The key idea
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'.
Understand it · 01
Follow a compression through air
A vibrating surface pushes nearby air particles together and then allows them to spread apart. The resulting pressure disturbances travel through the air. In a compression the particles are closer together and the pressure is higher than average; in a rarefaction they are farther apart and the pressure is lower. Each particle oscillates roughly parallel to the direction in which the sound carries energy.
The air particles do not travel all the way from the speaker to the listener. They interact with neighbours, passing on the disturbance. Sound therefore needs a material medium and cannot cross a vacuum. Light from a distant star can reach Earth through space, but an ordinary microphone on Earth could not directly hear a sound wave travelling through that vacuum from the star.
Understand it · 02
Interpret an oscilloscope trace carefully
Higher frequency normally produces a higher perceived pitch. Increasing amplitude increases the pressure variation and is associated with a louder sound under comparable listening conditions. A wave drawn taller is therefore not necessarily higher pitched. On an oscilloscope voltage–time trace, compare cycle spacing to compare frequency, and compare vertical size to compare the signal amplitude when the gain settings are unchanged.
Instrument settings are part of the evidence. If one trace uses a different time-base setting, visually closer crests do not automatically mean higher frequency. Work out the period from the number of horizontal divisions multiplied by the time per division. Similarly, a different vertical sensitivity changes the displayed height without necessarily changing the microphone signal. Always read the scales before comparing shapes.
Understand it · 03
Account for the return journey
An echo is reflected sound that returns to a listener after a delay. If a person sends a short sound towards a wall and receives the echo, the sound travels to the wall and back. The total distance is vt, but the one-way wall distance is vt/2. Draw the outward and return arrows first so the factor of two has a physical reason rather than becoming an unexplained rule.
The same geometry appears in depth sounding and ultrasonic distance measurement. Use the sound speed for the actual medium: the speed in water is different from the speed in air. A question may supply a value appropriate to its conditions. Do not use the vacuum speed of light simply because the signal is described as a wave.
Understand it · 04
Design a measurement that reduces timing error
To measure sound speed using two microphones, place them a measured distance apart along the direction of the pulse. Measure the time delay between corresponding features of their signals, then calculate speed as microphone separation divided by delay. In this arrangement the pulse travels from one microphone to the other once; there is no automatic factor of two. Repeating at several separations allows a distance–delay graph and helps identify timing offsets.
Ultrasound has a frequency above the upper limit of typical human hearing, conventionally about 20 kHz. It is still sound and requires a medium. Its shorter wavelength can help resolve smaller details in imaging, while reflection at boundaries can provide information about structure or distance. The name does not imply that it travels faster than audible sound in the same medium under the same conditions.
Key definitions
- 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.
Equations, units & conditions
Speed from an echo
v = 2d ÷ t
Units: m/s
distance doubled: there and back
Worked examples, step by step
Example 1
A student stands 660 m from a cliff, claps, and hears the echo 4.0 s later. Calculate the speed of sound.
Follow the method
- The sound travels to the cliff and back: total distance = 2 × 660 = 1320 m.
- v = distance ÷ time = 1320 ÷ 4.0.
v = 330 m/s
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Starter · 3 suggested marks
1. Find an echo distance
A student hears an echo 0.44 s after a clap. If sound speed is 330 m/s, calculate the distance to the reflecting wall.
Show solution and self-check: question 1
- The complete sound path is 330 × 0.44 = 145.2 m.
- The wall is halfway along the outward-and-return path.
- Distance = 145.2/2 = 72.6 m.
72.6 m (about 73 m).
Self-check · one mark per point
- Use distance = speed × time.
- Account for the outward and return path.
- Calculate 72.6 m.
Build · 3 suggested marks
2. Read an oscilloscope frequency
An oscilloscope shows six complete cycles in 15 ms. Calculate the frequency and state whether it is ultrasound.
Show solution and self-check: question 2
- Convert 15 ms to 0.015 s.
- f = number of cycles/time = 6/0.015 = 400 Hz.
- 400 Hz is below 20 000 Hz.
400 Hz; it is not ultrasound.
Self-check · one mark per point
- Convert milliseconds to seconds.
- Calculate 400 Hz.
- Compare with 20 kHz and conclude not ultrasound.
Common mistakes & how to avoid them
- 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.
- Saying sound travels through a vacuum. It cannot — sound needs particles. The bell-in-a-vacuum-jar experiment exists to test this sentence.
- 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.
- 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.
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
Why does sound travel faster in solids than in gases?
Sound often travels faster in solids because their stiffness allows disturbances to pass rapidly between interacting particles. Speed depends on both elasticity and density, not particle spacing alone. The statement is a useful trend for common materials, not a universal ranking of every substance.
