CIE IGCSE Physics (0625) Sound Study Notes - New Syllabus
CIE IGCSE Physics (0625) Sound Study Notes
LEARNING OBJECTIVE
- Understanding the concepts of Sound
Key Concepts:
- Sound Waves
- Transmission of Sound and Speed in Different Media
- Experiment: Measuring the Speed of Sound in Air
- Amplitude, Frequency, and Echoes in Sound Waves
- Ultrasound
Sound Waves
Sound Waves
1. Production of Sound by Vibrating Sources
Sound is produced when an object vibrates and causes the surrounding air particles to vibrate too.These vibrations travel through the air (or another medium) as mechanical waves.
Examples of vibrating sources that produce sound:
- Vocal cords in the throat
- Strings on a guitar
- Drum skins
- Speaker cones
2. Compression and Rarefaction
- As sound travels, it forms a repeating pattern of:
- Compressions: Regions where air particles are close together (high pressure)
- Rarefactions: Regions where air particles are spread apart (low pressure)
- This alternating compression and rarefaction allows the sound wave to propagate through the medium.
3. Longitudinal Nature of Sound Waves
- Sound waves are longitudinal waves, meaning:
- The vibrations of particles are in the same direction as the direction of the wave’s travel.
- Unlike transverse waves (where vibrations are at right angles), in longitudinal waves, air particles move back and forth along the path of the wave.
- Longitudinal waves need a medium (solid, liquid, or gas) to travel through.
4. Human Hearing Range
Humans can typically hear sound waves in the frequency range of: 20 Hz to 20,000 Hz (or 20 kHz)
- Sounds below 20 Hz are called infrasound.
- Sounds above 20,000 Hz are called ultrasound.
- The ability to hear high-frequency sounds decreases with age or hearing damage.
Example
A tuning fork is struck and held near a student’s ear. The student hears a clear tone. Explain how the sound is produced and travels to the student.
▶️ Answer/Explanation
Striking the tuning fork makes its prongs vibrate back and forth.
These vibrations cause the surrounding air particles to also vibrate in the same direction – forming a longitudinal wave.
This creates alternating regions of compression (high pressure) and rarefaction (low pressure).
These pressure waves travel through the air and reach the student’s ear, vibrating the eardrum and allowing them to hear the sound.
Conclusion: Sound is produced by a vibrating source and transmitted through air as a longitudinal wave.
Example
A loudspeaker cone vibrates forward and backward. Describe how this creates sound in the air.
▶️ Answer/Explanation
When the speaker cone moves forward, it pushes air particles together, forming a compression.
When the cone moves backward, it pulls air particles apart, creating a rarefaction.
This repeated motion generates a wave of compressions and rarefactions that travel outward from the speaker – a longitudinal sound wave.
As these pressure variations reach your ear, they make the eardrum vibrate, allowing you to hear the sound.
Conclusion: A speaker cone produces sound by causing alternating compressions and rarefactions in air, demonstrating the longitudinal nature of sound waves.
Transmission of Sound and Speed in Different Media
Transmission of Sound and Speed in Different Media
1. A Medium is Needed for Sound to Travel
Sound is a mechanical wave, which means it relies on particle vibrations to transfer energy.Therefore, sound cannot travel through a vacuum, since there are no particles to vibrate (unlike light, which can travel in space).
Examples:
- Sound can travel through air, water, metal, etc.
- Sound cannot travel through outer space (no medium).
2. Speed of Sound in Different Media
The speed of sound depends on the density and particle arrangement of the medium.
In general:
- Fastest in solids – particles are closely packed, allowing vibrations to pass quickly.
- Slower in liquids – particles are less tightly packed.
- Slowest in gases – particles are far apart, making vibrations slower to transfer.
