What is sound and how is it classified in Physics?

Sound is a type of mechanical wave that results from the vibration of particles in a medium such as air, water, or solids. In Physics, particularly in the Cambridge IGCSE curriculum, sound is classified as a longitudinal wave because the particles of the medium vibrate parallel to the direction of wave propagation.

How does the frequency of a sound wave affect its pitch?

The frequency of a sound wave determines its pitch. A higher frequency results in a higher pitch, while a lower frequency results in a lower pitch. In the context of the Cambridge IGCSE Physics syllabus, understanding the relationship between frequency and pitch is crucial for explaining how musical notes are produced and perceived.

What factors affect the speed of sound in different media?

The speed of sound is influenced by the medium through which it travels. In general, sound travels fastest in solids, slower in liquids, and slowest in gases. This is because particles in solids are more closely packed, allowing vibrations to transfer more efficiently. Temperature also affects sound speed; as temperature increases, the speed of sound in a gas typically increases due to increased particle energy and movement.

How does the amplitude of a sound wave relate to its loudness?

The amplitude of a sound wave is directly related to its loudness. A larger amplitude means the sound wave carries more energy, resulting in a louder sound. In the Cambridge IGCSE Physics curriculum, students learn that loudness is a subjective perception, but it is quantitatively related to amplitude, which can be measured in decibels (dB).

What is the Doppler Effect and how does it apply to sound waves?

The Doppler Effect refers to the change in frequency or wavelength of a wave in relation to an observer moving relative to the source of the wave. For sound waves, this effect is observed when a sound source moves towards or away from an observer, causing the sound to be perceived at a higher frequency (higher pitch) as it approaches and a lower frequency (lower pitch) as it moves away. This concept is an important part of the Cambridge IGCSE Physics syllabus, illustrating real-world applications such as radar and medical imaging.

Why is "Saying sound travels through a vacuum" a common mistake in Sound?

Why it happens: Confusing with light/EM. How to avoid it: Sound needs a medium. Light does not. Source: 0625/42 — recurring

Why is "Linking pitch to amplitude (or loudness to frequency)" a common mistake in Sound?

Why it happens: Mixing the two properties. How to avoid it: Pitch ↔ frequency. Loudness ↔ amplitude.

Why is "Forgetting to halve the round-trip distance for echoes" a common mistake in Sound?

Why it happens: Computing d = vt once. How to avoid it: An echo is a there-and-back signal — divide the total distance by 2 for the wall distance.

Why is "Saying sound is fastest in air" a common mistake in Sound?

Why it happens: Daily-life experience. How to avoid it: Sound is fastest in SOLIDS, slower in liquids, slowest in gases (closer particles → faster transfer).

Cambridge IGCSE Physics (0625)

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Short Notes - Sound

Detailed Study Notes

Detailed notes on Waves for Cambridge IGCSE Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Sound — Cambridge IGCSE 0625 Physics Extended (2026)

Longitudinal waves of compressions and rarefactions. Hearing range, speed comparison across media, echoes, ultrasound, and pitch vs loudness on an oscilloscope.

At a glance

Sound is a LONGITUDINAL wave: compressions and rarefactions.
Needs a medium — does NOT travel through vacuum.
Speed of sound: solid > liquid > gas (closer particles → faster transfer).
Speed in air: $\sim 340\,\text{m/s}$ at room temperature.
Human hearing: $20\,\text{Hz}$ to $20{,}000\,\text{Hz}$ (decreases with age).
Pitch = frequency. Loudness = amplitude.
Echoes: reflection of sound. $v = 2d/t$ for echo distance.
Ultrasound: $f > 20{,}000\,\text{Hz}$ . Used in pre-natal scans, sonar.

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

3.7 — Describe sound as a longitudinal wave.
3.7 — State the typical speed of sound in air and compare with solid/liquid.
3.7 — State the typical hearing range of humans.
3.7 — Describe an echo and use $v = 2d/t$ .
3.7 — Describe properties and uses of ultrasound.

Sound as a longitudinal wave

Compressions (close particles) and rarefactions (spread out) move along the direction of travel.

Sound is a mechanical, longitudinal wave. A vibrating source (vocal cords, speaker cone) pushes air molecules together, then pulls them apart. The pulse propagates as alternating regions of compression and rarefaction.

Particles vibrate along the direction of travel — bunched at compressions, spread at rarefactions.

No medium = no sound. A bell in a vacuum jar doesn't make audible sound — the air carrying the wave has been removed.

Drawing. Either:

Show particles bunched and spread (compression-rarefaction pattern), OR
Plot pressure vs position — a sine curve, but interpret peaks as compressions.

Speed of sound depends on medium.

Solid (steel): $\sim 5000\,\text{m/s}$ .
Liquid (water): $\sim 1500\,\text{m/s}$ .
Gas (air): $\sim 340\,\text{m/s}$ at $20\degree\text{C}$ .

Sound is FASTEST in solids — particles are closer, so collisions transfer energy more rapidly.

Sound is fastest in solids and slowest in gases — the closer the particles, the faster the wave travels.

Longitudinal wave.
Needs a medium.
Speed: solid > liquid > gas.
Speed in air: $\sim 340\,\text{m/s}$ .

Pitch and loudness — the oscilloscope view

Pitch comes from frequency. Loudness comes from amplitude.

Pitch depends on FREQUENCY.

Higher frequency → higher pitch (treble).
Lower frequency → lower pitch (bass).

Loudness depends on AMPLITUDE.

Larger amplitude → louder.
Smaller amplitude → quieter.

Oscilloscope traces. A microphone connected to an oscilloscope shows sound as a wave on screen.

Higher pitch: more peaks per second → more cycles per division.
Louder: peaks taller → larger amplitude.

More peaks → higher pitch (frequency); taller peaks → louder (amplitude). Time-base kept the same on all four.

Worked qualitative. Two oscilloscope traces:

Trace A: low amplitude, short wavelength on screen → quiet AND high pitch.
Trace B: high amplitude, long wavelength → loud AND low pitch.

Cambridge tip. Oscilloscope time-base controls horizontal scale. To compare pitches, the time-base must be the SAME on both traces.

Pitch from frequency.
Loudness from amplitude.
Oscilloscope: visualise both at once.
Same time-base when comparing two sounds.

