Chapter 4 of 8

All 8 chapters
  1. 1Kinds of waves, readunderstood
  2. 2Measuring waves, readunderstood
  3. 3Waves on water, readunderstood
  4. 4Sound and ultrasound, readunderstood
  5. 5Reflection and refraction, readunderstood
  6. 6Lenses, readunderstood
  7. 7The electromagnetic spectrum, readunderstood
  8. 8Waves in communication and health, readunderstood

S4.10.4 Chapter 4

Sound and ultrasound

Tariq claps in front of a tall wall 85 m away and hears an echo. How long after the clap does the echo reach him?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Measure the speed of sound using an echo from a wall
  • Explain how ultrasound is used to scan unborn babies and find sea depths

Sound is a longitudinal wave. Layers of the medium are pushed together and pulled apart, and the pattern of compressions and rarefactions runs outwards from the source.

Sound needs a medium to travel through. It can travel through a gas, a liquid or a solid. It is usually slowest in a gas, faster in a liquid and fastest in a solid. In a solid the particles are close and spring back quickly.

A table with the columns Medium and Speed of sound. Row 1: air, 340 m/s. Row 2: water, 1 500 m/s. Row 3: steel, 5 000 m/s.MediumSpeedair340 m/swater1 500 m/ssteel5 000 m/s
Sound is about 340 m/s in air at room temperature. It is a little slower when the air is colder and a little faster when it is warmer.

In a vacuum there are no particles, so there is nothing to push together. Sound cannot travel in a vacuum. Scientists show this with a bell in a jar. They cannot do it at home, because it needs a pump that takes the air out.

A glass jar with a bell hanging in it on a thread. The jar is joined by a pipe to a pump. The jar holds air, shown as dots. Curved lines outside the jar, labelled sound, show that the ringing is heard. The bell, the jar and the pump are labelled.belljarpumpsound
With air in the jar you hear the bell.
The same glass jar and bell, joined to a pump. This time the jar has no dots, because the air has been pumped out. There are no curved lines outside the jar. The bell, the jar and the pump are labelled.belljarpump
With the air pumped out, the bell still moves, but you hear almost nothing.

As the air is pumped out, the ringing gets fainter. When almost no air is left, it is almost silent. The bell hangs on a thin thread, so that the sound cannot run along something solid to the glass. You can still see the bell move, because light does cross a vacuum.

Light is far faster than sound. In a vacuum, light travels at 3 × 10⁸ m/s. That is nearly a million times the speed of sound in air. This is why a flash of lightning is seen long before the thunder is heard.

Now use an echo to measure the speed of sound. Stand in front of a large wall and clap. The sound travels to the wall and bounces back. You hear the echo a short time later.

Sound goes to the wall and comes back. So the distance it travels is twice the distance d to the wall. The speed is distance divided by time: v = 2d ÷ t.

A person on the ground on the left, labelled you, and a hatched wall on the right, labelled wall. A long arrow from the person to the wall is labelled sound goes. A second arrow of the same length from the wall back to the person is labelled echo. A mark under the ground between the person and the wall is labelled 85 m.sound goesecho85 myouwall
The sound covers the 85 m twice, there and back.

Worked example

A person stands 85 m from a wall. An echo comes back 0.5 s after a clap. Find the speed of sound.

  1. The sound goes to the wall and back, so it travels 2 × 85 m = 170 m.
  2. The speed is v = distance ÷ time = 170 m ÷ 0.5 s = 340 m/s.

Answer: 340 m/s

A hand held stopwatch cannot time 0.5 s well, because your reaction time is about 0.2 s. The trick is to clap again and again, so that each clap comes with the echo of the one before. Then the time between claps is the echo time. Count many claps and divide.

Try it

Measure the speed of sound with an echo

You need: a large flat wall, such as the side of a tall building, at least 50 m away across open ground, a long string or tape, or a known length of pace, to find the distance, a clock or phone that shows seconds, two flat pieces of wood to clap together.

Be safe: Stay well away from roads and from the edges of cliffs, roofs or deep water. Do not clap close to anyone's ears. Use a wall that belongs to nobody who would mind the noise.

