Physics · 3 Waves

3.4 Sound

Before you start

Waves transfer energy without transferring matter (3.1).
In a longitudinal wave the direction of vibration is parallel to the direction the wave travels; in a transverse wave it is at right angles (3.1).
Wavelength, frequency (in hertz, Hz) and amplitude, and the wave equation $v = f λ$ , which you must recall (3.1).
Speed is distance travelled per unit time, $v = s / t$ , which you must recall (1.2).
In solids and liquids the particles are close together; in a gas they are far apart (2.1).
Prefixes: 1 kHz = 1000 Hz and 1 ms = 0.001 s.

Syllabus objectives covered (12)

Describe the production of sound by vibrating sources
Describe the longitudinal nature of sound waves
State the approximate range of frequencies audible to humans as 20 Hz to 20 000 Hz
Know that a medium is needed to transmit sound waves
Know that the speed of sound in air is approximately 330–350 m / s
Describe a method involving a measurement of distance and time for determining the speed of sound in air
Describe how changes in amplitude and frequency affect the loudness and pitch of sound waves
Describe an echo as the reflection of sound waves
Define ultrasound as sound with a frequency higher than 20 kHz
Describe compression and rarefaction
Know that, in general, sound travels faster in solids than in liquids and faster in liquids than in gases
Describe the uses of ultrasound in non-destructive testing of materials, medical scanning of soft tissue and sonar including calculation of depth or distance from time and wave speed

Part 1 of 7

Every sound starts with something vibrating

A sound is always made by something vibrating: moving rapidly backwards and forwards. A guitar string, the prongs of a tuning fork, the cone of a loudspeaker and the vocal cords in your throat all vibrate when they make a sound.
Take the loudspeaker cone, because it is the easiest to picture. When the cone moves forward, it pushes on the air in front of it and squashes the air particles closer together.
When the cone moves back, it leaves more room, so the air particles in front of it spread out further apart than normal.
The cone does this over and over, many times each second. Each particle pushes on its neighbour, so each push and pull is passed on from one layer of air to the next. That travelling disturbance is the sound wave.
So an answer about how a sound is produced names two things: the source vibrates, and that makes the particles of the medium vibrate and pass the vibration on. 'The speaker makes a noise' describes nothing.
The frequency of the sound is the number of vibrations the source makes each second, in hertz (Hz). A tuning fork marked 256 Hz has prongs that vibrate 256 times every second.
Stop the vibration and the sound stops. Put your hand on a sounding guitar string and it goes quiet, because you have stopped it vibrating.

Part 2 of 7

Sound is longitudinal: compressions and rarefactions

Recall from 3.1: in a longitudinal wave, the direction of vibration is parallel to the direction the wave travels.
Sound is a longitudinal wave. The air particles vibrate backwards and forwards along the same line the sound is travelling. If the sound travels from left to right, each particle moves a little left and right about its own position.
Notice what does not happen: the particles do not travel from the loudspeaker to your ear. Each one vibrates about where it started and passes the push on. The wave carries energy to you; it does not carry the air.
A compression is a region where the particles are closer together than normal. The pressure there is a little higher than normal.
A rarefaction is a region where the particles are further apart than normal. The pressure there is a little lower than normal.
Each forward push of the source makes a compression and each backward pull makes a rarefaction. So a sound wave is a series of compressions and rarefactions, one after the other, moving away from the source.
apparatus diagram — not drawn yet
A loudspeaker on the left facing right, with a long row of air particles drawn as dots in front of it, bunched closely in evenly spaced regions labelled 'compression' and spread out in the regions between them labelled 'rarefaction'; one arrow pointing right labelled 'direction the wave travels', a short double-headed horizontal arrow on one particle labelled 'particle vibrates', and a bracket from the centre of one compression to the centre of the next labelled 'one wavelength'.
The particles bunch and spread along the direction of travel; one wavelength runs from the centre of one compression to the centre of the next.
The wavelength of a sound wave is the distance from the centre of one compression to the centre of the next, or from one rarefaction to the next. The distance from a compression to the next rarefaction is only half a wavelength.
When you describe these, give the word and what the particles are doing, compared with normal: 'compression: particles closer together than normal; rarefaction: particles further apart than normal'.