Close particles do not explain the whole result: two solids can have different sound speeds because they resist deformation differently. A disturbance spreads as one region exerts forces on the next, so the material’s restoring response matters. Greater density alone does not automatically mean faster sound. In an IGCSE comparison, explain the effective interactions in solids while recognising the question’s stated materials. Use a supplied value for calculations rather than treating the speed of sound in air as a constant for every medium.
How does sonar measure depth?
Sonar measures the time between emitting a sound pulse and receiving its reflection. For a vertical path to a seabed and back, depth equals sound speed multiplied by the measured time, divided by two. Use the speed in water, not air.
The division by two follows from the journey, not from a rule applied to all sound measurements. The pulse travels from the transmitter to the reflecting surface and returns to the receiver. If the receiver or reflector geometry differs, the path must be analysed accordingly. A short pulse makes its departure and return easier to separate. The sound speed can depend on water conditions, so real instruments need suitable assumptions or corrections; a school question usually provides the value to use.
Find the seabed depth
A pulse returns after 0.18 s and sound speed in water is 1500 m/s. Find the depth.
- Round-trip distance = 1500 × 0.18 = 270 m.
- Depth = 270/2.
The depth is 135 m.
What changes on an oscilloscope trace when a sound gets louder and higher-pitched?
With unchanged oscilloscope settings, a louder sound usually produces a larger signal amplitude and a higher-pitched sound produces more cycles in the same time. Compare height and cycle spacing separately. If the gain or time base changes, first convert divisions into the actual quantities.
A trace twice as tall is not automatically twice as loud to a listener, because perceived loudness is not a simple linear reading of graph height. At this level, use amplitude to compare the signal under controlled conditions. For pitch, find the period from horizontal spacing and calculate f = 1/T when needed. Distortion or complex waveforms can contain several frequencies, so a simple single-tone diagram is a model. Matching instrument scales are essential before a visual comparison of two traces is valid.
Why do I not divide by two when measuring sound speed with two microphones?
For two microphones separated along a pulse’s path, speed equals their separation divided by the delay between detections. There is no factor of two unless the measured signal makes a return journey. The geometry determines which distance corresponds to the recorded time.
Mark the first detection and the corresponding second detection of the same pulse. The delay is the time to travel from one microphone to the other, not the total time since the source was activated. Position the microphones along the direction of travel, since a sideways separation alone may not give the sound-path difference assumed by the formula. Repeating with several separations can reveal a consistent delay trend. Electronic timing avoids the large human reaction uncertainty associated with very short manually timed sound journeys.
Measure sound speed
Microphones 1.70 m apart detect a pulse 5.0 ms apart. Find sound speed.
- Convert 5.0 ms to 0.0050 s.
- v = 1.70/0.0050.
v = 340 m/s; no round-trip factor is needed.
Does ultrasound travel faster than audible sound?
Ultrasound means sound above the upper limit of typical human hearing, conventionally about 20 kHz. It has a higher frequency, not automatically a higher speed. In the same medium under comparable conditions, increasing frequency gives a shorter wavelength through v = fλ.
Ultrasound still consists of mechanical disturbances in a material and cannot cross a vacuum. Its short wavelength can help distinguish smaller details in suitable imaging applications, while reflections provide information about boundaries. The word “ultra” refers to its place beyond the audible frequency range; it does not turn the wave into electromagnetic radiation. Human hearing limits vary, especially with age and hearing conditions, so 20 kHz is a conventional boundary rather than a promise that every person hears exactly up to that frequency.
Why can light travel through space but sound cannot?
Sound requires particles in a medium to pass on a mechanical disturbance, so it cannot travel through a vacuum. Light is electromagnetic radiation and needs no material medium. An event in space can therefore be seen without its ordinary sound travelling directly to an observer through empty space.
A bell in a jar can illustrate the distinction: as air is removed, less sound is transmitted through the air even while the bell remains visible. Some vibration can still reach the surroundings through solid supports, so a real demonstration may not become perfectly silent. Describe the intended transmission route rather than claiming every remaining sound disproves the model. The explanation also distinguishes a vacuum from merely thin air: reducing the amount of gas changes the transmission conditions, whereas an ideal vacuum contains no particles to carry ordinary sound.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- 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
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
- Cambridge IGCSE Physics 0625 — 2026–2028 syllabus
- Pearson Edexcel International GCSE Physics 4PH1 — specification
Reference this page
Dr Desouky Physics Academy. IGCSE Physics: Sound. Updated 2026-09-25. https://www.drdesouky.com/revision/igcse-physics/sound
Use the arrow beside an explanation heading to link straight to that section. Spotted a problem? Contact the academy with the topic and the step you want us to check.