Echoes

Echo = sound reflected off a hard surface. $v = 2d/t$ — sound travels there AND back.

Echo. When sound hits a hard surface (cliff, wall, sea floor), it reflects. If you're far enough away, you hear two distinct sounds: the original and the reflection.

Distance from echo timing. $v = \dfrac{2d}{t},$ where $d$ is the distance to the reflector, $t$ is the time between making the sound and hearing the echo. The factor of $2$ accounts for the round trip.

Worked. Standing $170\,\text{m}$ from a cliff. Hear echo $1\,\text{s}$ after shouting.

$v = 2 \times 170 / 1 = 340\,\text{m/s}$ . ✓ (matches speed of sound in air).

The sound covers the distance twice — there and back — which is why the formula has the factor of 2.

Sonar. Same idea on the sea: a ship sends a pulse downward, measures the time before the echo returns. Calculates depth.

Worked. Sonar pulse echoes back from sea floor in $0.4\,\text{s}$ . Speed of sound in water $1500\,\text{m/s}$ .

$d = (v \times t)/2 = (1500 \times 0.4)/2 = 300\,\text{m}$ deep.

Echo = reflected sound.
$v = 2d/t$ (round trip).
Sonar uses this for sea floor depth.

Ultrasound

Sound above $20{,}000\,\text{Hz}$ — beyond human hearing. Used for safe imaging.

Ultrasound. Sound waves with frequency $> 20{,}000\,\text{Hz}$ . Can't be heard by humans.

Humans hear from about 20 Hz to 20 000 Hz; ultrasound lies just above the top of this range.

Uses.

Pre-natal scans: ultrasound waves bounce off internal tissues and a foetus; echoes are detected and converted into an image. Safe — no ionising radiation, unlike X-rays.
Industrial cleaning: ultrasound vibrations dislodge dirt from delicate parts (jewellery, electronics).
Echolocation by bats and dolphins: emit ultrasound and listen for echoes to navigate.
Sonar in ships and submarines: as above.
Quality control / flaw detection: ultrasound passes through metal castings; echoes from internal defects show up.

Why ultrasound for medical scans (not X-rays)?

Non-ionising → no risk of cell damage.
Soft-tissue contrast.

Why ultrasound for cleaning (not water alone)?

Vibrations cause cavitation (bubbles forming and collapsing) which physically dislodges contamination.

Ultrasound: $f > 20{,}000\,\text{Hz}$ .
Pre-natal scans: safe alternative to X-rays.
Industrial cleaning, echolocation, sonar.

Quick recap

Sound: longitudinal, needs medium.
Speed: solid > liquid > gas.
Air: $\sim 340\,\text{m/s}$ .
Hearing range: $20\,\text{Hz}$ to $20{,}000\,\text{Hz}$ .
Pitch ↔ frequency. Loudness ↔ amplitude.
$v = 2d/t$ for echoes.
Ultrasound: $> 20{,}000\,\text{Hz}$ , used in scans/sonar.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Sound wave — longitudinal pressure wave, requires a medium.
Pitch — perceived frequency of a sound.
Loudness — perceived amplitude of a sound.
Echo — reflection of sound from a hard surface.
Ultrasound — sound above $20{,}000\,\text{Hz}$ .

How it’s examined

Sound appears every Paper 2 (3-4 marks: hearing range, longitudinal/transverse) and many Paper 4s (5-7 marks: echo timing or oscilloscope trace comparisons). Examiner reports flag forgetting the factor of 2 in echo distance, and confusing pitch (frequency) with loudness (amplitude).

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (3.7); 0625/42 Oct/Nov 2024 — Q15 (echo, ultrasound); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

Take this whole topic with you

Step-by-step worked examples — Sound

Step-by-step solutions to past-paper-style questions on sound, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Sound

Almost every sound exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single find instruction using $v = s/t$ or $v = f\lambda$ with the sound travelling one way only.

How to approach it

Write the formula, substitute in SI units and solve — no halving needed when there is no echo.

Common trap

Examiner reports flag candidates halving a distance that was never a round trip — only do this for echoes and sonar.

Multi-step problem

Recognise it by

An echo or sonar pulse — sound travels there and back — or a scenario where you must justify an assumption before calculating.

How to approach it

Use $d = vt$ for the total round-trip path, then halve it for the one-way distance; deal with any justification step first.

Common trap

Examiner reports flag candidates forgetting to halve the round-trip distance — an echo signal covers twice the wall (or sea-bed) distance.

Identify & classify

Recognise it by

Compare, place in order or label instructions — rank the speed of sound in different media, or define a compression and a rarefaction.

How to approach it

Recall the fixed links: pitch ↔ frequency, loudness ↔ amplitude, and sound fastest in solids; then state the order or definition cleanly.

Common trap

Examiner reports flag candidates linking pitch to amplitude or loudness to frequency, and claiming sound is fastest in air — it is fastest in solids.

Show that / prove

Recognise it by

Explain why — for example why a bell in a vacuum chamber cannot be heard.

How to approach it

State that sound is a longitudinal wave needing particles to compress and rarefy, then link the absence of a medium to the absence of sound.

Common trap

Examiner reports flag candidates confusing sound with light — sound needs a material medium, whereas EM waves travel through a vacuum.

Graph or diagram

Recognise it by

An oscilloscope trace — find the frequency or compare pitch and loudness by reading the waveform.

How to approach it

Read the period from the wave width and the time-base ( $T =$ divisions $\times$ time/division), then use $f = 1/T$ ; compare cycle counts for pitch and peak heights for loudness.

Common trap

Examiner reports flag candidates forgetting to convert milliseconds to seconds before $f = 1/T$ , landing a factor of $1000$ out.