Part 3 of 7

Sound needs a medium, and its speed depends on the medium

A sound wave is passed on by particles pushing on their neighbours. With no particles there is nothing to pass the vibration on. So sound needs a medium: a solid, a liquid or a gas.
In a vacuum, such as space, sound cannot travel at all. Light can, because light is an electromagnetic wave and needs no medium.
The standard demonstration is an electric bell hung inside a sealed glass jar that is connected to a vacuum pump. With air in the jar, you hear the bell ringing clearly.
apparatus diagram — not drawn yet
An electric bell hanging inside a sealed glass bell jar that stands on a flat base plate, with its wires passing through the seal to a battery and switch outside, and a pipe from the base plate to a vacuum pump; labels 'glass jar', 'electric bell', 'to vacuum pump'.
As the pump removes the air, the ringing fades, but you can still see the hammer striking the gong.
As the pump removes the air, the sound becomes fainter and fainter. You can still see the hammer striking the gong, because light passes through the near-vacuum, so you know the bell is still vibrating.
Let the air back in and the sound returns. The only thing that changed was the amount of air, so the air was what carried the sound to you.
Do not claim the sound disappears completely. The pump never removes all the air, and some vibration passes out through the bell's mounting, so a faint sound may remain.
In air, sound travels at approximately 330–350 m/s. Learn that range. Sound is a little faster in warmer air, which is one reason it is given as a range and not as one number.
Compare that with light, at about $3.0 \times 10^{8}$ m/s. Over a few hundred metres, light arrives almost instantly and the sound arrives noticeably later. That is why you see a flash of lightning before you hear the thunder.
Sound also travels through liquids and solids, and the speed depends on the medium. In general, sound travels faster in solids than in liquids, and faster in liquids than in gases. For example, it travels at about 1500 m/s in water and around 5000 m/s or more in steel, against about 340 m/s in air.
Notice the words 'in general'. Learn the order, solid then liquid then gas, from fastest to slowest, as a general rule; do not claim it is true for every material. The syllabus asks you to know the value for air only, so a question that needs the speed in water or steel has to give it to you.
A picture that helps you remember the order: in a gas the particles are far apart, so each one must move some way before it pushes the next. In liquids and solids the particles are close together, and in a solid they are held by strong forces, so the push is passed along more quickly. The objective asks you to know the order, not to explain it.

Part 4 of 7

Measuring the speed of sound in air

Every method uses the speed equation. Paper 4 gives you no formula sheet, so recall it.
$v = \frac{s}{t}$
So you need two measurements: a distance the sound travels and the time it takes. The distance is easy: use a long tape measure or a trundle wheel. The time is the hard part, because sound crosses 100 m in only about 0.3 s.
A hand-held stopwatch has a delay at the start and at the stop: your reaction time. If the time you are measuring is short, that delay is a large fraction of it. A good method beats this problem in one of two ways: it makes the measured time longer, or it takes the human out of the timing.
Method 1: two observers. Two people stand a large measured distance apart in an open space, for example 500 m. One bangs two wooden blocks together above their head. The other starts a stopwatch on seeing the blocks hit and stops it on hearing the bang.
Seeing is the start signal because light covers the distance almost instantly. So the time on the stopwatch is the time the sound took to travel the distance, about 1.5 s for 500 m. Calculate $v = s / t$ .
Repeat several times and calculate the mean. Swap the two people's positions and repeat, which reduces the effect of any wind.
Method 2: echoes from a wall. Stand a measured distance, for example 80 m, from a large flat wall such as the side of a building, with open ground around you. Clap once and you hear the echo, which is the clap reflected from the wall.
apparatus diagram — not drawn yet
Plan view of an open space: a student clapping on the left and a large flat wall on the right, with a measured distance labelled '80 m' between them; an arrow from the student to the wall and a return arrow from the wall to the student, together labelled 'sound travels 2 × 80 m in each interval'; a second student beside the first holding a stopwatch.
Between one clap and the next, the sound travels to the wall and back, so it covers twice the measured distance.
Now clap repeatedly, adjusting the rhythm until each clap happens just as the echo of the one before arrives. A partner times a large number of intervals between claps, for example 20.
In each interval between claps, the sound goes to the wall and back. So in each interval it travels twice the distance to the wall.
$v = \frac{2 s}{time for one interval}$
Timing 20 intervals instead of one means the reaction-time error at the start and the stop is shared over 20 intervals, so it is a much smaller fraction of the measured time.
Method 3: electronic timing. Two microphones are placed a measured distance apart, for example 1.00 m, in line with a source of a sharp sound, and connected to an electronic timer. The sound reaching the first microphone starts the timer and the sound reaching the second stops it. No reaction time is involved, so a short distance is enough.
Whichever method you describe, include: what you measure the distance with, what starts and stops the timing, the equation you use (with $2 s$ for an echo), and one thing that improves the result, such as a long distance, timing many intervals, or repeating and finding a mean.

