Properties of, and experiments with, Sound Waves

including reflection, refraction and diffraction of sound waves, ultrasound uses and infrasound earthquake waves - Characteristics explained with diagrams, uses of sound waves

This page will answer many questions e.g.  Why is sound a longitudinal wave?  Why can't sound travel through a vacuum?  Know that and understand that sound waves and some mechanical waves are longitudinal, and cannot travel through a vacuum. Describe an experiment to measure the speed of sound. What are the uses of ultrasound?  Why can some earthquake waves be the same as sound waves?

Sub-index for this page

(a) The properties of sound waves - longitudinal waves, calculations using wave equation

(b) The reflection, refraction and diffraction of sound waves and wave model

(c) Sound & human hearing, displaying frequencies on a CRO, microphone & loudspeaker

(d) Echoes are reflections of sounds and calculations

(e) Experiments to measure the speed of sound in air and a solid

(f) Ultrasound waves - ultrasonic sound, its uses and calculations

(g) Infrasound waves - animals and earthquake waves

(h) Learning objectives and knowledge for sound and ultrasound

The above diagram gives an idea of a longitudinal wave of sound where the oscillations are in the direction the wave moves.

The oscillations in the same direction as the wave progression, can be considered as vibrations or disturbances in the medium through which the sound wave is travelling.

Reminder: Contrast this oscillation with transverse waves like water waves or electromagnetic radiation where the oscillations are at 90^o to the direction of wave movement.

These oscillations in longitudinal sound waves show areas of compression and rarefaction.

A compression is where the particles of the medium are compressed to a maximum pressure and a rarefaction is where the particles of the medium are spaced out the most to the minimum pressure.

When the particles get squeezed closer together or spaced apart more than 'normal' they will want to return to their rest position.

This they do, driving the wave in a forward direction. So the wave is a continuous series of compressions and decompressions (rarefactions) in which the particles pushed together and then spaced apart again.

In the diagram above, where the lines are close together, imagine the particles e.g. air molecules or atoms in a metal crystal, are also packed together - the opposite is true where the vertical lines are far apart.

In the diagram above for longitudinal sound waves, wave B has twice the frequency and half the wavelength of wave A.

Reminder of the general wave equation applied to sound:

speed of sound wave (m/s) = frequency of sound (Hz) x wavelength of sound wave (m)

in symbolic 'shorthand'    v = f x λ, rearrangements:  f = v ÷ λ   and   λ = v ÷ f

Sound waves that we hear travel at the same speed whatever the frequency, so, with reference to the diagram, if the speed stays constant and you halve the wavelength, you must then double frequency for the equation to be valid.

Where the vertical lines are close together you can imagine the particles in a material being compressed closer to one another (compression) and where the lines are spaced well apart, so are the particles of the medium (the rarefaction).

The frequency of a sound wave (or any wave) equals the number of compressions passing a point per second, and is perceived as the pitch eg of a musical note.

The frequency of sound is the number of vibrations per second (unit hertz, Hz).

The amplitude is the maximum compression with respect to the 'rest line' and is perceived as loudness.

The 'rest line' is effectively the point of no disturbance, zero amplitude - neither compression or rarefaction.

You can think of a sound wave as a pressure wave - a continuous variation of high and low pressure regions of the wave.

The diagram above illustrate the simulation of sound waves by push and pulling on a slinky spring to create pulses of energy being transmitted along the slinky spring (the 'medium'). Its a good simulation of the compression and rarefaction behaviour of a longitudinal wave.

NOTE: Not all frequencies of sound can be transmitted through an object or material.

The nature of the material can affect which frequencies can be transmitted.

The shape, size as well as the structure of an object affects the frequencies that can pass through.

When you hit a solid object that is 'sonorous' e.g. a metal block that rings when structure, it will vibrate-resonate most strongly with certain 'natural' frequencies - you will hear one particular note most loudly e.g. as with a tuning fork in music.

The frequency of sound doesn't change as it passes from one medium to another.

However, the speed does change and therefore the wavelength must change too.

 v = f x λ, rearrangements:  f = v ÷ λ   and   λ = v ÷ f

If the frequency f stays constant, then increase in speed v must be matched by an increase in the wavelength λ to keep the ratio speed / wavelength constant.

Unlike electromagnetic light waves, sound cannot travel through empty space (vacuum) because you need a material substance (gas, liquid or solid) which can be compressed and decompressed to transmit the wave vibration.

The more dense a material, the faster the sound wave travels (its the opposite for light in transparent materials).

