Sound, ultrasound and infrasound
What you'll learn
- How sound travels as a longitudinal wave.
- How vibrations in solids can become sound waves, and how sound can make solids vibrate.
- What ultrasound and infrasound are, and how they are used.
- How speed, frequency and wavelength are linked when sound enters a new medium.
In Edexcel GCSE Physics 1PH0, the points labelled “P” are for Separate Physics. Several points about the ear, ultrasound and infrasound are marked Higher Tier in the specification, so they are especially important if you are sitting Higher.
Sound is a mechanical wave
A sound wave is caused by something vibrating. For example, a loudspeaker cone moves back and forth, pushing and pulling the nearby air particles.
Medium and mechanical wave
A medium is the material that a wave travels through, such as air, water or metal. A mechanical wave is a wave that needs a medium to travel through. Sound is a mechanical wave, so it cannot travel through a vacuum.
In air, sound travels as a longitudinal wave. This means the particles vibrate backwards and forwards in the same direction that the wave travels.
The crowded regions are called compressions and the spread-out regions are called rarefactions.
Sound transfers energy, not matter
In a sound wave, particles vibrate about fixed positions. The particles do not travel all the way from the loudspeaker to your ear; the disturbance transfers energy through the medium.
Frequency, wavelength and wave speed
The frequency, symbol fff, is the number of complete vibrations each second. It is measured in hertz, Hz.
The wavelength, symbol λ\lambdaλ, is the distance from one point on a wave to the same point on the next wave. For a sound wave, you can measure it from one compression to the next compression.
The wave speed, symbol vvv, is how fast the wave disturbance travels through the medium. It is measured in metres per second, m/s.
The wave equation is:
v=fλv = f\lambdav=fλFor Edexcel Waves, you should be confident using this equation and rearranging it.
Calculating the wavelength of a sound wave
A sound wave has frequency 500 Hz and travels through air at 340 m/s. Calculate its wavelength.
- Use the wave equation and rearrange for wavelength: λ=vf\lambda = \frac{v}{f}λ=fv.
- Substitute the values with units: λ=340 m/s500 Hz\lambda = \frac{340\ \text{m/s}}{500\ \text{Hz}}λ=500 Hz340 m/s.
- Calculate the answer: λ=0.68 m\lambda = 0.68\ \text{m}λ=0.68 m.
Mixing up loudness and pitch
A louder sound has a bigger amplitude, meaning a bigger vibration or pressure change. A higher-pitched sound has a higher frequency. Loudness and pitch are not the same thing.
Changing between sound waves and vibrations in solids
A vibrating solid can produce sound. For example, a loudspeaker cone vibrates, pushing air particles to make compressions and rarefactions.
Sound can also make a solid vibrate. For example, sound waves hitting your eardrum cause it to vibrate.
Transducer
A transducer is a device that converts one type of signal or energy transfer into another. In this topic, transducers convert between sound waves, vibrations in solids and sometimes electrical signals.
This conversion only works well over a limited frequency range. Real objects have mass, stiffness and damping, so they cannot respond equally well to every frequency. Very high frequencies may be too fast for a part to follow properly, while very low frequencies may not cause a large enough useful vibration.
Explaining why a detector misses high-pitched sound
A microphone has a small solid diaphragm that should vibrate when sound reaches it. Explain why it may not detect very high-frequency sound well.
- The sound wave must first make the diaphragm vibrate at the same frequency as the sound.
- At very high frequency, the pressure changes direction extremely quickly, so the diaphragm’s mass makes it difficult to speed up, slow down and reverse fast enough.
- The diaphragm vibrates with a much smaller amplitude or gives a distorted signal, so the microphone works only over a limited frequency range.
How the human ear works
Your ear is a biological system for converting sound waves into electrical impulses that your brain can interpret.
The process is:
- Sound waves enter the ear canal, the tube leading into the ear.
- The eardrum, a thin membrane, vibrates.
- The ossicles, three tiny bones, pass on and amplify the vibrations.
- The cochlea, a fluid-filled spiral structure, contains tiny hair cells that vibrate.
- Different hair cells respond best to different frequencies.
- The auditory nerve carries electrical impulses to the brain.
Humans normally hear frequencies from about 20 Hz to 20,000 Hz. This useful hearing range is limited because the eardrum, ossicles and hair cells only convert vibrations effectively over a certain range.
Ultrasound and infrasound
Ultrasound and infrasound
- Ultrasound is sound with frequency greater than 20,000 Hz.
- Infrasound is sound with frequency less than 20 Hz.
Ultrasound is useful because high-frequency waves have short wavelengths, so they can give detailed information. Infrasound and low-frequency seismic waves are useful because they can travel very long distances and can reveal large-scale structures.
The diagram below summarises three important applications.
Sonar
Sonar uses pulses of ultrasound underwater. A pulse is sent out, reflects from an object or the seabed, and returns as an echo.