1Speed of sound in air
CoreDirect calculation• speed
Question
A clap is heard $0.4\,\text{s}$ after it is seen, from $136\,\text{m}$ away. Find the speed of sound.
Step-by-step solution
1. Step 1
  $v = s/t$ .
  $v = \dfrac{136}{0.4} = 340\,\text{m/s}$
Answer
$340\,\text{m/s}$
2Echo distance
ExtendedMulti-step problem• Adapted from 0625/42 May/Jun 2024 Q15• echo
Question
An echo of a clap returns after $1.2\,\text{s}$ . Take $v_{\text{sound}} = 340\,\text{m/s}$ . Find the distance to the wall.
Step-by-step solution
1. Step 1
  Sound travels there AND back. Total distance.
  $d = vt = 340 \times 1.2 = 408\,\text{m}$
2. Step 2
  One-way distance.
  $= \dfrac{408}{2} = 204\,\text{m}$
Answer
$204\,\text{m}$
3Pitch vs loudness
CoreIdentify & classify• pitch, loudness
Question
Two tuning forks. Fork A produces a higher-pitched sound than B; B produces a louder sound than A. Compare their frequency and amplitude.
Step-by-step solution
1. Step 1
  Higher pitch → HIGHER frequency.
  $f_{A} > f_{B}$
2. Step 2
  Louder → larger amplitude.
  $\text{amplitude}_{B} > \text{amplitude}_{A}$
Answer
$f_{A} > f_{B}$ , $\text{amp}_{B} > \text{amp}_{A}$
4Sound cannot travel in a vacuum
CoreShow that / prove• vacuum
Question
Explain why a bell ringing inside a vacuum chamber cannot be heard.
Step-by-step solution
1. Step 1
  Sound is a longitudinal wave needing PARTICLES to compress and rarefy.
2. Step 2
  A vacuum has no particles → no medium → no sound transmission.
Answer
Sound needs a material medium to propagate; a vacuum has no particles to oscillate.
5Speed of sound in air, water and steel
ExtendedIdentify & classify• Adapted from 0625/42 May/Jun 2023 Q15• speed, media
Question
Approximate speeds of sound: air $340\,\text{m/s}$ , water $1500\,\text{m/s}$ , steel $5000\,\text{m/s}$ . (a) Place them in increasing order. (b) Explain in particle terms why sound travels faster in solids than in gases.
Step-by-step solution
1. Step 1
  Increasing order.
  $v_{\text{air}} < v_{\text{water}} < v_{\text{steel}}$
2. Step 2
  In a solid, particles are CLOSE together and strongly bonded. A compression at one particle is passed almost instantly to the next.
3. Step 3
  In a gas, particles are far apart and only interact via collisions, so the compression travels much more slowly.
Answer
(a) air < water < steel. (b) Sound is fastest in solids because particles are closer together and more strongly bonded → quicker transfer of the compression.
6Find frequency from an oscilloscope trace
ExtendedGraph or diagram• period, oscilloscope
Question
An oscilloscope is set to a time-base of $2\,\text{ms}$ per division. One complete waveform of a pure tone fits in exactly $4$ divisions. Find the period and the frequency.
Step-by-step solution
1. Step 1
  Period = width of one wave in seconds.
  $T = 4 \times 2\,\text{ms} = 8\,\text{ms} = 8 \times 10^{-3}\,\text{s}$
2. Step 2
  $f = 1/T$ .
  $f = \dfrac{1}{8 \times 10^{-3}} = 125\,\text{Hz}$
Answer
$T = 8\,\text{ms}$ ; $f = 125\,\text{Hz}$
Examiner tip
The examiner report flags candidates often forget to convert milliseconds to seconds before computing $f = 1/T$ , ending up with $f = 125\,\text{kHz}$ instead of $125\,\text{Hz}$ .
7Ultrasound application — sonar depth
ExtendedMulti-step problem• ultrasound, sonar
Question
A ship's sonar emits an ultrasound pulse vertically downwards. The echo returns $0.40\,\text{s}$ later. The speed of sound in seawater is $1500\,\text{m/s}$ . Find the depth of the sea below the ship.
Step-by-step solution
1. Step 1
  Sound travels DOWN and back UP.
  $2d = v t = 1500 \times 0.40 = 600\,\text{m}$
2. Step 2
  One-way depth.
  $d = \dfrac{600}{2} = 300\,\text{m}$
Answer
$d = 300\,\text{m}$
8Pitch vs loudness from oscilloscope traces
ExtendedGraph or diagram• oscilloscope, pitch, loudness
Question
Two oscilloscope traces use the SAME time-base and gain. Trace P has 4 cycles in 10 divisions and peaks at $2$ divisions above zero. Trace Q has 2 cycles in 10 divisions and peaks at $3$ divisions above zero. Compare the pitches and loudnesses.
Step-by-step solution
1. Step 1
  More cycles per division → higher frequency → higher pitch. Trace P has TWICE as many cycles as Q, so $f_{P} = 2 f_{Q}$ .
2. Step 2
  Bigger peak height → larger amplitude → louder. Trace Q has 1.5× the amplitude of P, so Q is louder.
Answer
P is higher in pitch (twice the frequency of Q). Q is louder than P (1.5× amplitude).
9Compressions and rarefactions in a sound wave
ExtendedIdentify & classify• longitudinal
Question
A tuning fork vibrates and sends sound through air. (a) Label what is meant by a COMPRESSION and a RAREFACTION. (b) State the relationship between the wavelength of the sound and the spacing of the compressions.
Step-by-step solution
1. Step 1
  (a) Compression: a region where particles are closer together than average (high pressure).
2. Step 2
  (a) Rarefaction: a region where particles are further apart than average (low pressure).
3. Step 3
  (b) The wavelength of a sound wave is the distance from one compression to the next compression (or one rarefaction to the next).
Answer
(a) Compression = high-density region; rarefaction = low-density region. (b) $\lambda$ = distance between adjacent compressions.
10A* — Distance to a thunderstorm
ChallengeMulti-step problem• Adapted from 0625/42 Oct/Nov 2024 Q15• synoptic
Question
A student sees a flash of lightning, then hears the thunder $4.5\,\text{s}$ later. Take $v_{\text{light}} \approx 3.0 \times 10^{8}\,\text{m/s}$ and $v_{\text{sound}} = 340\,\text{m/s}$ . (a) Justify ignoring the travel time of light. (b) Find the distance to the lightning, in km.
Step-by-step solution
1. Step 1
  (a) For any reasonable storm distance (a few km), light takes microseconds while sound takes seconds. Light's travel time is negligible compared with sound's.
2. Step 2
  (b) All of the $4.5\,\text{s}$ is taken by sound travelling from the lightning to the student.
  $d = v t = 340 \times 4.5$
3. Step 3
  Compute and convert.
  $d = 1530\,\text{m} \approx 1.5\,\text{km}$
Answer
(a) Light is roughly $10^{6}$ times faster than sound → its travel time is negligible. (b) $d \approx 1.5\,\text{km}$ .