Therefore in general for sound: speed in solids > speed in liquids > speed in gases

This is borne out by the data of sound speeds quoted below:

Typically at room temperature, the speed of sound waves in various materials at ~20-25^oC

air 343 m/s (0.34 km/s),

that's why if a thunderstorm is 1 km away, you hear the thunderclap about 3 seconds after the flash of lightning - the speed of light is so much greater than the speed of sound that the flash is virtually instantaneous,

the speed of sound increases in air with temperature, an empirical formula (from experiment, no theoretical basis) I found on the internet is: speed of sound in air in m/s = 331 + 0.6T (where T = 0 to 100^oC).

water 1493 m/s (1.49 km/s),  sea water 1533 (1.53 km/s), high velocity useful in sonar scanning of sea bed

kerosene 1324 m/s (1.32 km/s) (liquid hydrocarbon)

ordinary glass 4540 m/s, pyrex glass 5640 m/s (5.64 m/s), much more dense than water

iron 5130 m/s (5.1 km/s), steel 5790 m/s (5.8 km/s)

rocks 2000 to 7000 m/s (2-7 km/s), the speed tends to be greater in the more dense igneous rocks compared to less dense sedimentary rocks.

for comparison:

longitudinal earthquake P-waves typically travel at 2 to 7 km/s depending on whether the wave is in the Earth's crust, mantle or core and the speed will also depend on density and temperature

Some sound wave calculations

speed of sound wave (m/s) = frequency of sound (Hz) x wavelength of sound wave (m)

 v = λ x f, rearrangements:  f = v ÷ λ   and   λ = v ÷ f

Q1 A typical audible high frequency sound might be 2 kHz.

Speed of sound = 340 m/s

(a) Calculate the wavelength of this sound wave.

λ = v ÷ f = 340 / 2000 = 0.17 m

(b) The musical note 'middle C' has a frequency of ~262 Hz

Speed of sound ~340 m/s

Therefore λ = v ÷ f = 262 / 340 = 0.77 m

(c) How is it that you can hear sounds from room to other rooms in the house?

The second wavelength is similar to the width of a doorway.

Sound will travel throughout a house by both reflection and refraction!

Q2 A musical note has a wavelength of 1.13 m.

If the speed of sound in air is 340 m/s calculate the frequency of the note.

 f = v ÷ λ = 340 / 1.13 = 301 Hz (3 s.f.)

Q3 Two people, 20 m apart, conduct an experiment using iron railings.

One taps the railings and the other places their ear on the iron railings.

(a) Why does the listener hear two sounds?

The listener hears the sound coming through the iron railings themselves and through the air.

(b) Which sound arrives first and why?

The sound coming through the internal vibration of the iron railings arrives first because iron is much more dense than air and sound travels faster the more dense the medium.

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(b) The reflection, refraction and diffraction of sound waves and wave model

1. Reflection of sound waves

When sound waves meet a barrier between two different media they can be reflected just like any other wave (diagram of wavefronts on the right). The angle of incidence will equal the angle of reflection (with respect to the normal at 90^o to the boundary). Any solid surface will reflect sound, though soft material will tend to absorb the sound wave energy.

A flat hard smooth surface is the best reflector of sound waves - think of echoes (reflected sounds) that permeate an empty house with no carpets laid down.

A soft rough surface is the best absorber of sound wave energy - the idea is used in recording studios to minimise unwanted sounds and in ear muffs to protect you from ear damage due to very loud sounds.

In reflection there is no change in speed, frequency or wavelength, only the direction of movement of the wave changes.

2. Refraction of sound waves 

(above right diagram, same diagram as wavefronts in reflection)

When waves, including sound, meet a boundary between two mediums, through which they can pass, there is a change in speed related to the difference in density.

The frequency stays the same, but both the speed and wavelength of the sound wave changes.

If the incident wave fronts are parallel to the boundary surface, there is no change in direction. → IIIIII│IIIIII→

BUT, if the incident angle is NOT 90^o, at the same time as reflection, and the sound waves can penetrate the 2nd medium of different density at a boundary, the wave will change speed, wavelength and direction, but no change in frequency.

Therefore if the sound waves have changed direction, the sound waves are refracted.

The number of waves passing through each medium per second is the same.

Since speed = wavelength x frequency (and frequency is constant),

if speed increases, wavelength must get larger,

if speed decreases, wavelength must get smaller.

Although they are longitudinal sound waves, this is normal wave behaviour just as you see with experiments with transverse light waves or water waves.

3. Diffraction of sound waves

Diffraction is the effect of waves spreading out when passing through a gap or passing by a barrier. In effect, waves bend round corners into the 'shadow zone'! and it doesn't matter if its sound, light or water waves - they all diffract and bend round corners!

You should appreciate that significant diffraction of sound waves only occurs when the wavelength of the sound wave is of the same order of magnitude as the size of the gap or obstacle.

A: There is a relatively small diffraction effect when sound waves pass through a wide gap that is much bigger than the wavelength of the sound wave - but the sound waves still bend round the 'corners' into the 'shadow zones'.

B: You get the maximum spreading or diffraction when the waves pass through a gap of similar size to the wavelength of the incident sound waves.

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(c) Sound and human hearing and displaying frequencies on a CRO

Your ear is designed to collect sound waves and cause the eardrum to vibrate.

Your ear drum resonates with a sound wave hitting it and via some bones and nerve receptors, the 'sound impulses' are transmitted to the brain.

Your ear is designed to collect sound waves - the outer part is a bit like a misshapen convex mirror!

When sound waves funnel down and hit your eardrums, the pressure variations cause them to vibrate and the vibrations are transferred to tiny bones in your ears called ossicles, then through the semicircular canals on to the cochlea.