Part 5 of 7

Echoes

An echo is a sound wave that has been reflected from a surface and comes back to you. It is reflection, which you met for all waves in 3.1; nothing new happens to the sound.
Large, hard, flat surfaces, such as cliffs and the walls of buildings, give clear echoes.
You hear the echo after a delay because the sound has to travel to the surface and back. So the total distance the sound travels is twice the distance to the surface.
$2 s = v \times t$
Rearrange for the distance to the surface: multiply speed by time, then halve.
$s = \frac{v \times t}{2}$
The trap is forgetting to halve. Every time, ask whether the time you were given is for a one-way trip or a there-and-back trip. An echo time is always there and back.
Check that the numbers make sense. If you stand 170 m from a cliff and shout, the sound travels 340 m in total. At 340 m/s, the echo comes back 1.0 s after you shout.

Part 6 of 7

Loudness, pitch, and the frequencies you can hear

Two properties of a sound wave decide what you hear. Amplitude decides loudness. Frequency decides pitch.
The amplitude of a sound wave is how far the particles of the medium move from their rest positions as they vibrate. When the source vibrates with a larger amplitude, the wave carries more energy and the sound is louder. A smaller amplitude gives a quieter sound.
Frequency is the number of vibrations per second. A higher frequency gives a higher pitch, a higher note. A lower frequency gives a lower pitch.
The two are independent. You can play a quiet high note or a loud low note. Change only the amplitude and only the loudness changes; change only the frequency and only the pitch changes.
Neither of them changes the speed. In air, loud and quiet sounds, high and low notes, all travel at the same speed. Since $v = f λ$ and $v$ stays the same, a higher frequency in the same medium means a shorter wavelength.
To compare sounds, you may be shown traces from a microphone connected to an oscilloscope. A trace is a graph of the vibration against time. It looks like a transverse wave even though sound is longitudinal, so read it as a graph, not as a picture of the wave.
graph diagram — not drawn yet
Three oscilloscope-style traces of a sine wave against time on identical square grids, one above another: trace A with 3 complete waves across the screen and peaks 2 squares high; trace B with 3 complete waves and peaks 4 squares high; trace C with 6 complete waves and peaks 2 squares high; labels A 'reference', B 'louder, same pitch', C 'higher pitch, same loudness'.
Taller peaks mean a larger amplitude and a louder sound; more waves across the same screen mean a higher frequency and a higher pitch.
On a trace, the height of the peaks shows the amplitude: taller peaks, louder sound. The number of complete waves across the screen shows the frequency: more waves fitted into the same time, higher pitch.
Human hearing has limits. The approximate range of frequencies audible to humans is 20 Hz to 20 000 Hz. Since 1 kHz = 1000 Hz, the top of the range is also written 20 kHz.
A vibration below 20 Hz or above 20 000 Hz still makes a sound wave in the medium; humans just cannot hear it. The range varies from person to person and the upper limit falls with age, which is why it is called approximate.
Use $v = f λ$ to get a feel for sizes. In air at 340 m/s, a 20 Hz sound has a wavelength of $340 \div 20 = 17$ m, and a 20 000 Hz sound has a wavelength of $340 \div 20 000 = 0.017$ m, which is 17 mm.

