IGCSE Physics revision · Waves
Sound
A short, friendly topic with one recurring trick: the echo calculation, where the sound travels to the wall and back — and half the class forgets the 'and back'.
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What the syllabus demands
- —Describe sound as a longitudinal wave produced by vibrating sources, requiring a medium
- —Relate pitch to frequency and loudness to amplitude
- —State the audible range: 20 Hz to 20,000 Hz; define ultrasound as above 20 kHz
- —Describe an echo method for measuring the speed of sound in air (~330–350 m/s)
- —Describe uses of ultrasound: sonar, medical imaging, cleaning
Definitions that earn marks
Clear definitions to practise — check your course mark scheme
- Sound wave
- A longitudinal wave consisting of compressions and rarefactions, produced by a vibrating source, requiring a medium to travel through.
- Ultrasound
- Sound with a frequency above 20,000 Hz (20 kHz) — beyond the upper limit of human hearing.
- Echo
- A reflection of sound from a hard surface, heard after the original sound.
The equations
More equations to practise: the IGCSE formula sheet.
Where the marks die
Common mistakes to check
- 01
Forgetting the factor of 2 in echo calculations. The sound travels to the wall and back, so the distance is 2d — or the time is halved. One or the other, never both.
- 02
Saying sound travels through a vacuum. It cannot — sound needs particles. The bell-in-a-vacuum-jar experiment exists to test this sentence.
- 03
Mixing up pitch and loudness: pitch is set by frequency, loudness by amplitude. On an oscilloscope trace, more waves per screen = higher pitch; taller waves = louder.
- 04
Reversing the media order: sound travels fastest in solids, slower in liquids, slowest in gases — because the particles are closer together, energy passes between them faster.
One worked example, done properly
Question
A student stands 660 m from a cliff, claps, and hears the echo 4.0 s later. Calculate the speed of sound.
Method
- 1.The sound travels to the cliff and back: total distance = 2 × 660 = 1320 m.
- 2.v = distance ÷ time = 1320 ÷ 4.0.
v = 330 m/s