CIE iGCSE Co-ordinated Sciences-P3.4 Sound- Study Notes- New Syllabus
CIE iGCSE Co-ordinated Sciences-P3.4 Sound – Study Notes
CIE iGCSE Co-ordinated Sciences-P3.4 Sound – Study Notes -CIE iGCSE Co-ordinated Sciences – per latest Syllabus.
Key Concepts:
Core
1. Describe the production of sound by vibrating sources
2. State the approximate range of frequencies audible to humans as 20 Hz to 20 kHz
3. Know that a medium is needed to transmit sound waves
4. Determine the speed of sound in air using a method involving a measurement of distance and time
5. Describe how changes in amplitude and frequency affect the loudness and pitch of sound waves
6. Describe an echo as the reflection of a sound wave
7. Define ultrasound as sound with a frequency higher than 20 kHz
Supplement
8. Describe the longitudinal nature of sound waves in air as a series of compressions and rarefactions
9. Describe, qualitatively, compressions as regions of higher pressure due to particles being closer together and rarefactions as regions of lower pressure due to particles being spread further apart
10. Know that, in general, sound travels faster in solids than in liquids and faster in liquids than in gases
CIE iGCSE Co-Ordinated Sciences-Concise Summary Notes- All Topics
Sound Waves
(a) Production of Sound by Vibrating Sources 
- Sound originates when a material body or source vibrates rapidly.
- These vibrations disturb the surrounding particles of the medium (usually air) and cause them to oscillate back and forth.
- The disturbance transfers energy away from the source in the form of sound waves.
- Different sources of sound:
- Musical instruments: Vibrating strings (guitar, violin) or air columns (flute, organ pipe).
- Human voice: Vocal cords vibrate, producing sound waves.
- Loudspeakers: A vibrating diaphragm pushes and pulls air particles.
- Everyday sounds: Striking a drum, clapping hands, ringing a bell.
Key idea: No medium → no sound. In a vacuum, sound cannot travel because there are no particles to vibrate.
(b) Longitudinal Nature of Sound Waves in Air
- Sound waves are mechanical longitudinal waves.
- Longitudinal wave: The particles of the medium vibrate parallel to the direction in which the wave travels.
- Contrast with transverse wave: In transverse waves, particles move perpendicular to wave direction (e.g. water waves, light waves).
- In sound waves:
- Particles oscillate back and forth.
- Energy is transmitted through pressure variations, not through particle transport.
(c) Compressions and Rarefactions
- A sound wave consists of alternating compressions and rarefactions that move through the medium.
- Compressions:
- Particles are squeezed closer together.
- Local density and pressure are higher than normal.
- Represented as the “crest” of a pressure wave diagram.
- Rarefactions:
- Particles are pulled further apart.
- Local density and pressure are lower than normal.
- Represented as the “trough” of a pressure wave diagram.
- Analogy: Think of compressions and rarefactions like crowded and empty sections in a moving traffic jam.
Example
A tuning fork vibrates at \( f = 256 \,\text{Hz} \). The speed of sound in air is \( v = 340 \,\text{m/s} \). Calculate the wavelength of the sound wave in air.
▶️Answer/Explanation
Step (1) – Formula:
\( v = f \lambda \)
Step (2) – Substitution:
\( 340 = 256 \times \lambda \)
Step (3) – Solve:
\( \lambda = \dfrac{340}{256} \approx 1.33 \,\text{m} \)
Final Answer:
The wavelength of the sound wave is \( 1.33 \,\text{m} \).
Example
Explain why astronauts cannot hear each other’s voices directly in outer space.
▶️Answer/Explanation
Key point: Sound waves require a medium (particles) to travel through.
In outer space there is a near-perfect vacuum → no air particles → no compressions and rarefactions can form.
Final Answer:
Astronauts cannot hear each other’s voices in space because sound cannot travel in a vacuum.
Audible Range of Human Hearing
- Humans can detect sound waves only within a certain range of frequencies.
- This range is called the audible frequency range.
(b) Approximate Range:
- Lower limit: \( 20 \,\text{Hz} \) → very low-pitched (infrasonic) sounds below this cannot be heard.
- Upper limit: \( 20{,}000 \,\text{Hz} \) or \( 20 \,\text{kHz} \) → very high-pitched (ultrasonic) sounds above this cannot be heard.