The time delay, symbol Δt\Delta tΔt, is the time between sending the pulse and receiving the echo. Because the pulse travels to the object and back, the one-way distance is half the total distance travelled:
d=vΔt2d = \frac{v\Delta t}{2}d=2vΔtFinding depth using sonar
A boat sends an ultrasound pulse to the seabed. The echo returns after 0.20 s. The speed of sound in water is 1500 m/s. Calculate the depth of the seabed.
- Calculate the total distance travelled by the pulse and echo: vΔt=1500×0.20=300 mv\Delta t = 1500 \times 0.20 = 300\ \text{m}vΔt=1500×0.20=300 m.
- Recognise that 300 m is the down-and-up distance, not just the depth.
- Divide by 2 to find the one-way depth: d=3002=150 md = \frac{300}{2} = 150\ \text{m}d=2300=150 m.
Foetal scanning
In foetal ultrasound scanning, a transducer sends ultrasound into the body and receives reflected echoes. Some ultrasound reflects when it reaches a boundary between different tissues, such as fluid and skin.
A computer uses the time taken for echoes to return and their strength to build an image. Ultrasound is used instead of X-rays because it is non-ionising, meaning it does not have enough energy to remove electrons from atoms in the way ionising radiation can.
Why gel is used in ultrasound scans
Ultrasound reflects strongly from air gaps. The gel removes air between the transducer and the skin, so more ultrasound enters the body.
Exploring the Earth’s core
Earthquakes produce seismic waves, which are vibrations that travel through the Earth. Some of these waves have very low frequencies.
Two important types are:
- P-waves, which are longitudinal waves and can travel through solids and liquids.
- S-waves, which are transverse waves and cannot travel through liquids.
Scientists compare where seismic waves are detected after earthquakes. Missing S-waves suggest the Earth has a liquid outer core. Changes in P-wave direction and speed produce shadow zones, which help scientists infer the structure of the core.
When sound enters a different medium
Transmission means a wave passing from one medium into another, such as from air into water.
When sound enters a new medium:
- The frequency stays the same because it is set by the source.
- The wave speed changes because speed depends on the medium.
- The wavelength changes so that v=fλv=f\lambdav=fλ still works.
If a wave enters the new medium at an angle and changes speed, it may change direction. This is called refraction.
Comparing wavelength in air and water
A 1000 Hz sound wave travels at 340 m/s in air and 1500 m/s in water. Compare its wavelength in each medium.
- In air, use λ=vf\lambda = \frac{v}{f}λ=fv, so λ=3401000=0.34 m\lambda = \frac{340}{1000} = 0.34\ \text{m}λ=1000340=0.34 m.
- The frequency stays 1000 Hz when the sound enters water because the source still vibrates at the same rate.
- In water, λ=15001000=1.5 m\lambda = \frac{1500}{1000} = 1.5\ \text{m}λ=10001500=1.5 m, so the wavelength increases because the wave speed is higher.
Saying frequency changes at a boundary
When a wave enters a different medium, its speed may change and its wavelength may change, but its frequency stays the same.
Core practical: measuring wave speed, frequency and wavelength
For the core practical, you need to investigate whether equipment is suitable for measuring the speed, frequency and wavelength of waves in a solid and a fluid.
A fluid means a liquid or a gas. Water and air are both fluids.
Possible approaches include:
- For water waves in a ripple tank, use a vibration generator to set the frequency, then measure the wavelength from the spacing between wavefronts.
- For sound in air, use a loudspeaker, signal generator, microphones and an oscilloscope or data logger. Electronic timing is much better than a stopwatch because sound is fast.
- For waves in a solid, such as a stretched string or spring, use a vibration generator or send a pulse along the material. Measure distance and time, or measure frequency and wavelength.
To improve reliability, measure several wavelengths and divide by the number of wavelengths. Repeat readings and calculate a mean.
Finding wave speed from repeated wavelengths
In a ripple tank, 10 wavelengths measure 0.25 m. The vibration generator frequency is 12 Hz. Calculate the wave speed.
- Find one wavelength by dividing the total distance by the number of wavelengths: λ=0.2510=0.025 m\lambda = \frac{0.25}{10} = 0.025\ \text{m}λ=100.25=0.025 m.
- Use the wave equation: v=fλv = f\lambdav=fλ.
- Substitute and calculate: v=12×0.025=0.30 m/sv = 12 \times 0.025 = 0.30\ \text{m/s}v=12×0.025=0.30 m/s.
Choosing suitable equipment
A stopwatch is usually unsuitable for very fast waves over short distances because human reaction time is too large. Use sensors, microphones, oscilloscopes or data loggers when timing needs to be precise.
In the exam
- If an echo is involved, remember the wave usually travels there and back, so divide the total distance by 2.
- When a wave changes medium, keep frequency the same and use v=fλv=f\lambdav=fλ to decide what happens to wavelength.
- For practical questions, explain why the equipment is suitable: mention precision, repeated readings, measuring several wavelengths, and avoiding echoes or reaction-time errors.
Check yourself
- What is the difference between ultrasound and infrasound?
- Why does the wavelength of sound change when it enters water from air?
- How does the ear convert a sound wave into an electrical signal?
