Sound
Sound waves are longitudinal. Unlike transverse waves (where particles move perpendicular to the wave direction), in a sound wave, the particles of the medium oscillate back and forth parallel to the direction of energy transfer.
Building on this particle motion, we define two key features:
- Compression: A region where particles are pushed close together (high pressure).
- Rarefaction: A region where particles are spread apart (low pressure).
The regions of compression and rarefaction are the physical manifestation of the longitudinal particle oscillation. As the wave travels, these regions move through the medium, but the individual particles only vibrate around a fixed position.
Speed of Sound in Air: The approximate speed of sound in air is 330–350 m/s. Note that this value changes with temperature and the medium.
Audible Range: The approximate range of frequencies audible to the human ear is 20 Hz to 20,000 Hz (20 kHz). Sounds below 20 Hz are called infrasound, and sounds above 20,000 Hz are called ultrasound.
Loudness is determined by the amplitude of the sound wave. Greater amplitude (more energy) results in a louder sound.
Echo: An echo is the reflection of sound waves from a surface (such as a wall or cliff). The sound travels to the surface and bounces back to the listener.
Medical Scanning: Ultrasound pulses are sent into soft tissue. Different tissues reflect different amounts of sound, creating an image. The depth of a structure is calculated using the time taken for the echo to return.
Sonar (Sound Navigation and Ranging): Used to measure the depth of the sea or locate objects underwater. A pulse is sent from a ship, reflects off the seabed, and returns.
Calculation Principle:
In all these applications, we use the formula:
v = \frac{d}{t}
Where:
- v is the speed of sound in the specific medium (e.g., v_{air}, v_{water}, or v_{tissue}). You must use the speed appropriate for the material the wave is traveling through.
- d is the total distance traveled by the wave.
- t is the time taken for the wave to travel that distance.
For echo problems (where the wave goes there and back), the total distance is 2 \times \text{depth} (or 2 \times \text{distance to wall}). Therefore:
v = \frac{2d}{t}
Rearranging for depth (d):
d = \frac{v \times t}{2}
Step 1: Identify the variables.
- Speed (v) = 1540 m/s (Note: We use the speed for tissue, not air).
- Time (t) = 0.002 s (This is the total time for the round trip).
Step 2: Choose the correct formula.
Since the wave travels to the organ and back, the distance traveled by the sound is 2d.
v = \frac{2d}{t}
Rearranging for depth (d):
d = \frac{v \times t}{2}
Step 3: Substitute and calculate.
d = \frac{1540 \times 0.002}{2}
d = \frac{3.08}{2}
d = 1.54 \text{ m}
Answer: The depth of the organ is 1.54 m.
- Two students stand at least 100 m apart (to minimize reaction time error).
- Student A makes a loud sound (e.g., clapping wooden blocks) while simultaneously starting a stopwatch.
- Student B sees the clap and stops their stopwatch when they hear the sound.
- Measure the distance (d) between them with a tape measure or trundle wheel.
- Calculate speed using v = \frac{d}{t}.
Why this works: The light from the clap reaches Student B almost instantly, so we assume the time recorded is purely for sound to travel distance d.
Correct Understanding: Pitch is directly proportional to frequency. Higher frequency means higher pitch. This is a common conceptual error because students sometimes confuse 'frequency' with 'wavelength' (where higher frequency means shorter wavelength). Remember: High Frequency = High Pitch.
Correct Understanding: All sound waves (including ultrasound) are longitudinal. The particles oscillate parallel to the direction of energy transfer. Transverse waves (like light) have perpendicular oscillations.
When to use: When asked to state the approximate speed of sound in air.
Why examiners accept this: Examiners look for the specific range 330 m/s to 350 m/s. Values outside this range may be marked incorrect depending on the precision required. Stating '340 m/s' is also widely accepted as it falls within the range.
Example Usage:
- Question: State the approximate speed of sound in air.
- Correct Answer: 330 m/s to 350 m/s (or any single value within this range, e.g., 340 m/s).
When to use: When asked why a student's calculated value for the speed of sound is different from the accepted value (usually lower).
Why examiners accept this: The most common source of error in school experiments is human reaction time. If the time measured is too short, the denominator in v=d/t is small, but if the start/stop delay is significant relative to the travel time, it introduces large percentage errors. Also, wind can affect results.
Example Usage:
- Question: Explain why the students' value for the speed of sound might be different from the accepted value.
- Correct Answer: Reaction time of the students causes an error in measuring the time interval. OR The distance/time interval is too small, making reaction time errors significant.
Step 2: Use formula for echo: d = \frac{v \times t}{2}.
Step 3: Calculate: d = \frac{1500 \times 2.2}{2} = \frac{3300}{2} = 1650 m.
Answer: 1650 m
Step 2: Rearrange for frequency: f = \frac{v}{\lambda}.
Step 3: Substitute: f = \frac{340}{0.28}.
Step 4: Calculate: f \approx 1214 Hz (or 1200 Hz if using 330 m/s or rounding appropriately as per markscheme tolerance).
Answer: 1214 Hz (Accept range based on speed used, e.g., 1200 Hz).
Explanation: The normal human hearing range is 20 Hz to 20,000 Hz (20 kHz). Since 14 kHz (or 14,000 Hz) lies within this range (20 < 14000 < 20000), it is audible.