The cochlea converts the vibrations into electrical nerve signals which transmitted to the brain.

The brain interprets the nerve signals (your sense of hearing) relating to the different frequencies ('pitches') and amplitudes ('volumes').

In some ways the effect is similar to a microphone works!

Human hearing is limited by the size and shape of the eardrum and the structural features of the parts that make the ear's vibration sensing mechanism.

A typical frequency range of human hearing is 20 Hz to 20 kHz, frequencies outside this range would be beyond many people's hearing.

This is clear example of yourself appreciating energy transfer by sound waves - greatly appreciated by somebody who is deaf in one ear!

A higher frequency sound is perceived as a higher pitch (lower frequency = lower pitch).

A sound wave of greater amplitude is perceived as a louder sound (lower amplitude is a softer sound).

 

We experience longitudinal waves as sound, but we can only hear a relatively narrow range of frequencies.

What sound frequencies can we hear and why?

What we can hear as human beings is limited by the size and shape of the eardrum and anything else that is connected and vibrates - resonating with the eardrum. The ossicles, the bones of the middle ear only function well over a limited frequency range. We cannot hear very low pitched or very high pitched sounds.

The bones are most efficient at transmitting frequencies of around 1000 Hz to 3000 Hz (1-3 kHz)

Younger people have a much greater hearing range which can be as wide as 20 Hz (0.02 kHz) to 20 000 Hz (20 kHz).

Unfortunately, as you get older, the upper frequency limit decreases AND your sensitivity decreases - you become harder at hearing - sounds like speech need to be louder.

This is often due to unavoidable wear and tear of the cochlea or auditory nerve.

A personal note (if you pardon the pun!)

The cochlea of my left ear never developed correctly, and so, although all the bones are there and presumably vibrate, no nerve signals are generated, so I've always been deaf in my left ear. My deafness was spotted by a primary school teacher when I was 10 and duly tested to confirm I was indeed deaf in my left ear. I didn't know anything different to monophonic sound, so I've never known what stereophonic sound sounds like! My loving parents didn't seem to realise it either, even though my deafness got me into trouble! One line in my school report, as regards homework, read "plays on his deafness", brilliant eh! It has had very amusing consequences for my classroom teaching (many years ago!). If there was a bit of nonsense on the left at the back of the lab, I always enquired to the right and entirely blamed the wrong group. The students thought this most amusing with many giggles and sniggers and I was regarded as a bit eccentric. Since I couldn't resolve the problem, I once more 'played on my deafness' and accepted at times I'd never find the culprits and sought 'diplomatic' and 'amicable' solutions and survived to teach in comprehensive schools for over 28 years!

 

Sound is important to humans - a means of communication via speech, enjoyment of music etc.

Another good example, which I'm glad to say I have not encountered, is the enormous power of earthquake waves - a huge amounts of energy can be conveyed many miles through the Earth's crust, mantle and even through the core.

Echo sounding is important to bats, they can generate and listen to sounds from 30 Hz to 20 kHz.

Sound waves are produced by mechanical vibrations (e.g. musical instruments) and travel through any medium, gas, liquid or solid, but not vacuum, where there is nothing to vibrate!

In music, if a middle C tuning fork is struck, the two prongs vibrate from side to side 262 times every second ie middle C has a frequency or pitch of 262 Hz.

The pitch of a sound is determined by its frequency and loudness by its amplitude.

The rest line is represented by the horizontal red line on the CRO diagrams below.

The four pictures could represent the sound waves of musical notes recorded by a microphone, converted to an electronic signal and displayed in wave form on an oscilloscope screen (CRO).

You can produce a wide range of frequencies using a signal generator and they can be converted into sound waves.

Note that ...

The shorter the wavelength the higher the frequency (or pitch) of the sound.

The higher the waveform (greater the amplitude) at the point of maximum compression, the louder the sound - and conveying more energy.

So, we can interpret the four signals as follows:

(1) has the smallest amplitude, the softest note (opposite of loudest) - just a whisper!

(2) has the largest amplitude, the loudest note - a good shout out loud!

(3) has the longest wavelength, lowest frequency, lowest pitch e.g. a low note sung by a base singer.

(4) has the shortest wavelength, highest frequency, highest pitch, e.g. a treble note or a squeaky animal.

Some more examples - imagine some musical sounds from a microphone displayed on a CRO

The height of the wave above the baseline (0) gives the amplitude.

In this case, two of the amplitudes, for waves A and D are double that of waves B and C, in other words the height of the wave at the peak is double when measured from the zero horizontal base line.

Wave A will transfer more energy than wave B, and, wave D will transfer more energy than wave C.

(You don't need to know the maths, but the energy in a wave is proportional to the amplitude squared).

I've made the time frame 0.02 seconds so that we can do some simple calculations

Frequency = oscillations per second (Hz)

For waves A and B there are 10 complete cycles of the wave in 0.02 s. Frequency = 10 / 0.02 = 500 Hz.

For waves C and D there are 5 complete cycles of the wave in 0.02 s. Frequency = 5 / 0.02 = 250 Hz.

Assume sketched of the same width of a CRO screen and note the number of waves and their height.