- So, the audible range of humans is approximately \( 20 \,\text{Hz} \) to \( 20 \,\text{kHz} \).
(c) Notes:
- The exact range varies with age and health.
- Children and young people can usually hear higher frequencies (close to 20 kHz).
- With age, the upper limit decreases, often falling below 15 kHz in adults.
Example
A person hears a sound at a frequency of \( 15{,}000 \,\text{Hz} \). Is this sound audible to humans?
▶️Answer/Explanation
Step (1) – Recall human audible range:
\( 20 \,\text{Hz} \leq f \leq 20{,}000 \,\text{Hz} \).
Step (2) – Compare:
Given \( f = 15{,}000 \,\text{Hz} \), which lies within the range.
Final Answer:
Yes, the sound at \( 15{,}000 \,\text{Hz} \) is audible to humans.
Transmission and Speed of Sound
(a) Medium Requirement:
- Sound waves are mechanical longitudinal waves.
- They require a medium (air, liquid, or solid) to travel through, because the vibration of particles is necessary for wave propagation.
- No medium → No sound, which is why sound cannot travel in a vacuum.
(b) Speed of Sound in Air – Experimental Determination:
- The speed of sound can be calculated using the relation:
\( v = \dfrac{d}{t} \)
- Method:
- Measure a known distance \( d \) between a sound source and an observer.
- Use a stopwatch or electronic sensors to measure the time \( t \) taken for the sound to travel.
- Calculate the speed \( v \) by dividing distance by time.
(c) Relative Speeds in Different Media:
- Sound travels at different speeds depending on particle density and bonding strength.
- General Rule: Speed is fastest in solids, slower in liquids, and slowest in gases.
- It is easier for sound waves to go through solids than through liquids because the molecules are closer together and more tightly bonded in solids. Similarly, it is harder for sound to pass through gases than through liquids, because gaseous molecules are farther apart. The speed of sound is faster in solid materials and slower in liquids or gases.
| Medium | Approximate Speed of Sound | Reason |
|---|---|---|
| Air (Gas) | \( 340 \,\text{m/s} \) | Particles far apart → weak collisions |
| Water (Liquid) | \( 1500 \,\text{m/s} \) | Particles closer → faster transfer of vibrations |
| Steel (Solid) | \( 5000 \,\text{m/s} \) | Particles tightly packed → strongest bonding |
Example
An experiment is carried out to measure the speed of sound. A person stands \( 170 \,\text{m} \) away from a wall and claps. The echo is heard after \( 1.0 \,\text{s} \). Calculate the speed of sound in air.
▶️Answer/Explanation
Step (1) – Recall formula:
\( v = \dfrac{d}{t} \).
Step (2) – Distance travelled:
The sound travels to the wall and back, so distance = \( 2 \times 170 = 340 \,\text{m} \).
Step (3) – Time taken:
\( t = 1.0 \,\text{s} \).
Step (4) – Calculate speed:
\( v = \dfrac{340}{1.0} = 340 \,\text{m/s} \).
Final Answer:
Speed of sound in air = \( 340 \,\text{m/s} \).
Effects of Amplitude and Frequency on Sound Waves
(a) Amplitude and Loudness:
- The amplitude of a sound wave corresponds to the maximum displacement of air particles from their equilibrium position.
- A larger amplitude means the particles vibrate more strongly → the sound is louder.
- A smaller amplitude means weaker vibrations → the sound is softer.
- Note: Loudness is a perceived property, but it is directly related to the energy carried by the wave, which is proportional to amplitude squared.
(b) Frequency and Pitch:
- The frequency of a sound wave is the number of vibrations per second, measured in hertz (Hz).
- A higher frequency means more vibrations per second → the sound has a higher pitch (e.g., a whistle).
- A lower frequency means fewer vibrations per second → the sound has a lower pitch (e.g., a drumbeat).