3. Approximate Speed of Sound in Air
The speed of sound in dry air at room temperature (20°C) is about:
\( \boxed{340 \, \text{m/s}} \)
It can vary slightly based on:
- Temperature (higher temperature → faster speed)
- Humidity (more moisture → faster speed)
Typical Sound Speeds in Different Media:
| Medium | Speed of Sound (m/s) |
|---|---|
| Air (gas) | ~340 |
| Water (liquid) | ~1500 |
| Steel (solid) | ~5000 |
Example
In a science experiment, a ringing alarm clock is placed inside a sealed glass jar. Air is gradually pumped out of the jar using a vacuum pump. After a while, the sound of the ringing can no longer be heard, even though the clock is still ringing. Explain why.
▶️ Answer/Explanation
Sound requires a medium (like air) to travel, as it is a mechanical wave.
When air is removed from the jar, a vacuum is created — meaning no particles are present to carry the vibrations.
Without air, the sound produced by the alarm cannot reach the observer’s ears.
Conclusion: This demonstrates that sound cannot travel in a vacuum because it needs a medium for transmission.
Example
In a metal rod, sound travels at a speed of \( 5000 \, \text{m/s} \). How long will it take for a sound pulse to travel through a 2.5 m long steel rod?
▶️ Answer/Explanation
Use the formula:
\( \text{Time} = \dfrac{\text{Distance}}{\text{Speed}} \)
Substitute the values:
\( \text{Time} = \dfrac{2.5}{5000} \, \text{s} \)
Calculate:
\( \text{Time} = \boxed{0.0005 \, \text{s}} = 0.5 \, \text{ms} \)
Conclusion: It takes \( \boxed{0.5 \, \text{milliseconds}} \) for sound to travel through the 2.5 m long metal rod.
Experiment: Measuring the Speed of Sound in Air
Experiment: Measuring the Speed of Sound in Air
Equipment Needed:
- Two people with stopwatches (or one person with a stopwatch and a wall)
- Large open space (e.g., school field)
- Starting signal (e.g., a wooden block and hammer or clap)
- Measuring tape (to measure distance)
Method 1: Two-Person Timing Method
- Person A stands at a known distance (e.g. 100 m) from Person B.
- Person A hits two wooden blocks together to make a loud, sharp sound and raises the blocks visibly at the same moment.
- Person B starts the stopwatch when they see the blocks collide and stops it when they hear the sound.
- The time recorded is the time taken for the sound to travel the known distance.
Method 2: Echo Method (One Person)
- Stand a known distance (e.g. 50 m) from a large flat wall.
- Clap or make a loud sound.
- Start the stopwatch when you make the sound and stop it when the echo is heard.
- The sound travels to the wall and back, so the total distance is twice the measured distance.
Formula Used:
\( \text{Speed} = \dfrac{\text{Distance}}{\text{Time}} \)
For the echo method: \( \text{Speed} = \dfrac{2 \times \text{Distance to wall}}{\text{Echo time}} \)
Example
A student stands 80 m away from a large wall. They clap and hear the echo 0.48 seconds later. Calculate the speed of sound in air based on this measurement.
▶️ Answer/Explanation
Total distance travelled by the sound is:
\( \text{Distance} = 2 \times 80 = 160 \, \text{m} \)
Time taken = 0.48 s
Use the formula:
\( \text{Speed} = \dfrac{160}{0.48} = \boxed{333.3 \, \text{m/s}} \)
Conclusion: The measured speed of sound in air is approximately \( \boxed{333 \, \text{m/s}} \), which is consistent with the known range of 330–350 m/s.
Amplitude, Frequency, and Echoes in Sound Waves
Amplitude, Frequency, and Echoes in Sound Waves
1. Effect of Amplitude and Frequency on Sound
Amplitude:
Amplitude is the height of the wave from the rest position to the crest or trough.It determines the loudness of a sound.
- Greater amplitude → louder sound.
- Smaller amplitude → softer sound.
Frequency:
Frequency is the number of vibrations (or waves) per second, measured in hertz (Hz).It determines the pitch of a sound.
- Higher frequency → higher pitch (e.g. a whistle).
- Lower frequency → lower pitch (e.g. a drum).
2. Echo – Reflection of Sound
An echo is heard when sound waves are reflected from a hard surface (e.g., a wall or mountain).The reflected sound reaches your ears after a delay, making it sound separate from the original sound.