A class of students were listening to single music notes played into a microphone and the result displayed on a cathode ray oscilloscope (CRO).

The students where asked to sketch pictures for each sound produced.

A selection of their sketches is shown above and interpreted below.

FRED is producing a loud highly pitched note (many waves, big amplitude).

JO is producing a soft low pitched note (few waves, small amplitude).

TANYA is producing a loud low pitched note (few waves, large amplitude).

RICKY is producing a highly pitched soft note (many waves, small amplitude)

The microphone and loudspeaker

The diagram above illustrate the principle of a microphone e.g. for a vocalist or a telephone mouthpiece.

A microphone converts the energy of the pressure variation of sound waves into an electrical energy signal in an ac current.

The oscillation of the sound waves vibrates the diaphragm which generates an oscillating signal in the electrical circuit.

In some ways the effect is similar to how your ear works!

The electrical signal could in turn be used to re-generate sound in a loudspeaker.

In loudspeakers, an ac current conducting coil is moved by in a magnetic field to convert electrical energy into sound energy by way of a vibrating cone (diaphragm).

The cone vibrates the air and the oscillations produce the sound waves you hear.

Sound and health and safety

 

The graph shows the safe time limits for levels of sound measured in decibels (Db).

If your work involves noisy machinery, you should wear good quality ear defenders.

Industrial ear muffs have built in layers of soft material to absorb high amplitude sound waves.

We can tolerate loud sounds for a short period, though ear drum damage can still occur.

Apparently the biggest danger of loss of hearing in later life comes from many hours of repetitive fairly loud sound, which at the time doesn't seem harmful - beware of discotheque experience!!!

The frequency limits of hearing of some animals

TYPICAL FREQUENCY HEARING RANGES (Hertz, Hz)

Many animals can hear much higher frequency sounds than us humans.

Bats use high frequency echo location waves to sense the world around them.

Dolphins (and whales) use high frequency sound for communication.

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(d) Echoes are reflections of sounds - calculations

Sound waves are reflected of hard flat surfaces eg walls, but tend to be absorbed by rough soft surfaces eg like foam -used in ear protectors.

Note the difference in echoes between an empty bare room in a house and when it is carpeted and filled with furniture and curtains etc.

The sound is dispersed by both reflection (bouncing off surfaces) and diffraction (bending round corners)

Echoes are heard when you shout towards a hard flat surface and you then hear the reflected sound waves impacting on your inner ear drum.

The further away a reflecting surface is, the longer the time interval between your shout and hearing the echo.

If the wall or side of a mountain or valley is 340 m away, its 2 seconds before you hear the echo (speed of sound 340 m/s).

If the reflecting surface is a km (1000 m) away, its about 6 seconds before hear the echo.

speed = total distance / total time, time = distance/speed, time = 2000/340 = 5.9 s

You can hear sounds from some distance throughout a building or even a wide area outside because the sound waves are reflected and bounced around by all hard flat surfaces BUT sound waves are also diffracted and can therefore bend round corners into your ear!

The further away you are from the sound source, the fainter it will sound for two reasons:

(i) on every reflection some of the sound wave energy is absorbed

(ii) waves naturally spread out from a central source.

Echoes can be used to measure the speed of sound in air, next section (e)

and see also uses of ultrasound in section (f).

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(e) Experiments to measure the speed of sound in air and a solid

(1) Using synchronised microphones to measure the speed of sound in air

The experiment is performed by connecting a loud speaker to a signal generator to generate the sound to be picked up by the microphones.

You select a particular known specific frequency e.g. 250 Hz (f in Hz).

Two microphones are connected to an oscilloscope which pick up the sound from the speaker and which is converted to an electrical signal by the microphone and displayed as a trace on the cathode ray oscilloscope screen.

You can secure the speaker and two microphones with stands and clamps making sure they are aligned at the same height.

You set up the oscilloscope to detect the sound wave signals from both microphones - to give you two traces on the screen.

You start with the two microphones next to each other near the speaker.

You then slowly move one microphone away from the other.

When the two microphones are first exactly one wavelength apart, the two signal traces on the oscilloscope are exactly aligned - synchronised, as in the diagram above.

The trace from the microphone 2, furthest away from the speaker, will show a smaller amplitude - the diagram does not take this into account!

You then measure the distance between the microphones and this gives you the wavelength of the sound.

This is because the sound waves are aligned in phase and just one wavelength apart.

speed of sound wave (m/s) = frequency of sound (Hz) x wavelength of sound wave (m)

in 'shorthand'    v = f x λ

you know the frequency in Hz from the signal generator setting

and the wavelength is the distance between the microphones in cm ==> m

 

You repeat the experiment to calculate the average wavelength to give statistically the best result.

You can then repeat the experiments with other frequencies from the signal generator and you should find the speed stays the same, but, as the frequency is increased the wavelength of the sound wave should get shorter.

(2) Echo method to measure the speed of sound in air

You need two people to do the experiment.

Measure a distance d, e.g. 50 m from a tall wall or a building with a broad flat wall that will act as a sound wave reflector.