Model Answers — Sound

High-scoring sample answers for sound on the Cambridge IGCSE 0625 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 2/4 short-answer style1 mark
Explain why sound cannot travel through a vacuum.
Model answer
Sound is a wave that needs particles of a medium to travel (it is passed on by particles vibrating/colliding). A vacuum has no particles, so there is nothing to carry the vibrations and the sound cannot travel.
Why this scores
One mark for sound needing a material medium / particles to travel, which a vacuum lacks. The contrast with light (which does cross a vacuum) is worth noting.
Question 2
Paper 2/4 style2 marks
A student stands $170\,\text{m}$ from a wall and bangs two blocks together. The sound of each bang is heard $0.50\,\text{s}$ after it is made, by a distant observer at the same distance. Calculate the speed of sound.
Model answer
$v = \dfrac{\text{distance}}{\text{time}} = \dfrac{170}{0.50} = \mathbf{340\,\text{m/s}}.$
Why this scores
One mark for $v = s/t$ , one for $340\,\text{m/s}$ — the typical speed of sound in air. (No halving here, because the sound travels one way only.)
Question 3
Paper 4 structured style3 marks
Two notes are played. Note X has a higher pitch than note Y, and note Y is louder than note X. Compare the two notes in terms of their frequency and their amplitude.
Model answer
Pitch depends on frequency: since X has the higher pitch, X has the higher frequency than Y. Loudness depends on amplitude: since Y is louder, Y has the larger amplitude than X. (So X = higher frequency, smaller amplitude; Y = lower frequency, larger amplitude.)
Why this scores
Three marks: pitch linked to frequency so X has higher frequency (1); loudness linked to amplitude so Y has larger amplitude (1); a clear, correctly matched comparison (1). Linking pitch to amplitude (or loudness to frequency) is the standard error.
Question 4
Paper 4 structured style4 marks
A person claps their hands and hears the echo from a cliff $0.30\,\text{s}$ later. The speed of sound in air is $340\,\text{m/s}$ . Calculate the distance from the person to the cliff.
Model answer
The sound travels to the cliff and back, so the total path length is

$d_{\text{total}} = vt = 340 \times 0.30 = 102\,\text{m}.$

This is twice the distance to the cliff, so the distance to the cliff is

$d = \dfrac{102}{2} = \mathbf{51\,\text{m}}.$
Why this scores
Four marks: recognise the sound makes a round trip (1); total distance $vt = 102\,\text{m}$ (1); halve it (1); $51\,\text{m}$ (1). Forgetting to halve the round-trip distance is the recurring echo/sonar error.
Question 5
Paper 6 / Paper 4 extended style5 marks
(a) Explain, in terms of particles, why sound travels faster in a solid such as steel than in air. (b) Describe a method to measure the speed of sound in air using an echo, stating the measurements taken and how the speed is calculated.
Model answer
(a) In a solid the particles are very close together and strongly bonded, so a vibration (compression) is passed on from one particle to the next almost immediately. In a gas the particles are far apart and interact only when they collide, so the vibration is passed on more slowly — hence sound is faster in steel than in air.

(b) Stand a measured distance $d$ (e.g. $50\,\text{m}$ ) from a large wall. Make a sharp sound (clap or bang two blocks) and start a stopwatch, stopping it when the echo is heard; record the time $t$ for the sound to travel to the wall and back. The sound covers $2d$ , so the speed is $v = \dfrac{2d}{t}$ . Repeat and average to reduce timing (reaction-time) error.
Why this scores
Five marks: solid particles close/strongly bonded pass the vibration quickly (1) vs gas particles far apart (1); measure distance to wall (1); time the echo (1); $v = 2d/t$ with repeat/average (1).
Question 6
Paper 4 multi-part structured style6 marks
(a) State what is meant by ultrasound. (b) A ship's sonar sends an ultrasound pulse straight down and detects the echo from the sea bed $0.18\,\text{s}$ later. The speed of sound in seawater is $1500\,\text{m/s}$ ; calculate the depth of the sea. (c) State two other uses of ultrasound.
Model answer
(a) Ultrasound is sound with a frequency above the upper limit of human hearing — above about $20\,000\,\text{Hz}$ (20 kHz).

(b) The pulse travels down and back, a total distance $2d = vt = 1500 \times 0.18 = 270\,\text{m}$ . So the depth is

$d = \dfrac{270}{2} = \mathbf{135\,\text{m}}.$

(c) Any two of: medical scanning (e.g. imaging a fetus), cleaning delicate objects, checking/detecting flaws in metals, or physiotherapy.
Why this scores
Six marks: ultrasound is sound above ~20 kHz (1); recognise the round trip $2d = vt$ (1) and value $270\,\text{m}$ (1); halve to $135\,\text{m}$ (1); two further uses (1 each). For sonar and scanning, always halve the round-trip distance.

Key Formulae — Sound

The formulae you need to memorise for sound on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Speed of sound
$v_{\text{air}} \approx 340\,\text{m/s} \text{ (varies with temperature, humidity)}$
When to use
Air at room temperature.
Wave equation (sound)
$v = f\lambda$
When to use
Connect frequency, wavelength and speed of sound.

Key Definitions and Keywords — Sound

Definitions to memorise and the exact keywords mark schemes credit for sound answers — sharpened from recent examiner reports for the 2026 0625 sitting.

Sound wave
Examiner keyword
A longitudinal mechanical wave caused by the vibration of particles in a medium.
Pitch
Examiner keyword
Determined by frequency. Higher frequency → higher pitch.
Loudness
Examiner keyword
Determined by amplitude. Larger amplitude → louder sound.
Audible range
Examiner keyword
Approximately $20\,\text{Hz}$ to $20{,}000\,\text{Hz}$ for a healthy young human.
Ultrasound
Examiner keyword
Sound with frequency above $20{,}000\,\text{Hz}$ . Used in medical imaging and sonar.