Part 7 of 7

Ultrasound and its uses

Ultrasound is sound with a frequency higher than 20 kHz. That is the definition to write. It is above the upper limit of human hearing, so we cannot hear it.
Ultrasound is still sound. It is longitudinal, it is made by vibrations, it needs a medium, and it reflects to give echoes. Only its frequency is different.
Every use in this objective rests on one idea: send a short pulse of ultrasound, detect its echo, and use the time between them to find a distance.
$s = \frac{v \times t}{2}$
Non-destructive testing of materials. A probe sends pulses of ultrasound into a metal object such as a rail, a pipe or a weld. The pulse travels to the far side of the object and reflects back to the probe. If there is a crack or flaw inside, part of the pulse reflects from it and returns earlier than the echo from the far side.
So an extra, earlier echo shows that there is a flaw, and its time tells you how deep the flaw is. 'Non-destructive' means the object is tested without being cut open or damaged, so it can still be used afterwards.
graph diagram — not drawn yet
Detector signal against time for an ultrasound pulse sent into a metal block: a peak at time zero labelled 'pulse sent', a second peak part-way along the time axis labelled 'echo from crack', and a third, later peak labelled 'echo from far side of block'; the time axis labelled 'time' with no numbers.
The echo from the crack arrives before the echo from the far side; its time gives the depth of the crack.
Medical scanning of soft tissue. A probe placed on the skin sends pulses of ultrasound into the body. At each boundary between different tissues, some of the ultrasound is reflected and the rest travels on. The probe detects the echoes, and a computer uses their times to work out how deep each boundary is and builds an image.
This is how a developing fetus is checked during pregnancy. Ultrasound is not ionising, unlike X-rays, which is an important reason it is used for this.
A gel is spread between the probe and the skin. Without it, a layer of air between them would reflect almost all of the ultrasound before it entered the body.
Sonar. A ship sends pulses of ultrasound down through the water. They reflect from the seabed, or from a shoal of fish or a submarine, and a detector on the ship picks up the echo. The time between sending a pulse and receiving its echo gives the depth or distance.
A sonar calculation, step by step. Sound travels through seawater at about 1500 m/s; a question that needs this will state it. If the echo returns after 0.40 s, the pulse has travelled $1500 \times 0.40 = 600$ m there and back, so the seabed is $600 \div 2 = 300$ m below the ship.
The same three steps work for all three uses: distance travelled = speed × time; halve it, because the pulse went there and back; give the answer with its unit.

Worked examples

Example 1. A student stands $80.0$ m from a large flat wall. She claps and adjusts her rhythm until each clap coincides with the echo of the one before. Her partner times 25 intervals between claps and the stopwatch reads $11.8$ s. Calculate the speed of sound in air from these results, and comment on whether the value is reasonable.

Show the working

Find the time for one interval. 25 intervals took 11.8 s, so divide by 25.
$t = \frac{11.8}{25} = 0.472 s$
Find the distance the sound travels in one interval. It goes to the wall and back, so it is twice the distance to the wall.
$2 s = 2 \times 80.0 = 160 m$
Recall $v = s / t$ and substitute the distance and time for one interval.
$v = \frac{160}{0.472} = 338.98 \dots m/s$
The data are given to 3 significant figures, so give $339$ m/s.
Comment: the speed of sound in air is approximately 330–350 m/s, and 339 m/s lies inside that range, so the result is reasonable.

Answer: $339$ m/s (from $160 \div 0.472$ ). This lies inside the expected range of 330–350 m/s, so it is a reasonable value.

Example 2. A student presses her ear against one end of a steel railing $170$ m long. Her friend hits the far end once with a hammer, and she hears two bangs. The speed of sound in air is $340$ m/s and in steel is $5000$ m/s. Explain why she hears two bangs, and calculate the time between them.

Show the working

Sound needs a medium, and here there are two: the steel railing and the air beside it. The bang travels to her through both.
Sound travels faster in a solid than in a gas, so the bang through the steel arrives first and the bang through the air arrives later. That is why there are two.
Time through the steel: recall $v = s / t$ , so $t = s / v$ .
$t_{steel} = \frac{170}{5000} = 0.034 s$
Time through the air, with the same distance.
$t_{air} = \frac{170}{340} = 0.50 s$
The time between the bangs is the difference.
$0.50 - 0.034 = 0.466 s \approx 0.47 s$

Answer: The sound travels through both the steel and the air, and it is faster in the solid steel, so the bang through the steel arrives first. Time between them $= 0.50 - 0.034 = 0.466$ s, which is $0.47$ s to 2 significant figures.

Example 3. A ship's sonar sends a pulse of ultrasound straight down. An echo from a shoal of fish is detected $0.10$ s after the pulse is sent, and an echo from the seabed is detected $0.36$ s after the pulse is sent. The speed of sound in seawater is $1500$ m/s. Calculate the depth of the shoal and the depth of the seabed.

Show the working

Both echoes are there-and-back trips, so each distance travelled must be halved.
Shoal of fish: distance travelled = speed × time.
$1500 \times 0.10 = 150 m$
Halve it to get the depth.
$150 \div 2 = 75 m$
Seabed: the same steps with the longer time.
$\frac{1500 \times 0.36}{2} = \frac{540}{2} = 270 m$
Check the order: the fish are between the ship and the seabed, so their echo must arrive first and give the smaller depth. It does.