You then clap two pieces of flat wood together and adjust the rate of clapping until the sound of the claps are synchronised with the return of the echo.

Use a stopwatch to find the time interval between the claps e.g. measure the time of 10 claps and compute average.

Calculation of speed of sound in air.

If d = distance to wall (m), if t = average time interval between claps (s)

v = 2d/t (m/s)

Note that the distance is doubled because the sound is 'going there and back' in the time interval t.

This is not an accurate method, the clapping can't be done in perfect harmony, but its a bit of fun doing it.

Example of speed of sound calculation based on an echo

Suppose two students measured the following data.

The side of the school sports hall was 80 m away from the clapper.

50 claps took a total time of 24.2 s

v = 2d/t (m/s), d = 2 x 80 = 160 m, average clap time = 24.2 / 50 = 0.485 s

speed of sound v = 160 / 0.485 = 330 m/s (3 sf)

 

(3) An experiment to measure the speed of sound in air using a stretched string or wire.

In this method you use a mechanical vibrator (vibration transducer) to vibrates a tensioned (stretched) steel wire or elastic cord.

The vibrator, whose frequency is controlled by the signal generator, continuously transfers energy to the wire/cord making it vibrate.

This sends transverse waves down the wire/cord and produces a particular note when the wire/cord vibrates with specific number of wavelengths along the length of the wire.

The experimental set-up is shown below.

The wire/cord is fixed to the vibration transducer and stretched horizontally over a pulley and tensioned with weights on the end.

You switch on the signal generator and observe the vibrations in terms of numbers of wavelength as you slowly increase the frequency.

You note the frequency when the vibrations seem 'stable' and a number of wavelengths can be clearly observed from the stable wave 'pattern'.

You count the number of waves along the wire/cord and the frequency displayed on the signal generator.

Things are a bit blurred so you need to take care in your observations and note that when stable, the wire seems to vibrate up and down and there seems to be points on the wire which don't seem to move up and down (these points are called nodes where the amplitude is at a minimum amplitude).

e.g. from the diagram 3 wavelengths = 60 cm, one wavelength = 20 cm or 0.20 m at a frequency of 1650 Hz.

speed = wavelength x frequency

speed of sound in air = 0.20 x 1650 = 330 m/s

You can repeat the experiments with different frequencies to produce ½, 1, 1½, 2, 2½, 3, 3½, 4 etc. wavelengths as you increase the frequency from the signal generator.

You can measure the number of wavelengths for each frequency, and you should get the same speed of sound in air, even though the note you hear changes.

You can experiment with different 'string' materials and different tension weights on the end of the wire.

You can also employ a bridge ↑ over which the wire is stretched, so you can vary the length of the vibrating wire.

A simple home experiment to show a standing wave in an elastic cord

I stretch a 30 cm (0.30 m) thick rubber band above a wooden base and clamped it in position on wooden blocks (upper photographs) and illuminated with a couple of suspended LED torches.

You can see, in the lower photographs, the two extremes (maximum amplitude) the rubber band reaches when tugged back and released to vibrate at its fundamental frequency.

The sound produced was rather 'dull', but using the speed of sound (v) as 330 m/s, you can calculate the frequency of oscillation of the rubber band.

The wavelength (λ) = 2 x 0.30 = 0.60 m

This is elastic cord length is doubled because the half-wave doubles back on itself to give a complete wavelength of the standing wave - the natural wavelength and frequency resulting from the particular mass, length and tension of the elastic cord (see animation below).

v = f x λ,  f = v / λ = 330 / 0.60 = 550 Hz  (hence the blurred photograph).

This calculation oversimplifies the situation, but it is correct in principle!

 The note was not clearly heard, but on stretching the same elastic band over an empty circular tin can of ~30 cm diameter, with a similar tension, a note could be clearly heard!

 The note was theoretically between C₅ and D₅ on the musical frequency scale.

 Standing waves are illustrated by the wave 'pictures on the left'.

 They correspond in length to ½, 1, 1½, 2, 2½ and 3 wavelengths.

 The animation is from https://en.wikipedia.org/wiki/String_vibration

(4) A simple experiment to measure the speed of sound in a solid

Introduction to the experiment

You can measure the speed of sound waves in a solid by measuring the frequency of sound waves produced when you hit a solid.

The method works best by hitting metal rods that will resonate strongly, mainly with one particular 'note' called the fundamental natural frequency - think of a tuning fork or musical triangle in music.

When you hit the rod, longitudinal sound waves are induced in the metal and they will also vibrate the surrounding air.

These sound wave frequencies can be picked up by a microphone and displayed on the screen of an oscilloscope.

You can pick out this fundamental natural frequency from other frequencies because it should give the highest amplitude signal.

The experimental set-up

A uniform rod of metal of known length is suspended from its mid-point in a horizontal position.

The rod (length L) needs to at least 50 cm long and a few cm in diameter made of e.g. aluminium, brass or iron.

Near one end of the rod is placed a microphone connected to an oscilloscope and a hammer at the ready!

The oscilloscope is used to monitor both the amplitude and frequency of the sound waves produced when the rod is hit with the hammer.

Experimental procedure

This is a very technical experiment to do!