Common Mistakes and Misconceptions — Sound

The traps other students keep falling into on sound questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Saying sound travels through a vacuum
0625/42 — recurring
Why it happens
Confusing with light/EM.
How to avoid it
Sound needs a medium. Light does not.
✕Linking pitch to amplitude (or loudness to frequency)
Why it happens
Mixing the two properties.
How to avoid it
Pitch ↔ frequency. Loudness ↔ amplitude.
✕Forgetting to halve the round-trip distance for echoes
Why it happens
Computing $d = vt$ once.
How to avoid it
An echo is a there-and-back signal — divide the total distance by 2 for the wall distance.
✕Saying sound is fastest in air
Why it happens
Daily-life experience.
How to avoid it
Sound is fastest in SOLIDS, slower in liquids, slowest in gases (closer particles → faster transfer).

Sound — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge IGCSE Physics (0625)

Sound

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Short Notes - Sound

Detailed Study Notes

Detailed notes on Waves for Cambridge IGCSE Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Sound — Cambridge IGCSE 0625 Physics Extended (2026)

Longitudinal waves of compressions and rarefactions. Hearing range, speed comparison across media, echoes, ultrasound, and pitch vs loudness on an oscilloscope.

At a glance

Sound is a LONGITUDINAL wave: compressions and rarefactions.
Needs a medium — does NOT travel through vacuum.
Speed of sound: solid > liquid > gas (closer particles → faster transfer).
Speed in air: $\sim 340\,\text{m/s}$ at room temperature.
Human hearing: $20\,\text{Hz}$ to $20{,}000\,\text{Hz}$ (decreases with age).
Pitch = frequency. Loudness = amplitude.
Echoes: reflection of sound. $v = 2d/t$ for echo distance.
Ultrasound: $f > 20{,}000\,\text{Hz}$ . Used in pre-natal scans, sonar.

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

3.7 — Describe sound as a longitudinal wave.
3.7 — State the typical speed of sound in air and compare with solid/liquid.
3.7 — State the typical hearing range of humans.
3.7 — Describe an echo and use $v = 2d/t$ .
3.7 — Describe properties and uses of ultrasound.

Sound as a longitudinal wave

Compressions (close particles) and rarefactions (spread out) move along the direction of travel.

Particles vibrate along the direction of travel — bunched at compressions, spread at rarefactions.

No medium = no sound. A bell in a vacuum jar doesn't make audible sound — the air carrying the wave has been removed.

Drawing. Either:

Show particles bunched and spread (compression-rarefaction pattern), OR
Plot pressure vs position — a sine curve, but interpret peaks as compressions.

Speed of sound depends on medium.

Solid (steel): $\sim 5000\,\text{m/s}$ .
Liquid (water): $\sim 1500\,\text{m/s}$ .
Gas (air): $\sim 340\,\text{m/s}$ at $20\degree\text{C}$ .

Sound is FASTEST in solids — particles are closer, so collisions transfer energy more rapidly.

Sound is fastest in solids and slowest in gases — the closer the particles, the faster the wave travels.

Longitudinal wave.
Needs a medium.
Speed: solid > liquid > gas.
Speed in air: $\sim 340\,\text{m/s}$ .

Pitch and loudness — the oscilloscope view

Pitch comes from frequency. Loudness comes from amplitude.

Pitch depends on FREQUENCY.

Higher frequency → higher pitch (treble).
Lower frequency → lower pitch (bass).

Loudness depends on AMPLITUDE.

Larger amplitude → louder.
Smaller amplitude → quieter.

Oscilloscope traces. A microphone connected to an oscilloscope shows sound as a wave on screen.

Higher pitch: more peaks per second → more cycles per division.
Louder: peaks taller → larger amplitude.

More peaks → higher pitch (frequency); taller peaks → louder (amplitude). Time-base kept the same on all four.

Worked qualitative. Two oscilloscope traces:

Trace A: low amplitude, short wavelength on screen → quiet AND high pitch.
Trace B: high amplitude, long wavelength → loud AND low pitch.

Cambridge tip. Oscilloscope time-base controls horizontal scale. To compare pitches, the time-base must be the SAME on both traces.

Pitch from frequency.
Loudness from amplitude.
Oscilloscope: visualise both at once.
Same time-base when comparing two sounds.

Echoes

Echo = sound reflected off a hard surface. $v = 2d/t$ — sound travels there AND back.

Echo. When sound hits a hard surface (cliff, wall, sea floor), it reflects. If you're far enough away, you hear two distinct sounds: the original and the reflection.

Worked. Standing $170\,\text{m}$ from a cliff. Hear echo $1\,\text{s}$ after shouting.

$v = 2 \times 170 / 1 = 340\,\text{m/s}$ . ✓ (matches speed of sound in air).

The sound covers the distance twice — there and back — which is why the formula has the factor of 2.

Sonar. Same idea on the sea: a ship sends a pulse downward, measures the time before the echo returns. Calculates depth.

Worked. Sonar pulse echoes back from sea floor in $0.4\,\text{s}$ . Speed of sound in water $1500\,\text{m/s}$ .

$d = (v \times t)/2 = (1500 \times 0.4)/2 = 300\,\text{m}$ deep.

Echo = reflected sound.
$v = 2d/t$ (round trip).
Sonar uses this for sea floor depth.

Ultrasound

Sound above $20{,}000\,\text{Hz}$ — beyond human hearing. Used for safe imaging.

Ultrasound. Sound waves with frequency $> 20{,}000\,\text{Hz}$ . Can't be heard by humans.

Humans hear from about 20 Hz to 20 000 Hz; ultrasound lies just above the top of this range.

Uses.

Pre-natal scans: ultrasound waves bounce off internal tissues and a foetus; echoes are detected and converted into an image. Safe — no ionising radiation, unlike X-rays.
Industrial cleaning: ultrasound vibrations dislodge dirt from delicate parts (jewellery, electronics).
Echolocation by bats and dolphins: emit ultrasound and listen for echoes to navigate.
Sonar in ships and submarines: as above.
Quality control / flaw detection: ultrasound passes through metal castings; echoes from internal defects show up.

Why ultrasound for medical scans (not X-rays)?

Non-ionising → no risk of cell damage.
Soft-tissue contrast.

Why ultrasound for cleaning (not water alone)?