Answer: Shoal of fish: $75$ m below the ship. Seabed: $270$ m below the ship.

Example 4. A steel block is $0.25$ m thick. A probe on its top surface sends a pulse of ultrasound down into it. The speed of ultrasound in the steel is $5000$ m/s. (a) Calculate the time after which the echo from the bottom of the block reaches the probe. (b) An extra echo is detected $0.060$ ms after the pulse is sent. Deduce what this shows, and calculate the depth at which it was produced.

Show the working

(a) The pulse goes down to the bottom and back up, so it travels twice the thickness.
$2 \times 0.25 = 0.50 m$
Recall $v = s / t$ and rearrange to $t = s / v$ .
$t = \frac{0.50}{5000} = 1.0 \times 10^{- 4} s = 0.10 ms$
(b) The extra echo arrives at 0.060 ms, earlier than the echo from the bottom at 0.10 ms. So something inside the block reflected part of the pulse before it reached the bottom: there is a crack or flaw.
Convert the time to seconds first: $0.060$ ms $= 6.0 \times 10^{- 5}$ s. Then distance travelled = speed × time.
$5000 \times 6.0 \times 10^{- 5} = 0.30 m$
Halve it, because the pulse went down to the flaw and back.
$0.30 \div 2 = 0.15 m$

Answer: (a) $1.0 \times 10^{- 4}$ s ( $0.10$ ms). (b) The earlier echo shows a crack or flaw inside the block, at a depth of $0.15$ m below the top surface.

Where people go wrong

Mistake: Saying the air particles travel from the source to your ear.

Instead: Each air particle vibrates backwards and forwards about its own position and passes the vibration on to its neighbours.

A wave transfers energy without transferring matter. The compressions and rarefactions move away from the source; the particles do not.

Mistake: Using the whole echo time as if it were the time to reach the surface, and so getting twice the true depth.

Instead: Calculate speed × time, then divide by 2.

An echo is a reflection: the sound goes to the surface and comes back, so the time is for twice the distance you want.

Mistake: Thinking a louder sound, or a higher note, travels faster.

Instead: In a given medium, all sounds travel at the same speed, whatever their loudness or pitch.

Amplitude changes loudness and frequency changes pitch, but the speed is set by the medium. With $v$ fixed, $v = f λ$ means a higher frequency has a shorter wavelength, not a higher speed.

Mistake: Deciding that sound is transverse because its oscilloscope trace looks like a transverse wave.

Instead: Sound is longitudinal. The trace is a graph of the vibration against time.

The up-and-down line on the screen comes from the electrical signal of the microphone plotted against time. The air particles themselves vibrate along the direction the sound travels.

Mistake: Thinking ultrasound is a different kind of wave from sound, or that it needs no medium.

Instead: Ultrasound is sound with a frequency higher than 20 kHz.

It is produced by vibrations, it is longitudinal, it needs a medium and it reflects to give echoes, exactly like audible sound. Only the frequency is different, which is why humans cannot hear it.