The rod is hit at one end with the hammer so it vibrates continuously making sound.

Tune in the oscilloscope to the frequency range of greatest amplitudes.

Record the frequency as best you can that corresponds to the highest amplitude on the screen.

Repeat several times to get the average frequency - the best value you can obtain.

A bit of theory before the speed calculation

When the rod is hit, its vibrations produce lots of difference frequencies.

However, all objects have a natural vibrational frequency that sets up a longitudinal standing wave that should give the maximum amplitude of sound.

This particular frequency is called the fundamental mode of vibration.

A standing wave does not vary its amplitude profile i.e. it doesn't appear to move - stationary.

What you see is one wave occupying twice the length of the rod.

I've shown this on the diagram in the cyan inset box below the rod.

The wavelength of this fundamental standing wave is equal to double the length of the rod.

(A point where the amplitude is zero is called a node - don't need to know this point for GCSE physics.)

An example of calculating the speed of sound in the rod

Suppose the rod is made of aluminium, diameter 2 cm and length 65 cm.

If the maximum amplitude was found to be at ~4.0 kHz, calculate the speed of sound in aluminium.

wavelength = 2 x L = 65 x 2 = 130 cm = 1.3 m

frequency = 4.0 kHz = 4000 Hz.

speed = frequency x wavelength = 4000 x 1.3 = ~5200 m/s

Note that the speed of sound in solids is much greater than in gases like air (~340 m/s).

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(f) ULTRASOUND - ultrasonic sound, its uses and calculations

As already mentioned, you can use a signal generators to produce electrical oscillations of any frequency.

The electrical oscillations are converted to mechanical vibrations producing sound waves.

The sound waves can be produced to well beyond the range of human hearing.

The high frequencies are above 20 kHz (20000 Hz) and are known as ultrasound.

Ultrasound waves behave like any other waves, they can be absorbed, reflected or refracted.

The reflection and refraction effects can be used to measure distances and 3D scanning with sonar searching and medical imaging applications.

 

Medical uses of ultrasound

Ultrasound of very high frequency sound waves are used in scanning pregnant women to monitor the progress of unborn baby.

The ultrasound waves enter the woman's body and the echoes-reflections are picked up by a microphone and converted into electronic signals from a which an internal picture of the womb can be constructed.

Ultrasound is considered a safe technique for pre-natal scanning of a foetus, various soft-tissue organs and is much safer than using dangerous X-rays.

Tissues e.g. muscle, stomach, womb or fluids of different density give different intensities of reflection and so differentiation of the structure of the womb, foetus or baby can be seen - modern ultrasound scanners can produce quite high resolution images.

The speed of ultrasound is different in different tissues and the ultrasound scanner is able to work out the distance between different boundaries and construct a '3D' image of the developing foetus in the womb.

Ultrasound can also be used in the 'medical imaging' of soft tissue organs like the bladder, kidneys and liver.

The scans can detect changes in the structure of these organs and help diagnose medical conditions associated with them.

Although medical imaging using ultrasound is quite safe, the images are not sharp enough to replace the use of X-rays for investigating bone structure.

Note that a gel is placed on the patient's body between the ultrasound probe and their skin.

The gel ensures that most of the ultrasound would be refracted at the skin and a good image of the internal structure of the body could not be obtained.

 

More on how does ultrasound scanning works

When sound waves (in this case ultrasound) are passing from one medium to another (solid or liquid - body fluid, tissue, bone, organs etc.) some are partially reflected, and some are transmitted and refracted at the boundary interface (see the diagram on the right of wavefronts meeting a boundary between two media).

In foetal scanning, the two media would be the fluid in the woman's womb and the skin and tissue of the growing foetus.

The time for reflections to take place at a particular angle is measured i.e. the time it takes for the sound impulses to be emitted from the transmitter and reflected back off a boundary and the return signals picked up by the detector.

From the reflections, it is the echo time intervals that allows a distance to be calculated by the computer and their distribution allows a 3D image to be built up.

The recorded data is processed by a computer to build up a 3D video on the screen from which individual images can be viewed and even printed out (for medical checks-investigation and hopefully for the delight of expectant parents).

This form of medical imaging works because ultrasound waves can pass through the body but if they meet a boundary e.g. between fluid and tissue some of the waves are reflected.

It is the distribution of these echo signals that enables the computer to build up the image.

Treatment of kidney stones

The patient lies on a water-filled cushion, and the surgeon uses X-rays or ultrasound tests to precisely locate the kidney stone.

Ultrasound shock waves are then sent to the stone from a machine to break it into smaller pieces that can be carried in your urine and pass through the urinary tract and out of your body - job done!

 

Industrial uses of ultrasound

Ultrasound is used to detect flaws in manufactured products such metal castings or pipes.

Ultrasound waves, after entering a material are usually reflected back by the far side of a object.

If there is a flaw in the casting or the welding of an object, when scanned with ultrasound the flaws show up as some of the waves are reflected-deflected back where you might expect them to go right through the object to the other side.

In other words if there is an internal flaw in the object e.g. a weld joint, the ultrasound is reflected back sooner than expected.