Vibrations cause cavitation (bubbles forming and collapsing) which physically dislodges contamination.

Ultrasound: $f > 20{,}000\,\text{Hz}$ .
Pre-natal scans: safe alternative to X-rays.
Industrial cleaning, echolocation, sonar.

Quick recap

Sound: longitudinal, needs medium.
Speed: solid > liquid > gas.
Air: $\sim 340\,\text{m/s}$ .
Hearing range: $20\,\text{Hz}$ to $20{,}000\,\text{Hz}$ .
Pitch ↔ frequency. Loudness ↔ amplitude.
$v = 2d/t$ for echoes.
Ultrasound: $> 20{,}000\,\text{Hz}$ , used in scans/sonar.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Sound wave — longitudinal pressure wave, requires a medium.
Pitch — perceived frequency of a sound.
Loudness — perceived amplitude of a sound.
Echo — reflection of sound from a hard surface.
Ultrasound — sound above $20{,}000\,\text{Hz}$ .

How it’s examined

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (3.7); 0625/42 Oct/Nov 2024 — Q15 (echo, ultrasound); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

Take this whole topic with you

Step-by-step worked examples — Sound

Step-by-step solutions to past-paper-style questions on sound, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Sound

Almost every sound exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single find instruction using $v = s/t$ or $v = f\lambda$ with the sound travelling one way only.

How to approach it

Write the formula, substitute in SI units and solve — no halving needed when there is no echo.

Common trap

Examiner reports flag candidates halving a distance that was never a round trip — only do this for echoes and sonar.

Multi-step problem

Recognise it by

An echo or sonar pulse — sound travels there and back — or a scenario where you must justify an assumption before calculating.

How to approach it

Use $d = vt$ for the total round-trip path, then halve it for the one-way distance; deal with any justification step first.

Common trap

Examiner reports flag candidates forgetting to halve the round-trip distance — an echo signal covers twice the wall (or sea-bed) distance.

Identify & classify

Recognise it by

Compare, place in order or label instructions — rank the speed of sound in different media, or define a compression and a rarefaction.

How to approach it

Recall the fixed links: pitch ↔ frequency, loudness ↔ amplitude, and sound fastest in solids; then state the order or definition cleanly.

Common trap

Examiner reports flag candidates linking pitch to amplitude or loudness to frequency, and claiming sound is fastest in air — it is fastest in solids.

Show that / prove

Recognise it by

Explain why — for example why a bell in a vacuum chamber cannot be heard.

How to approach it

State that sound is a longitudinal wave needing particles to compress and rarefy, then link the absence of a medium to the absence of sound.

Common trap

Examiner reports flag candidates confusing sound with light — sound needs a material medium, whereas EM waves travel through a vacuum.

Graph or diagram

Recognise it by

An oscilloscope trace — find the frequency or compare pitch and loudness by reading the waveform.

How to approach it

Read the period from the wave width and the time-base ( $T =$ divisions $\times$ time/division), then use $f = 1/T$ ; compare cycle counts for pitch and peak heights for loudness.

Common trap

Examiner reports flag candidates forgetting to convert milliseconds to seconds before $f = 1/T$ , landing a factor of $1000$ out.