  1. Find the distance d from where you stand to the wall. Walk it and count paces, after you have measured the length of one pace on a string.
  2. Clap the two pieces together. Listen for the echo.
  3. Clap again so that your clap comes just as the echo of the last clap arrives. Keep a steady rhythm.
  4. When the rhythm is steady, count 20 claps and time them. Divide the time by 20. This is the echo time t.
  5. Work out v = 2d ÷ t.

What you will see: Your answer will be close to 340 m/s, but not exactly. Timing 20 claps instead of one clap makes the answer much better. The speed of sound also changes a little with the temperature of the air.

If you cannot try it, this is what happens.
A table with the columns Measure and Value. Row 1: distance to wall, 102 m. Row 2: 20 claps take, 11.8 s. Row 3: t = 11.8 ÷ 20, 0.59 s. Row 4: v = 2 × 102 ÷ 0.59, 346 m/s.MeasureValuedistance to wall102 m20 claps take11.8 st = 11.8 ÷ 200.59 sv = 2 × 102 ÷ 0.59346 m/s

The same idea gives the distance to a reflector. If you know the speed of sound and the echo time, then v × t is the whole way there and back. Half of that is the distance.

A person on the ground on the left, labelled you, and a hatched wall on the right, labelled cliff. A long arrow from the person to the cliff is labelled sound goes. A second arrow of the same length from the cliff back to the person is labelled echo. A mark under the ground between the person and the cliff is labelled ? m.sound goesecho? myoucliff
An echo gives the distance to a cliff when the speed and the time are known.

Worked example

A shout echoes from a cliff and the echo returns 1.2 s later. Take the speed of sound as 340 m/s. How far away is the cliff?

  1. The sound goes there and back in 1.2 s.
  2. The whole way is v × t = 340 m/s × 1.2 s = 408 m.
  3. The cliff is half of the whole way: 408 ÷ 2 = 204 m.

Answer: 204 m

Ordinary sound is used in this way, but sound that we can hear is a nuisance for it. For many jobs, scientists use sound that is too high for us to hear. Sound with a frequency above 20 000 Hz is called ultrasound.

Humans hear from about 20 Hz to about 20 000 Hz. The top of the range falls as people grow older. Sound below 20 Hz is called infrasound. Elephants use it. Many bats and dolphins use ultrasound.

A bar on a scale from 1 to 100 000 hertz, with each ten times further along the scale. The bar has three parts. The first part, hatched, is labelled infrasound. The middle part, the thick one, is labelled human hearing, with 20 Hz at its left end and 20 000 Hz at its right end. The last part, cross hatched, is labelled ultrasound. The scale numbers are 10, 1 000 and 100 000. The axis is labelled frequency (Hz).human hearing20 Hz20 000 Hzinfrasoundultrasound101 000100 000frequency (Hz)
The scale goes up in tens: each tick is ten times the one before. The top of the human range is 1 000 times the bottom.

Ultrasound is the same kind of wave as ordinary sound. It is longitudinal, and in a given medium it travels at the same speed as ordinary sound. It differs only in having a higher frequency, and so a shorter wavelength.

A ship uses ultrasound to find the depth of the sea. This is sonar. The ship sends a short pulse of ultrasound down. It reflects from the sea floor, or from a shoal of fish, and comes back up. The ship times the echo.

A ship on the sea surface, labelled ship, with the water shown by dots and the sea floor hatched, labelled sea floor. A downward arrow, labelled pulse, and an upward arrow of the same length, labelled echo. A mark at the right from the surface to the sea floor is labelled ? m.? mpulseechoshipsea floor
Sonar. The pulse goes down and the echo comes up, so the ship halves the whole way.

Sound in sea water travels at about 1 500 m/s. The speed changes a little with temperature, salt and depth, so the ship uses a value from measurements. The depth is half the distance there and back: d = v × t ÷ 2.

Worked example

A ship's sonar pulse returns 0.8 s after it was sent. Sound travels at 1 500 m/s in the sea. Find the depth.

  1. The echo sounder times the pulse as it goes down and comes back.
  2. Going down alone takes half the time: 0.8 s ÷ 2 = 0.4 s.
  3. The depth is speed × time = 1 500 m/s × 0.4 s = 600 m.
  4. In kilometres: 600 m ÷ 1 000 = 0.6 km.