The same technique is used on test trains to check for flaws in railway tracks e.g. cracks in train rails.

An ultrasound beam can be used to measure the thickness of a material by detecting the reflection from both surfaces.

Use of ultrasound for sonar - echo sounding

Ultrasound systems are used by small boats, ships and submarines for echo sounding.

You need a transmitter and receiver and from the echo signals you can get the distance to the seabed.

With more sophisticated systems you can get an 'underwater picture' of what's there e.g. shoal of fish, sunken wreck, dangerous underwater rocks.

The signal time of the echo can be used to measure the depth of water beneath a boat.

If d = depth of water (m), if t = time of echo signal 'there and back' (s), v = speed of sound wave (m/s)

v = d / t, therefore depth d = v x (t / 2) (m/s)

BUT, note that the time is halved because the sound is 'going there and back' in the time interval t.

Bats and ultrasound!

Bats are amongst other animals, such as cats and dogs,  that can hear high frequency sounds.

Bats emit pulses of sound of 20 000 Hz to 100 00 Hz to find their way around using echo location.

Their large ears pick up reflected sounds and their brain builds up a 'picture' of the 3D world in front of them.

 

Calculations involving ultrasound waves and echo sounding

Be able to use both the equations below, which apply to sound waves (and their rearrangements):

appropriate units used in ()

a) sound wave speed (metre/second, m/s) = frequency (hertz, Hz) x wavelength (metre, m)

in 'shorthand'    v = f x λ

rearrangements:  f = v ÷ λ   and   λ = v ÷ f

b) sound wave speed (metre/second, m/s) = distance (metre, m) / time (second, s)

in 'shorthand'    v = d ÷ t

rearrangements:  d = v x t   and   t = d ÷ v

This is the general formula for the speed or velocity of anything moving.

Equation (a) and equation (b)

Q1 A pulse of ultrasound from a fishing boat takes 1.40 seconds to travel from the boat down to the seabed and back to the 'microphone' detector.

If the speed of sound in seawater is 1530 m/s calculate the depth of the water at that point.

speed = distance / time, rearranging gives d = s x t,

but you halve the time t to 0.70 seconds because of the double journey (there and back) of the wave.

Therefore depth = 1530 x 0.70 = 1070 m (3 s.f.)

 

Q2 The speed of sound in seawater is 1530 m/s.

How long will it take an ultrasound signal to be transmitted and received after reflecting back from a shoal of fish swimming at a depth of 200 m? (give your answer to two significant figures)

speed = distance /time, time = distance / speed

time = (200 x 2) / 1530 = 0.26 s (2 sf)

 

Q3 In misty weather, a ship sounds a foghorn warning when near a coastline of cliffs.

If the echo is heard 5.0 seconds later, how far away are the reflecting cliffs if the seed of sound is 340 m/s?

v = d / t, so d = v x t, d = distance to cliffs, v = 340 m/s, t = 5.0 / 2 = 2.5 s,

distance = 340 x 2.5 = 850 m

Q4 Suppose the time difference is 150 µs between two pulses from either side of the head of a fetus in a woman's womb.

If the speed of ultrasound is 1540 m/s, calculate the size of the baby's head in cm.

s = d / t, rearranging gives d = s x t, but d = twice the width of the head ('there and reflection back'),

size of bay's head = (1540 x 150 x 10^-6 ) ÷ 2 =  0.1155 m = 11.6 cm (3 sf)

(note that µs = microseconds, 10^-6 factor)

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(g) INFRASOUND WAVES - animals and earthquake waves

Infrasound is sound waves with frequencies less than 20 Hz and they are inaudible to the human ear.

We can't hear these infrasound frequencies of such long wavelength.

However, some animals can hear infrasonic sound e.g. like whales, elephants, rhinos, hippos, giraffes, alligators, squid/cuttlefish/octopi, and even pigeons.

What sort of wavelengths are we talking about?

From the wave equation: λ = v ÷ f

(i) Using the speed of sound in air as 340 m/s, and a frequency of 20 Hz

wavelength = speed / frequency = 340 / 20 = 17 m  (our little eardrums can't cope with this !!!)

(ii) The speed of sound in seawater is ~1520 m/s.

Whales can transmit and receive infrasonic sounds down to 10 Hz.

λ = v ÷ f  = 1520 / 10 = 152 m 

This is a very long wavelength and low frequency sounds can travel up to 10,000 miles!

(iii) Blue whales can communicate with each other over distances up to ~500 miles (~750 km).

How long goes it take for the 'message' to travel 750 km?

s = d / t,  rearranging t = d / s = (750 x 1000) / 1520 = 493 s (3 sf, 8.2 minutes)

 

In some case the infrasounds are used by animals to communicate with each other.

It is possible for scientists to use infrasound to track some animals to help in conservation projects.

There are several natural events that produce sound waves that travel through the Earth.

Volcanic eruptions release huge amounts of energy, some of it as infrasound waves travelling through the Earth's crust - some infrasound waves are detected prior to eruption and can be used to help in the prediction of eruptions

The fall of large quantities of snow in avalanches sets of infrasound waves.