1Speed of sound in air
CoreDirect calculation• speed
Question
A clap is heard $0.4\,\text{s}$ after it is seen, from $136\,\text{m}$ away. Find the speed of sound.
Step-by-step solution
1. Step 1
  $v = s/t$ .
  $v = \dfrac{136}{0.4} = 340\,\text{m/s}$
Answer
$340\,\text{m/s}$
2Echo distance
ExtendedMulti-step problem• Adapted from 0625/42 May/Jun 2024 Q15• echo
Question
An echo of a clap returns after $1.2\,\text{s}$ . Take $v_{\text{sound}} = 340\,\text{m/s}$ . Find the distance to the wall.
Step-by-step solution
1. Step 1
  Sound travels there AND back. Total distance.
  $d = vt = 340 \times 1.2 = 408\,\text{m}$
2. Step 2
  One-way distance.
  $= \dfrac{408}{2} = 204\,\text{m}$
Answer
$204\,\text{m}$
3Pitch vs loudness
CoreIdentify & classify• pitch, loudness
Question
Two tuning forks. Fork A produces a higher-pitched sound than B; B produces a louder sound than A. Compare their frequency and amplitude.
Step-by-step solution
1. Step 1
  Higher pitch → HIGHER frequency.
  $f_{A} > f_{B}$
2. Step 2
  Louder → larger amplitude.
  $\text{amplitude}_{B} > \text{amplitude}_{A}$
Answer
$f_{A} > f_{B}$ , $\text{amp}_{B} > \text{amp}_{A}$
4Sound cannot travel in a vacuum
CoreShow that / prove• vacuum
Question
Explain why a bell ringing inside a vacuum chamber cannot be heard.
Step-by-step solution
1. Step 1
  Sound is a longitudinal wave needing PARTICLES to compress and rarefy.
2. Step 2
  A vacuum has no particles → no medium → no sound transmission.
Answer
Sound needs a material medium to propagate; a vacuum has no particles to oscillate.
5Speed of sound in air, water and steel
ExtendedIdentify & classify• Adapted from 0625/42 May/Jun 2023 Q15• speed, media
Question
Approximate speeds of sound: air $340\,\text{m/s}$ , water $1500\,\text{m/s}$ , steel $5000\,\text{m/s}$ . (a) Place them in increasing order. (b) Explain in particle terms why sound travels faster in solids than in gases.
Step-by-step solution
1. Step 1
  Increasing order.
  $v_{\text{air}} < v_{\text{water}} < v_{\text{steel}}$
2. Step 2
  In a solid, particles are CLOSE together and strongly bonded. A compression at one particle is passed almost instantly to the next.
3. Step 3
  In a gas, particles are far apart and only interact via collisions, so the compression travels much more slowly.
Answer
(a) air < water < steel. (b) Sound is fastest in solids because particles are closer together and more strongly bonded → quicker transfer of the compression.
6Find frequency from an oscilloscope trace
ExtendedGraph or diagram• period, oscilloscope
Question
An oscilloscope is set to a time-base of $2\,\text{ms}$ per division. One complete waveform of a pure tone fits in exactly $4$ divisions. Find the period and the frequency.
Step-by-step solution
1. Step 1
  Period = width of one wave in seconds.
  $T = 4 \times 2\,\text{ms} = 8\,\text{ms} = 8 \times 10^{-3}\,\text{s}$
2. Step 2
  $f = 1/T$ .
  $f = \dfrac{1}{8 \times 10^{-3}} = 125\,\text{Hz}$
Answer
$T = 8\,\text{ms}$ ; $f = 125\,\text{Hz}$
Examiner tip
The examiner report flags candidates often forget to convert milliseconds to seconds before computing $f = 1/T$ , ending up with $f = 125\,\text{kHz}$ instead of $125\,\text{Hz}$ .
7Ultrasound application — sonar depth
ExtendedMulti-step problem• ultrasound, sonar
Question
A ship's sonar emits an ultrasound pulse vertically downwards. The echo returns $0.40\,\text{s}$ later. The speed of sound in seawater is $1500\,\text{m/s}$ . Find the depth of the sea below the ship.
Step-by-step solution
1. Step 1
  Sound travels DOWN and back UP.
  $2d = v t = 1500 \times 0.40 = 600\,\text{m}$
2. Step 2
  One-way depth.
  $d = \dfrac{600}{2} = 300\,\text{m}$
Answer
$d = 300\,\text{m}$
8Pitch vs loudness from oscilloscope traces
ExtendedGraph or diagram• oscilloscope, pitch, loudness
Question
Two oscilloscope traces use the SAME time-base and gain. Trace P has 4 cycles in 10 divisions and peaks at $2$ divisions above zero. Trace Q has 2 cycles in 10 divisions and peaks at $3$ divisions above zero. Compare the pitches and loudnesses.
Step-by-step solution
1. Step 1
  More cycles per division → higher frequency → higher pitch. Trace P has TWICE as many cycles as Q, so $f_{P} = 2 f_{Q}$ .
2. Step 2
  Bigger peak height → larger amplitude → louder. Trace Q has 1.5× the amplitude of P, so Q is louder.
Answer
P is higher in pitch (twice the frequency of Q). Q is louder than P (1.5× amplitude).
9Compressions and rarefactions in a sound wave
ExtendedIdentify & classify• longitudinal
Question
A tuning fork vibrates and sends sound through air. (a) Label what is meant by a COMPRESSION and a RAREFACTION. (b) State the relationship between the wavelength of the sound and the spacing of the compressions.
Step-by-step solution
1. Step 1
  (a) Compression: a region where particles are closer together than average (high pressure).
2. Step 2
  (a) Rarefaction: a region where particles are further apart than average (low pressure).
3. Step 3
  (b) The wavelength of a sound wave is the distance from one compression to the next compression (or one rarefaction to the next).
Answer
(a) Compression = high-density region; rarefaction = low-density region. (b) $\lambda$ = distance between adjacent compressions.
10A* — Distance to a thunderstorm
ChallengeMulti-step problem• Adapted from 0625/42 Oct/Nov 2024 Q15• synoptic
Question
A student sees a flash of lightning, then hears the thunder $4.5\,\text{s}$ later. Take $v_{\text{light}} \approx 3.0 \times 10^{8}\,\text{m/s}$ and $v_{\text{sound}} = 340\,\text{m/s}$ . (a) Justify ignoring the travel time of light. (b) Find the distance to the lightning, in km.
Step-by-step solution
1. Step 1
  (a) For any reasonable storm distance (a few km), light takes microseconds while sound takes seconds. Light's travel time is negligible compared with sound's.
2. Step 2
  (b) All of the $4.5\,\text{s}$ is taken by sound travelling from the lightning to the student.
  $d = v t = 340 \times 4.5$
3. Step 3
  Compute and convert.
  $d = 1530\,\text{m} \approx 1.5\,\text{km}$
Answer
(a) Light is roughly $10^{6}$ times faster than sound → its travel time is negligible. (b) $d \approx 1.5\,\text{km}$ .

Model Answers — Sound

High-scoring sample answers for sound on the Cambridge IGCSE 0625 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 2/4 short-answer style1 mark
Explain why sound cannot travel through a vacuum.
Model answer
Sound is a wave that needs particles of a medium to travel (it is passed on by particles vibrating/colliding). A vacuum has no particles, so there is nothing to carry the vibrations and the sound cannot travel.
Why this scores
One mark for sound needing a material medium / particles to travel, which a vacuum lacks. The contrast with light (which does cross a vacuum) is worth noting.
Question 2
Paper 2/4 style2 marks
A student stands $170\,\text{m}$ from a wall and bangs two blocks together. The sound of each bang is heard $0.50\,\text{s}$ after it is made, by a distant observer at the same distance. Calculate the speed of sound.
Model answer
$v = \dfrac{\text{distance}}{\text{time}} = \dfrac{170}{0.50} = \mathbf{340\,\text{m/s}}.$
Why this scores
One mark for $v = s/t$ , one for $340\,\text{m/s}$ — the typical speed of sound in air. (No halving here, because the sound travels one way only.)
Question 3
Paper 4 structured style3 marks
Two notes are played. Note X has a higher pitch than note Y, and note Y is louder than note X. Compare the two notes in terms of their frequency and their amplitude.
Model answer
Pitch depends on frequency: since X has the higher pitch, X has the higher frequency than Y. Loudness depends on amplitude: since Y is louder, Y has the larger amplitude than X. (So X = higher frequency, smaller amplitude; Y = lower frequency, larger amplitude.)
Why this scores
Three marks: pitch linked to frequency so X has higher frequency (1); loudness linked to amplitude so Y has larger amplitude (1); a clear, correctly matched comparison (1). Linking pitch to amplitude (or loudness to frequency) is the standard error.
Question 4
Paper 4 structured style4 marks
A person claps their hands and hears the echo from a cliff $0.30\,\text{s}$ later. The speed of sound in air is $340\,\text{m/s}$ . Calculate the distance from the person to the cliff.
Model answer
The sound travels to the cliff and back, so the total path length is

$d_{\text{total}} = vt = 340 \times 0.30 = 102\,\text{m}.$

This is twice the distance to the cliff, so the distance to the cliff is

$d = \dfrac{102}{2} = \mathbf{51\,\text{m}}.$
Why this scores
Four marks: recognise the sound makes a round trip (1); total distance $vt = 102\,\text{m}$ (1); halve it (1); $51\,\text{m}$ (1). Forgetting to halve the round-trip distance is the recurring echo/sonar error.
Question 5
Paper 6 / Paper 4 extended style5 marks
(a) Explain, in terms of particles, why sound travels faster in a solid such as steel than in air. (b) Describe a method to measure the speed of sound in air using an echo, stating the measurements taken and how the speed is calculated.
Model answer
(a) In a solid the particles are very close together and strongly bonded, so a vibration (compression) is passed on from one particle to the next almost immediately. In a gas the particles are far apart and interact only when they collide, so the vibration is passed on more slowly — hence sound is faster in steel than in air.