Check yourself

A tuning fork is struck and held near your ear. Describe how it produces the sound you hear.
Answer
The prongs of the fork vibrate. They push and pull on the air next to them, making the air particles vibrate. The vibration is passed from particle to particle through the air, as compressions and rarefactions, until it reaches your ear.
A sound wave travels due east. In which direction do the air particles vibrate, and what is a region of the wave called where the particles are further apart than normal?
Answer
They vibrate east and west, backwards and forwards along the direction the wave travels (parallel to it), because sound is longitudinal. A region where the particles are further apart than normal is a rarefaction.
Four sounds have frequencies of 12 Hz, 440 Hz, 18 kHz and 35 kHz. State which of them a human can hear, and identify the one that is ultrasound.
Answer
440 Hz and 18 kHz (18 000 Hz) lie inside the range 20 Hz to 20 000 Hz, so they can be heard. 12 Hz is below 20 Hz, so it cannot be heard. 35 kHz is higher than 20 kHz, so it is ultrasound and cannot be heard.
An astronaut floating some distance away from a space station sees a piece of equipment strike its outside wall, but hears nothing. Explain why.
Answer
Space is a vacuum, so there are no particles between the station and the astronaut to carry the vibration. Sound needs a medium, so it cannot reach her. Light needs no medium, so she can still see the impact.
Two students stand $510$ m apart. One bangs two blocks together; the other starts a stopwatch on seeing the blocks hit and stops it on hearing the bang. The reading is $1.52$ s. Calculate the speed of sound, and explain why the stopwatch is started on seeing the blocks hit.
Answer
$v = s / t = 510 \div 1.52 = 335.5$ m/s, so $336$ m/s, which is within 330–350 m/s. The stopwatch is started on seeing the blocks because light travels the 510 m almost instantly, so the time measured is the time the sound takes.
You shout towards a cliff and hear the echo $1.4$ s later. The speed of sound in air is $340$ m/s. Calculate how far away the cliff is.
Answer
Distance travelled $= 340 \times 1.4 = 476$ m, there and back. Distance to the cliff $= 476 \div 2 = 238$ m.
Two oscilloscope traces are drawn on the same settings. Trace Q has peaks twice as tall as trace P and the same number of complete waves across the screen. Compare the two sounds.
Answer
Q has the larger amplitude, so Q is louder than P. They have the same frequency, so they have the same pitch.
A fishing boat's sonar detects an echo from a shoal of fish $0.12$ s after a pulse of ultrasound is sent down. The speed of sound in seawater is $1500$ m/s. Calculate the depth of the fish.
Answer
Distance travelled $= 1500 \times 0.12 = 180$ m, there and back. Depth $= 180 \div 2 = 90$ m.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A loudspeaker is connected to a signal generator. The cone of the loudspeaker vibrates, and a sound wave travels away from it through the air.

(a) Describe how the vibrating cone produces a sound wave in the air. 2 marks No calculator

Mark scheme

B1 the cone moves backwards and forwards and makes the air particles next to it vibrate / pushes and pulls on the air
B1 any one of: the air particles pass the vibration on to neighbouring particles / the air particles vibrate parallel to the direction the wave travels (longitudinal) / compressions and rarefactions are produced and travel away from the cone, owtte

(b) The sound wave consists of compressions and rarefactions. Describe what is meant by a compression and by a rarefaction. 2 marks No calculator

Mark scheme

B1 compression: a region where the air particles are closer together than normal / pressure higher than normal
B1 rarefaction: a region where the air particles are further apart than normal / pressure lower than normal

(c) The frequency of the signal generator is increased and the amplitude of the vibration of the cone is decreased. State the effect on (i) the pitch of the sound and (ii) the loudness of the sound. 2 marks No calculator

Mark scheme

B1 (i) the pitch is higher / increases (in answer (i), 'higher' alone is accepted)
B1 (ii) the sound is quieter / less loud / the loudness decreases

Question 2. A student stands 102 m from a large flat wall. She claps, and adjusts the rhythm of her claps until each clap coincides with the echo of the one before. Her partner starts a stopwatch on one clap and stops it 20 intervals later. The stopwatch reads 12.0 s.

(a) Calculate the speed of sound in air given by these results. 3 marks

Mark scheme

C1 distance travelled in one interval = 2 × 102 = 204 m (or total distance = 20 × 204 = 4080 m)
C1 time for one interval = 12.0 ÷ 20 = 0.60 s (or v = 4080 ÷ 12.0 seen)
A1 340 m/s; unit required

(b) Suggest why her partner times 20 intervals instead of a single interval. 1 mark No calculator

Mark scheme

B1 one interval is a short time, so reaction time is a large fraction of it; timing 20 intervals makes the reaction-time error a much smaller fraction of the measured time, owtte

(c) Describe a different method that two students could use to determine the speed of sound in air in a large open field, using two wooden blocks, a stopwatch and a trundle wheel. 3 marks No calculator

Mark scheme

B1 measure a large distance between the two students (e.g. hundreds of metres) with the trundle wheel
B1 one student bangs the blocks together; the other starts the stopwatch on seeing the blocks hit and stops it on hearing the sound
B1 speed = distance ÷ time (repeating and calculating the mean is good practice but is not needed for this mark)

Question 3. A fishing boat uses sonar to find the depth of the sea. A pulse of ultrasound is sent straight down from the boat. The echo from the seabed is detected 0.24 s after the pulse is sent. The speed of sound in seawater is 1500 m/s.