The pounding of large waterfalls, meteor strikes and the breaking up of large icebergs all cause the emission of infrasound waves.

The biggest source of infrasound waves is earthquakes. seismic waves are transmitted through all the sections of the Earth - crust, mantle and core. These are discussed in detail in the next section.

Earthquake waves - seismic waves passing through the concentric layers of the Earth

When an earthquake happens in the Earth's crust it results in the spreading out of seismic waves ('shock waves').

Seismic waves result from the huge amounts of potential energy stored in the stressed layers of rock being released by tectonic plate movement.

These earthquake waves can be detected all around the world using an instrument called a seismometer.

Some earthquake waves are infrasonic, meaning their frequencies are less than 20 Hz.

The frequencies can be as low as 0.1 Hz and the speed of earthquake waves varies from 1800 to 7200 m/s (1.8 - 7.2 km/s) depending on the physical state and chemical composition of the medium e.g. type of rock.

The speed of earthquake/seismic waves depends on the material they are travelling through, in particular the density of the rock layers.

When the waves meet a boundary they may be partly/completely reflected or partly/completely absorbed, they may continue in a direct line with a different speed or the waves might change direction and speed (and wavelength) - refraction - so things get pretty complicated.

Because the density of the rock changes gradually in a particular layer, so does the speed of the wave. If refracted, the waves follow curved paths (see the diagram below).

However, at a boundary, the speed may change more abruptly giving a bigger change in direction (just as you see with light ray experiments with prisms.

Scientists (seismologists) study the properties and pathways of seismic waves to deduce the internal structure of the Earth.

From scientific studies of where the different types of waves are detected, or not detected, due to absorption, reflection or refraction you can work out the structure of the layers of the Earth the seismic waves pass through.

These seismologists calculate the time it takes for these shockwaves to reach every seismometer around the world and, important to work out a specific earthquake below the Earth's surface.

There are three types of earthquake waves (seismic waves)

P-waves are longitudinal waves and so can travel right through the Earth to the other side of the world.

They are also called primary pressure or compression waves and are identical to sound waves but with much longer wavelengths - see calculation below. The above diagram was used to illustrate a sound wave!

The animation of a P-wave simulation is shown on the right (from https://en.wikipedia.org/wiki/P-wave ).

P-waves from an earthquake travel at ~5 to 8 km/s in the crust, mantle or core and ~1.5 km/s (the same as 'sound') in water!

Weak to moderate earthquake waves have a frequency of 0.1 to 2 Hz on the surface.

If a seismic wave has a speed of 5 km/s (5000 m/s) and a frequency of 0.5 Hz

the wavelength = λ = v ÷ f = 5000 / 0.5 = 10 000 m (10 km), ~1000 x more than our audible sounds

S-waves are transverse waves and can only travel through solids, so they cannot travel through the core.

S-waves cause an 'up and down' shearing movement of the rock layers at 90^o to the direction of the wave. Secondary shear waves.

On the right is an animation of the movement of an S-wave of an earthquake.

(from https://en.wikipedia.org/wiki/S-wave )

L-waves are transverse move along the surface of the Earth moving the ground up and down.

 

W = atmosphere  : From seismic studies we can say ...

X = the relatively thin Earth's crust - mainly consisting of hard rock 92/3rds overlaid with water).

Y = about half the radius of the Earth's is the mantle, consisting of rock that is almost solid, it can flow very slowly under stress or in a convection current - great plumes of rock can rise producing bands of volcanoes like the 'Ring of Fire' in the Pacific Ocean - lots of earthquakes too!

Z = the inner solid metal core and an outer liquid metal core of the Earth, mostly iron and some nickel and smaller amounts of other metals - the iron is the source of the Earth's magnetic field.

For more details see earthquake (seismic) wave analysis notes in Earth Science section

Earthquake waves spread out in all directions from the epicentre of an earthquake in the Earth's crust.

These vibrations caused by these waves are detected by instruments called seismometers, which are positioned in many locations on the Earth's surface.

P-waves (primary waves) and S-waves take curved paths because of the ever changing density of the Earth's layers producing a gradual refraction effect.

The longitudinal P-waves can pass right through the centre of the Earth but due to refraction give two small shadow zones (marked black on the diagram). They travel faster than S-waves.

The transverse S-waves are absorbed by the liquid outer core and give one much larger shadow zone (marked blue + black on the diagram). They travel slower than P-waves.

All the wave paths are curved because the density is changing continuously with depth (increase in pressure), so they are continuously being refracted (change in direction).

BUT, you get much greater refraction effects at the boundaries between the crust/mantle, mantle/outer core and outer core/inner core.

Boundaries are created by the different physical states and densities of the rock and metal layers.

Seismometers pick up the vibrations of earthquake waves from many seismographic stations around the world (over 2000 locations).

Analysis of the paths of waves in terms of velocity and direction data has enabled geologists to work out the basic layered structure of the Earth.

From the speed, absorption and refraction of seismic waves scientists have worked out the number and depth of the four layers of the internal structure of the Earth.

e.g. from the shadow zones you can work out the depth of the mantle and the inner and outer layers of the core.

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