- To hear a clear echo, the reflected sound must arrive at least 0.1 seconds after the original sound.
- This typically requires the reflecting surface to be at least 17 m away in air.
- Used in:
- Measuring distance using sonar or echolocation
- Detecting objects underwater (submarines, fish finders)
Example:
A student shouts loudly in a canyon and hears an echo 1.2 seconds later. Then the student whistles softly and hears no echo. Explain why the echo is heard in the first case and not the second, and how loudness and pitch differ in both sounds.
▶️ Answer/Explanation
1. Shouting:
The shout has a large amplitude, which means it is loud. The sound wave travels to the canyon wall, reflects back, and the echo is heard after 1.2 seconds.
A shout also usually has a lower frequency, resulting in a lower pitch.
2. Whistling:
The whistle has a small amplitude, so it is quiet. Although the sound still reflects off the canyon wall, it may be too quiet to be heard as an echo.
A whistle has a higher frequency, so it has a higher pitch than a shout.
3. Echo Requirement:
For a distinct echo to be heard, the reflected sound must be:
- Loud enough to be heard (i.e. large amplitude)
- Delayed by at least 0.1 seconds
Conclusion:
The shout created a loud, low-pitched sound that reflected clearly and was heard as an echo. The soft, high-pitched whistle produced a weak sound that was not heard after reflection.
Ultrasound
Ultrasound
Ultrasound is sound with a frequency greater than 20,000 Hz (20 kHz).
- This is above the upper limit of human hearing.
- Ultrasound waves are longitudinal mechanical waves that require a medium to travel through.
Uses of Ultrasound
Medical Scanning (Ultrasound Imaging):
- Ultrasound is used to scan soft tissues in the body, especially for fetal monitoring during pregnancy.
- It is safe as it does not involve ionizing radiation (unlike X-rays).
- The waves are reflected at boundaries between different tissues, producing an image.
Non-Destructive Testing (NDT):
- Ultrasound can detect flaws (cracks, air gaps) inside solid materials like metal or plastic without damaging them.
- The wave reflects off flaws inside the object and returns to a detector.
Sonar (Sound Navigation and Ranging):
- Used by submarines and ships to detect objects or measure ocean depth.
- Ultrasound pulses are sent underwater, and the echo from the seabed or object is measured.
- The time for the pulse to return is used to calculate depth.
Formula for Depth or Distance
\( \text{Distance} = \dfrac{\text{Speed} \times \text{Time}}{2} \)
We divide by 2 because the measured time is for the wave to travel to the object and back (round trip).
Example :
A sonar device on a boat sends an ultrasound pulse down to the seabed. The echo returns after 1.5 seconds. If the speed of sound in water is \( 1500 \, \text{m/s} \), calculate the depth of the sea.
▶️ Answer/Explanation
Use the formula:
\( \text{Distance} = \dfrac{\text{Speed} \times \text{Time}}{2} \)
Substitute values:
\( \text{Distance} = \dfrac{1500 \times 1.5}{2} \)
Calculate:
\( \text{Distance} = \dfrac{2250}{2} = \boxed{1125 \, \text{m}} \)
Conclusion: The depth of the sea is approximately \( \boxed{1125 \, \text{m}} \).
Example:
In a prenatal ultrasound scan, a gel is applied to the mother’s abdomen and an ultrasound probe sends sound waves into the body. Some waves are reflected back from inside the body and used to form an image. Explain why ultrasound reflects at certain points and why gel is used.
▶️ Answer/Explanation
Ultrasound waves travel through soft tissues in the body.
When the waves reach a boundary between two different tissues (e.g., muscle and fluid, or tissue and bone), part of the wave is reflected, and the rest continues deeper.
The reflected waves (echoes) are detected by the probe and used to build a digital image of internal structures.
The gel is used to eliminate air between the probe and the skin, because ultrasound cannot travel well through air – air reflects most of the wave, reducing signal quality.
Conclusion: Reflections occur at boundaries between tissues of different densities. The returning echoes help form a real-time image. The gel ensures proper transmission of sound into the body by removing air gaps.