(a) Define ultrasound. 1 mark No calculator

Mark scheme

B1 sound with a frequency higher than 20 kHz / above 20 000 Hz

(b) Calculate the depth of the sea below the boat. 3 marks

Mark scheme

C1 distance travelled = speed × time or 1500 × 0.24 (= 360 m)
C1 ÷ 2 because the pulse travels there and back, or 360 ÷ 2 seen
A1 180 m; unit required

(c) Ultrasound is also used to check steel rails for cracks without damaging them. Explain how a crack inside a rail is detected. 2 marks No calculator

Mark scheme

B1 a pulse of ultrasound is sent into the rail and part of it is reflected by the crack
B1 an extra echo is detected, arriving before the echo from the far side of the rail (its time gives the depth of the crack)

Question 4. A student presses her ear against one end of a long steel railing. Her friend, at the far end of the railing, hits it once with a hammer.

(a) The student hears two separate bangs. Explain why, and state which bang reaches her first. 2 marks No calculator

Mark scheme

B1 the sound travels to her through both the steel railing and the air
B1 sound travels faster in a solid / steel than in a gas / air, so the bang through the steel arrives first

How this is examined

Describing how to measure the speed of sound in air: a three- or four-mark Describe of a method, which may name the equipment available, such as a stopwatch and a trundle wheel Measure a large distance with a tape measure or trundle wheel; start the timer on seeing the source (or time many echo intervals from a wall); stop it on hearing the sound; speed = distance ÷ time, using twice the distance for an echo; repeat and take the mean. Each of these is a separate point, so write each one.
Finding a depth or distance from an echo time: a two- or three-mark Calculate about sonar, a crack in a metal block or an echo from a cliff, with the speed of sound given in the question Distance travelled = speed × time, then ÷ 2 because the pulse went there and back, and the answer with its unit, in m. If you forget to halve, the final answer mark is lost, although the marks for the equation and substitution may still be scored.
How a vibrating source makes a longitudinal wave of compressions and rarefactions: a two- or three-mark Describe about a loudspeaker or a tuning fork, which may come with a diagram of particles to label The source vibrates and makes the air particles vibrate; the particles vibrate parallel to the direction the wave travels; compression = particles closer together than normal (higher pressure); rarefaction = particles further apart than normal (lower pressure).
What changing the amplitude or the frequency does to a sound: a two-mark State or Describe, which may give two traces to compare Larger amplitude → louder; higher frequency → higher pitch. On a trace, the height of the peaks is the amplitude and the number of waves across the screen shows the frequency. Write 'higher pitch', not just 'higher', so that it cannot be read as 'louder'.
Ultrasound: the definition and how each use works: a one-mark Define, then a one- or two-mark Explain or Describe of a use Sound with a frequency higher than 20 kHz (20 000 Hz). For a use, say what the ultrasound reflects from: a crack inside the metal (an echo arriving earlier than the far-side echo), boundaries between tissues, or the seabed or fish.
Why sound needs a medium, and which medium is fastest: a one- or two-mark State or Explain about a bell in a vacuum jar, sound in space, or two bangs heard through a railing and through the air Sound is passed on by vibrating particles, so it cannot travel through a vacuum; light needs no medium, so the bell can still be seen. In general: faster in solids than liquids, faster in liquids than gases. In air: approximately 330–350 m/s. Human hearing: 20 Hz to 20 000 Hz.

Words to know

Vibration: A rapid backwards-and-forwards movement. Every sound is produced by a vibrating source.
Medium: The solid, liquid or gas that a sound wave travels through. Sound cannot travel through a vacuum.
Longitudinal wave: A wave in which the direction of vibration is parallel to the direction the wave travels. Sound is longitudinal.
Compression: A region of a sound wave where the particles are closer together than normal, so the pressure is higher than normal.
Rarefaction: A region of a sound wave where the particles are further apart than normal, so the pressure is lower than normal.
Amplitude: The maximum distance the particles move from their rest positions as they vibrate. A larger amplitude gives a louder sound.
Frequency: The number of vibrations per second, in hertz (Hz). A higher frequency gives a higher pitch.
Pitch: How high or low a note sounds. It is decided by the frequency.
Echo: A sound wave reflected from a surface back to the listener.
Ultrasound: Sound with a frequency higher than 20 kHz, above the upper limit of human hearing.
Sonar: Using pulses of ultrasound and the timing of their echoes to find the depth of the sea or the distance to objects under water.
Non-destructive testing: Checking a material for cracks or flaws without cutting or damaging it, for example by timing echoes of ultrasound from inside it.

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