Answer: 600 m, which is 0.6 km

Doctors use ultrasound to look inside the body without cutting it. The scanner has a probe that sends in pulses of ultrasound, with frequencies of several million hertz. A clear gel on the skin keeps air from lying between the probe and the skin. Air reflects almost all the ultrasound.

Echoes come back from every boundary where one kind of tissue meets another: skin, fluid, the baby and bone. The scanner times each echo, and notes how strong it is. A computer turns these into a picture. Sound in soft tissue travels at about 1 540 m/s, so the echoes come back in about a tenth of a millisecond.

A cut through the body seen from the side. A probe sits on the skin at the top, with a thick line under it labelled gel. A pulse arrow, labelled pulse, goes down towards a dotted oval labelled baby, and an echo arrow, labelled echo, comes back up. A mark at the right from the skin to the top of the baby is labelled 8 cm.8 cmpulseechoprobebabygel
A scan. The echo from the baby comes back to the probe, and the time says how deep it is.

Ultrasound does not ionise, which means it does not knock electrons off atoms. At the low powers used for scans, no harm to the mother or the baby has been found. This is why it is used to look at unborn babies, in place of X-rays. Scans are still only done when they are useful.

A wrong idea many people have

The echo time is the time the sound takes to reach the wall.

Why it feels right: We think of the wall as the place the sound is going to, and the echo as the end of the journey.

The test that breaks it: Take a wall 85 m away and the speed 340 m/s. Reaching the wall alone takes 85 ÷ 340 = 0.25 s. The echo reaches you after 0.5 s, twice as long, because it must come back.

What is true: The echo time covers the way there and the way back. The distance to the wall is half of v × t.

A wrong idea many people have

Light and sound travel at about the same speed.

Why it feels right: A clap seems to be seen and heard together. A flash and a bang from a nearby firework seem to come at once.

The test that breaks it: Watch a firework across a wide field or a far storm. The flash comes first, and the bang follows seconds later. At 1 km, light takes about 3 millionths of a second, and sound takes about 3 s.

What is true: Light is nearly a million times faster than sound. Sound is slow, about 340 m/s in air, and it can never be faster than light.

A wrong idea many people have

Ultrasound travels faster than ordinary sound, because it is higher.

Why it feels right: Higher frequency sounds like "more" and "quicker", and we call the sound "ultra".

The test that breaks it: In air at 340 m/s, a note of 400 Hz and an ultrasound of 40 000 Hz both travel at 340 m/s. The ultrasound has 100 times the frequency, so with v = fλ it has one hundredth of the wavelength.

What is true: The speed is set by the medium. Ultrasound has the same speed as ordinary sound in the same medium. Only its frequency and wavelength differ.

Key words

vacuum
a space with no particles in it. Sound cannot cross it, and light can.
echo
sound that reflects from a surface and returns. It travels the distance there and back.
ultrasound
sound with a frequency above 20 000 Hz, too high for people to hear.
infrasound
sound with a frequency below 20 Hz, too low for people to hear.
sonar
a way to find depths and objects in water by timing the echo of a pulse of sound.

Back to your guess

The answer is B: About 0.5 s, because the sound goes to the wall and back.

The sound has to go to the wall and come back, a total of 170 m. At about 340 m/s that takes 0.5 s. An echo always covers the distance twice.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. A clap is echoed from a wall, and the echo comes back 0.8 s later. The speed of sound is 340 m/s. How far away is the wall?

    A person on the ground on the left, labelled you, and a hatched wall on the right, labelled wall. A long arrow from the person to the wall is labelled sound goes. A second arrow of the same length from the wall back to the person is labelled echo. A mark under the ground between the person and the wall is labelled ? m.sound goesecho? myouwall
    Tap your answer.

    1. Not this one. That is the whole way there and back, 340 × 0.8. The wall is half that far away. Try another.

    2. Not this one. You divided 340 by 0.8. Distance is speed times time, and the sound went there and back. Try another.

    3. Yes, that is right. The sound travels 340 m/s × 0.8 s = 272 m there and back. The wall is half of that: 272 ÷ 2 = 136 m.

