutionHide solution

Method: Each echo is a round trip.

Distances = 102 m and 238 m; total = 340 m
43
ChallengeCore[4 marks]

A tuning fork of 512 Hz is struck harder. Describe the change in the sound wave and what is heard, compared with a 256 Hz fork struck gently.

Show solutionHide solution

Method: Amplitude and frequency.

Harder: larger amplitude, louder, same pitch; the 512 Hz fork has twice the frequency and a higher pitch than 256 Hz
44
ChallengeCore[3 marks]

Explain why sound cannot travel through space but light can.

Show solutionHide solution

Method: Medium.

Sound needs particles to vibrate; light is an electromagnetic wave and needs no medium
45
ChallengeCore[4 marks]

Lightning is seen and thunder heard 4.5 s later. Calculate the distance to the strike and explain why the light's travel time can be ignored.

Show solutionHide solution

Method: Light very fast.

d = 340 × 4.5 = 1530 m; light takes about 5 μs, negligible
46
ChallengeExtended[4 marks]

A sonar pulse of frequency 50 kHz returns from the sea floor after 0.40 s (speed in water 1500 m/s). Calculate the depth and the wavelength of the pulse in water.

Show solutionHide solution

Method: d = vt/2; λ = v/f.

Depth = 300 m; λ = 0.0300 m
47
ChallengeExtended[4 marks]

An ultrasound scan of a foetus: the pulse returns from the front and back of the head after 26 μs and 156 μs. Speed in tissue = 1540 m/s. Calculate the width of the head.

Show solutionHide solution

Method: Difference in time ÷ 2.

Width = 1540 × 130×10⁻⁶ ÷ 2 = 0.100 m
48
ChallengeExtended[4 marks]

Describe compressions and rarefactions and explain how they show that sound is longitudinal.

Show solutionHide solution

Method: Particle motion parallel.

Compressions are regions where particles are pushed together, rarefactions where they spread apart; particles vibrate parallel to the direction of travel
49
ChallengeExtended[4 marks]

Explain how ultrasound can find a crack in a steel beam, and why a pulse from a crack returns sooner than one from the far side.

Show solutionHide solution

Method: Reflection timing.

Pulses reflect from any boundary; a crack is closer than the far side, so its echo has a shorter round trip
50
ChallengeExtended[3 marks]

Explain why sound travels faster in steel than in water and faster in water than in air.

Show solutionHide solution

Method: Particle spacing and bonding.

Particles are closer and more strongly bonded in solids, so vibrations pass on more quickly; gases have widely spaced particles
All IGCSE Physics topics← 3.3 Electromagnetic Spectrum4.1 Simple Phenomena of Magnetism →

Book a free trial class

Tell us a little about your requirement.