(b) Stand a measured distance $d$ (e.g. $50\,\text{m}$ ) from a large wall. Make a sharp sound (clap or bang two blocks) and start a stopwatch, stopping it when the echo is heard; record the time $t$ for the sound to travel to the wall and back. The sound covers $2d$ , so the speed is $v = \dfrac{2d}{t}$ . Repeat and average to reduce timing (reaction-time) error.
Why this scores
Five marks: solid particles close/strongly bonded pass the vibration quickly (1) vs gas particles far apart (1); measure distance to wall (1); time the echo (1); $v = 2d/t$ with repeat/average (1).
Question 6
Paper 4 multi-part structured style6 marks
(a) State what is meant by ultrasound. (b) A ship's sonar sends an ultrasound pulse straight down and detects the echo from the sea bed $0.18\,\text{s}$ later. The speed of sound in seawater is $1500\,\text{m/s}$ ; calculate the depth of the sea. (c) State two other uses of ultrasound.
Model answer
(a) Ultrasound is sound with a frequency above the upper limit of human hearing — above about $20\,000\,\text{Hz}$ (20 kHz).

(b) The pulse travels down and back, a total distance $2d = vt = 1500 \times 0.18 = 270\,\text{m}$ . So the depth is

$d = \dfrac{270}{2} = \mathbf{135\,\text{m}}.$

(c) Any two of: medical scanning (e.g. imaging a fetus), cleaning delicate objects, checking/detecting flaws in metals, or physiotherapy.
Why this scores
Six marks: ultrasound is sound above ~20 kHz (1); recognise the round trip $2d = vt$ (1) and value $270\,\text{m}$ (1); halve to $135\,\text{m}$ (1); two further uses (1 each). For sonar and scanning, always halve the round-trip distance.

Key Formulae — Sound

The formulae you need to memorise for sound on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Speed of sound
$v_{\text{air}} \approx 340\,\text{m/s} \text{ (varies with temperature, humidity)}$
When to use
Air at room temperature.
Wave equation (sound)
$v = f\lambda$
When to use
Connect frequency, wavelength and speed of sound.

Key Definitions and Keywords — Sound

Definitions to memorise and the exact keywords mark schemes credit for sound answers — sharpened from recent examiner reports for the 2026 0625 sitting.

Sound wave
Examiner keyword
A longitudinal mechanical wave caused by the vibration of particles in a medium.
Pitch
Examiner keyword
Determined by frequency. Higher frequency → higher pitch.
Loudness
Examiner keyword
Determined by amplitude. Larger amplitude → louder sound.
Audible range
Examiner keyword
Approximately $20\,\text{Hz}$ to $20{,}000\,\text{Hz}$ for a healthy young human.
Ultrasound
Examiner keyword
Sound with frequency above $20{,}000\,\text{Hz}$ . Used in medical imaging and sonar.

Common Mistakes and Misconceptions — Sound

The traps other students keep falling into on sound questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Saying sound travels through a vacuum
0625/42 — recurring
Why it happens
Confusing with light/EM.
How to avoid it
Sound needs a medium. Light does not.
✕Linking pitch to amplitude (or loudness to frequency)
Why it happens
Mixing the two properties.
How to avoid it
Pitch ↔ frequency. Loudness ↔ amplitude.
✕Forgetting to halve the round-trip distance for echoes
Why it happens
Computing $d = vt$ once.
How to avoid it
An echo is a there-and-back signal — divide the total distance by 2 for the wall distance.
✕Saying sound is fastest in air
Why it happens
Daily-life experience.
How to avoid it
Sound is fastest in SOLIDS, slower in liquids, slowest in gases (closer particles → faster transfer).

Sound — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Sound

Short Notes - Sound

Detailed Study Notes

At a glance

What you’ll learn

Sound as a longitudinal wave

Pitch and loudness — the oscilloscope view

Echoes

Ultrasound

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Sound

1Speed of sound in air

2Echo distance

3Pitch vs loudness

4Sound cannot travel in a vacuum

5Speed of sound in air, water and steel

6Find frequency from an oscilloscope trace

7Ultrasound application — sonar depth

8Pitch vs loudness from oscilloscope traces

9Compressions and rarefactions in a sound wave

10A* — Distance to a thunderstorm

Speed of sound

Wave equation (sound)

Sound wave

Pitch

Loudness

Audible range

Ultrasound

✕Saying sound travels through a vacuum

✕Linking pitch to amplitude (or loudness to frequency)

✕Forgetting to halve the round-trip distance for echoes

✕Saying sound is fastest in air

Sound — frequently asked questions

Sound

Short Notes - Sound

Detailed Study Notes

At a glance

What you’ll learn

Sound as a longitudinal wave

Pitch and loudness — the oscilloscope view

Echoes

Ultrasound

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Sound

1Speed of sound in air

2Echo distance

3Pitch vs loudness

4Sound cannot travel in a vacuum

5Speed of sound in air, water and steel

6Find frequency from an oscilloscope trace

7Ultrasound application — sonar depth

8Pitch vs loudness from oscilloscope traces

9Compressions and rarefactions in a sound wave

10A* — Distance to a thunderstorm

Speed of sound

Wave equation (sound)

Sound wave

Pitch

Loudness

Audible range

Ultrasound

✕Saying sound travels through a vacuum

✕Linking pitch to amplitude (or loudness to frequency)

✕Forgetting to halve the round-trip distance for echoes

✕Saying sound is fastest in air

Sound — frequently asked questions