    4. Not this one. You halved twice. Halve the whole way once, and no more. Try another.

    See the answer

    136 m

    The sound travels 340 m/s × 0.8 s = 272 m there and back. The wall is half of that: 272 ÷ 2 = 136 m.

    • If you chose A: That is the whole way there and back, 340 × 0.8. The wall is half that far away.
    • If you chose B: You divided 340 by 0.8. Distance is speed times time, and the sound went there and back.
    • If you chose D: You halved twice. Halve the whole way once, and no more.

  2. A ship's sonar pulse returns after 0.4 s. Sound travels at 1 500 m/s in sea water. How deep is the sea?

    Tap your answer.

    1. Not this one. That is the whole way down and up. The depth is half of it. Try another.

    2. Not this one. You divided 1 500 by 0.4. Multiply the speed by the time, then halve. Try another.

    3. Not this one. You halved twice. The time is already the there and back time, so halve once. Try another.

    4. Yes, that is right. The pulse travels 1 500 m/s × 0.4 s = 600 m down and up. The depth is half of that: 600 ÷ 2 = 300 m.

    See the answer

    300 m

    The pulse travels 1 500 m/s × 0.4 s = 600 m down and up. The depth is half of that: 600 ÷ 2 = 300 m.

    • If you chose A: That is the whole way down and up. The depth is half of it.
    • If you chose B: You divided 1 500 by 0.4. Multiply the speed by the time, then halve.
    • If you chose C: You halved twice. The time is already the there and back time, so halve once.

  3. Why does a doctor put gel on the skin before an ultrasound scan?

    Tap your answer.

    1. Not this one. Ultrasound cannot be heard at all. Its frequency is above the range of human hearing. Try another.

    2. Yes, that is right. Air would reflect almost all the ultrasound at the skin, so little would get into the body. The gel fills the gap so the ultrasound enters the body.

    3. Not this one. The speed depends on the tissue, not on the gel. The gel helps the ultrasound get in. Try another.

    See the answer

    The gel keeps air out between the probe and the skin, because air reflects nearly all the ultrasound.

    Air would reflect almost all the ultrasound at the skin, so little would get into the body. The gel fills the gap so the ultrasound enters the body.

    • If you chose A: Ultrasound cannot be heard at all. Its frequency is above the range of human hearing.
    • If you chose C: The speed depends on the tissue, not on the gel. The gel helps the ultrasound get in.

  4. What is ultrasound?

    Tap your answer.

    1. Yes, that is right. Ultrasound has a frequency above 20 000 Hz, which is higher than people can hear. It is the same kind of wave as ordinary sound.

    2. Not this one. In the same medium, ultrasound travels at the same speed as ordinary sound. It has a higher frequency. Try another.

    3. Not this one. That is infrasound. Ultrasound has a frequency above 20 000 Hz, above the top of human hearing. Try another.

    4. Not this one. Loudness depends on the amplitude. Ultrasound is defined by its frequency, and people cannot hear it however large it is. Try another.

    See the answer

    Sound with a frequency above 20 000 Hz

    Ultrasound has a frequency above 20 000 Hz, which is higher than people can hear. It is the same kind of wave as ordinary sound.

    • If you chose B: In the same medium, ultrasound travels at the same speed as ordinary sound. It has a higher frequency.
    • If you chose C: That is infrasound. Ultrasound has a frequency above 20 000 Hz, above the top of human hearing.
    • If you chose D: Loudness depends on the amplitude. Ultrasound is defined by its frequency, and people cannot hear it however large it is.

  5. Lightning flashes, and the thunder is heard 6 s later. What does this show?

    Tap your answer.

    1. Not this one. It is the other way round. The flash reaches you first because light is much faster than sound. Try another.

    2. Not this one. The flash and the thunder begin together. Light arrives at once and sound is slow, which makes the gap. Go to S3.9.7. Try another.

    3. Yes, that is right. The flash and the bang start together. Light reaches you almost at once, and sound takes about 3 s for each kilometre, so you hear it later.

    See the answer

    Light travels much faster than sound, so the flash arrives first.

    The flash and the bang start together. Light reaches you almost at once, and sound takes about 3 s for each kilometre, so you hear it later.