- Note: Human audible range ≈ \( 20 \,\text{Hz} \) to \( 20{,}000 \,\text{Hz} \). Outside this range, sounds cannot be heard even if amplitude is large.
| Wave Property | Effect on Sound | Example |
|---|---|---|
| Amplitude ↑ | Sound becomes louder | Turning up the volume of a speaker |
| Amplitude ↓ | Sound becomes softer | Whisper compared to shouting |
| Frequency ↑ | Higher pitch | Violin string sound vs. cello |
| Frequency ↓ | Lower pitch | Bass drum vs. flute |
Example
A tuning fork produces a sound of frequency \( 440 \,\text{Hz} \) with a small amplitude. If the amplitude is doubled but the frequency remains the same, what changes will be noticed in the sound?
▶️Answer/Explanation
Step (1) – Frequency:
The frequency remains \( 440 \,\text{Hz} \), so the pitch does not change.
Step (2) – Amplitude:
Doubling the amplitude means the sound wave carries more energy.
Step (3) – Perception:
The sound will be perceived as louder, but still have the same pitch.
Final Answer:
The sound becomes louder, but the pitch remains unchanged.
Echo – Reflection of Sound Waves
- An echo is a sound that is heard after being reflected from a surface such as a wall, building, or mountain.
- It is simply the reflection of sound waves back to the listener’s ears.
- The same law of reflection applies: angle of incidence = angle of reflection.
(b) Conditions for an Echo:
- The reflecting surface must be large, smooth, and hard (e.g., cliff, wall, tunnel).
- The reflected sound must reach the ear at least \( 0.1 \,\text{s} \) after the original sound to be heard as a separate sound.
- Since the speed of sound in air is approximately \( 340 \,\text{m/s} \), the minimum distance required for an echo is: \( d = \dfrac{v \times t}{2} = \dfrac{340 \times 0.1}{2} = 17 \,\text{m} \). So, the reflecting surface must be at least 17 m away.
(c) Applications of Echoes:
- Used in sonar (sound navigation and ranging) to measure water depth or locate underwater objects.
- Used by animals such as bats and dolphins for navigation and hunting (echolocation).
- Echo sounding is used in geology to locate oil and gas reserves underground.
Example
A person shouts near a large wall and hears an echo after \( 0.5 \,\text{s} \). If the speed of sound in air is \( 340 \,\text{m/s} \), calculate the distance of the wall from the person.
▶️Answer/Explanation
Step (1) – Echo path:
The sound travels to the wall and back. So, total distance = \( 2d \).
Step (2) – Formula:
\( v = \dfrac{\text{distance}}{\text{time}} \). \( 340 = \dfrac{2d}{0.5} \).
Step (3) – Solve:
\( 2d = 340 \times 0.5 = 170 \). \( d = \dfrac{170}{2} = 85 \,\text{m} \).
Final Answer:
The wall is \( 85 \,\text{m} \) away.
Ultrasound
- Ultrasound is sound with a frequency higher than 20 kHz (\( 20,000 \,\text{Hz} \)).
- This is above the upper limit of the human hearing range (\( 20 \,\text{Hz} \) to \( 20 \,\text{kHz} \)).
- While humans cannot hear ultrasound, many animals (e.g., bats, dogs, dolphins) can detect these higher frequencies.
(b) Properties of Ultrasound Waves: 
- They are longitudinal waves (like normal sound), consisting of compressions and rarefactions.
- They can travel through solids, liquids, and gases, but need a medium.
- They reflect strongly at boundaries between two different media (e.g., tissue and bone).
(c) Applications of Ultrasound:
- Medical imaging: Used in prenatal scans to create images of unborn babies.
- Industrial testing: Used to detect cracks or faults in metals and machinery parts.
- Sonar: Ships use ultrasound waves to measure sea depth and locate underwater objects.
- Cleaning: High-frequency vibrations used in ultrasonic cleaners to remove dirt from delicate items.
Example
An ultrasound wave of frequency \( 5.0 \times 10^6 \,\text{Hz} \) travels through human tissue at a speed of \( 1500 \,\text{m/s} \). Calculate its wavelength.
▶️Answer/Explanation
Step (1) – Formula:
\( v = f \lambda \) → \( \lambda = \dfrac{v}{f} \).
Step (2) – Substitution:
\( \lambda = \dfrac{1500}{5.0 \times 10^6} \).
Step (3) – Solve:
\( \lambda = 3.0 \times 10^{-4} \,\text{m} \) = \( 0.30 \,\text{mm} \).
Final Answer:
The wavelength of the ultrasound wave is \( 0.30 \,\text{mm} \).