Properties and Applications of Sound
Sound waves are longitudinal vibrations that transfer energy through a medium, allowing us to hear. This topic covers how sound is produced and perceived, and how its reflective properties are used in technologies like sonar and medical ultrasound.
Part of the ESAT Physics syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- Sound waves are generated by vibrating sources and require a medium (solid, liquid, or gas) to propagate; they cannot travel through a vacuum.
- Sound is a longitudinal wave, where particles of the medium oscillate parallel to the direction of energy transfer, creating regions of compression and rarefaction.
- The perceived loudness of a sound is related to the wave's amplitude, while the pitch is determined by its frequency.
- The typical range of human hearing is from 20 Hz to 20,000 Hz (or 20 kHz).
- An echo is a sound wave that has been reflected off a surface. This phenomenon can be used to measure distance.
- Ultrasound refers to sound with a frequency above 20 kHz. It is used in medical imaging and sonar systems because its high frequency allows for detailed resolution.
Diagram
› Why does this happen?
Why can't sound travel through space?
Sound is a mechanical wave, which means it needs a medium (particles) to travel through. A vibrating object, like a guitar string, makes the particles next to it vibrate. These vibrations are passed on, creating a series of compressions (areas where particles are bunched up) and rarefactions (areas where they are spread out) that travel through the medium. This process transfers energy. The vacuum of space is empty – there are no particles to compress or pass the vibration along. Without a medium, sound cannot travel.
Why do you divide by 2 when calculating distance with echoes?
The time delay measured is for the sound's entire round trip: from the source, to the object, and all the way back again. The basic formula `distance = speed × time` would calculate this total distance travelled. Since we usually want to know the distance just to the object (a one-way trip), we must divide the total round-trip distance by two.
Why is high-frequency ultrasound used for detailed scans?
The level of detail (or resolution) in an image from a wave depends on its wavelength. To see very small objects, you need a wave with a very short wavelength; a wave can't detect details smaller than its own wavelength. The wave equation is `wavespeed = frequency × wavelength`. Since the speed of sound is constant in a specific medium (like body tissue), this equation tells us that to get a very short wavelength, we must use a very high frequency. Ultrasound uses high frequencies to produce the short wavelengths needed to resolve small internal structures and create a detailed image.
Formulae
distancetoobject = (speed × timedelay) / 2 To calculate the one-way distance to a reflecting surface when you know the speed of sound and the time taken to hear the echo (the round-trip time).
Definitions
- Longitudinal Wave
- A wave in which the oscillations of the medium's particles are parallel to the direction of energy propagation. Sound is a key example.
- Echo
- A reflected sound wave that is heard after the original sound. The time delay is due to the extra distance the reflected wave travels.
- Ultrasound
- Sound waves with frequencies above the upper limit of human hearing (> 20 kHz).
Worked examples
A bat emits an ultrasound pulse to locate an insect. The pulse travels at 340 m/s in air. The echo from the insect is detected 0.05 seconds after the pulse was emitted. How far away is the insect from the bat?
- 1
Identify the given values:
speed of sound (v) = 340 m/s, and the total time for the echo (t) = 0.05 s.
- 2
Recognise that the time given is for the sound to travel to the insect and return.
The distance to the insect is based on the one-way travel time.
- 3
Calculate the one-way travel time by halving the total time:
0.05 s / 2 = 0.025 s - 4
Use the formula distance = speed × time to find the distance to the insect:
Distance = 340 m/s × 0.025 s - 5
To calculate 340 × 0.025 without a calculator, note that 0.025 is the same as 1/40.
So, the calculation is 340 / 40.
- 6
This simplifies to 34 / 4, which is 17 / 2 = 8.5 m.
Answer: 8.5 m
A sound wave in air has a frequency of 220 Hz. The distance between the centre of a compression and the centre of the next rarefaction is 0.75 m. What is the speed of the sound wave?
- 1
The distance from a compression to the next rarefaction is half a wavelength.
So lambda / 2 = 0.75 m, which gives lambda = 1.5 m.
- 2
Use the wave equation linking speed, frequency and wavelength:
v = f × lambda - 3
Substitute the values:
v = 220 Hz × 1.5 m - 4 v = 330 m/s
Answer: 330 m/s
Common mistakes
- ×In echo calculations, forgetting to halve the total measured time. The time delay is for the sound's round trip (there and back), so the one-way distance requires the one-way time.
- ×Misinterpreting an oscilloscope trace of a sound wave. The trace is a graph of pressure vs. time and looks transverse, but it represents a longitudinal sound wave.
- ×Confusing loudness with pitch. Loudness is determined by amplitude, while pitch is determined by frequency. A high-frequency, low-amplitude wave is quiet and high-pitched.
No-calculator tips
- ✓When using `d = (v × t) / 2`, it is often easier to halve the speed or the time first, whichever is an even number, before multiplying.
- ✓The speed of sound in air is about 340 m/s. For quick estimates, you can think of it as roughly 1 km every 3 seconds.
- ✓To handle decimal multiplication, convert decimals to fractions. For instance, in the worked example, multiplying by 0.025 is the same as dividing by 40.