    • If you chose A: It is the other way round. The flash reaches you first because light is much faster than sound.
    • If you chose B: The flash and the thunder begin together. Light arrives at once and sound is slow, which makes the gap. Go to S3.9.7.

  6. A bell rings in a jar, and a pump takes the air out. What do you hear, and why?

    Tap your answer.

    1. Not this one. Sound needs the air. Without air, nothing carries the vibrations to the glass and to your ears. Try another.

    2. Yes, that is right. Sound needs a medium. As the air goes, the sound fades. Light still crosses the empty jar, so you can see the bell move.

    3. Not this one. Sound is a wave of the air. With the air taken out, nothing is left to pass on the sound. Go to S4.10.1. Try another.

    See the answer

    The ringing gets fainter, because there is less air to carry the sound.

    Sound needs a medium. As the air goes, the sound fades. Light still crosses the empty jar, so you can see the bell move.

    • If you chose A: Sound needs the air. Without air, nothing carries the vibrations to the glass and to your ears.
    • If you chose C: Sound is a wave of the air. With the air taken out, nothing is left to pass on the sound. Go to S4.10.1.

  7. An ultrasound echo comes back from a boundary 0.12 ms after the pulse was sent. The speed of sound in the body is 1 540 m/s. How deep is the boundary, to one decimal place? Note: 1 ms = 0.001 s.

    Tap your answer.

    1. Yes, that is right. First change the time: 0.12 ms = 0.00012 s. The whole way is 1 540 × 0.00012 = 0.185 m. Half of it is 0.0924 m, which is 9.2 cm.

    2. Not this one. That is the whole way there and back. The boundary is half that deep. Try another.

    3. Not this one. You may have used 0.12 s instead of 0.12 ms. That gives 1 540 × 0.12 ÷ 2 = 92.4 m. 0.12 ms is only 0.00012 s. Try another.

    See the answer

    9.2 cm

    First change the time: 0.12 ms = 0.00012 s. The whole way is 1 540 × 0.00012 = 0.185 m. Half of it is 0.0924 m, which is 9.2 cm.

    • If you chose B: That is the whole way there and back. The boundary is half that deep.
    • If you chose C: You may have used 0.12 s instead of 0.12 ms. That gives 1 540 × 0.12 ÷ 2 = 92.4 m. 0.12 ms is only 0.00012 s.

  8. From an earlier chapter

    Sound of frequency 800 Hz travels at 340 m/s in air. What is its wavelength?

    Tap your answer.

    1. Not this one. You multiplied 340 by 800. For the wavelength divide the speed by the frequency: λ = v ÷ f. Go to S4.10.2. Try another.

    2. Not this one. You divided 800 by 340. The wavelength is the speed divided by the frequency, not the other way round. Go to S4.10.2. Try another.

    3. Yes, that is right. λ = v ÷ f = 340 m/s ÷ 800 Hz = 0.425 m.

    See the answer

    0.425 m

    λ = v ÷ f = 340 m/s ÷ 800 Hz = 0.425 m.

    • If you chose A: You multiplied 340 by 800. For the wavelength divide the speed by the frequency: λ = v ÷ f. Go to S4.10.2.
    • If you chose B: You divided 800 by 340. The wavelength is the speed divided by the frequency, not the other way round. Go to S4.10.2.

  9. A sonar pulse returns after 0.5 s. Sound travels at 1 500 m/s in the sea. How deep is the sea?

    See the answer

    375 m

    The whole way is 1 500 × 0.5 = 750 m. The depth is half of it: 375 m.

  10. You clap, and the echo from a cliff comes back 1.5 s later. The speed of sound is 340 m/s. How far away is the cliff?

    See the answer

    255 m

    The whole way is 340 × 1.5 = 510 m. The cliff is half of it: 255 m.

Keep it

Copy this into your exercise book.

Sound needs a medium and cannot cross a vacuum. Echo: v = 2d ÷ t. Depth = v × t ÷ 2. Ultrasound is above 20 000 Hz. Light is far faster than sound.

Draw: A ship, the sea and the sea floor. Draw a pulse arrow going down and an echo arrow coming up. Mark the depth.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

No internet tomorrow? Save the whole book as one file (951 KB). It opens on any phone.