Applied chapter: ultrasound imaging and earthquake waves
Why can an ultrasound machine draw a picture of the inside of the body using nothing but echoes, and why is gel spread under the probe first?
Prior knowledge
- The speed, frequency and wavelength of a wave are related by the equation $v = f\lambda$, and a wave can be transverse or longitudinalchapter How waves form and travel
- A wave is partly reflected from the boundary of two substances, and an obstacle bends a wave more the greater the wavelength is compared with the obstaclechapter A wave meets an obstacle or a gap
- Sound is a longitudinal wave, and a person hears frequencies from about $20\,\text{Hz}$ to $20\,\text{kHz}$chapter Sound as a wave: pitch, timbre and the Doppler effect
Think about it
In the examination room a patient lies on their back. A nurse spreads cold, translucent gel on the skin of the abdomen and presses a probe the size of a palm on top of it. A grey fan appears on the screen, showing the edges of organs. The probe sends out neither light nor X-rays. It sends sound that the ear cannot hear, and listens for echoes.
How can the depth of a structure in the body be worked out from echoes alone, and what is the gel needed for?
Review: wavelength
Predict
Make your predictions before investigating 2 questions
Prediction 1: how deep?
Prediction 2: why gel?
Investigate
Investigate: how much is reflected?
Values calculated by a model
The table shows what fraction of the energy of an ultrasound pulse is reflected from different boundaries when it arrives perpendicularly. The percentages have been calculated from the typical acoustic properties of the substances, and they are approximate. The table also gives the speed of sound in the substances.
| Boundary | Reflected about |
|---|---|
| gel – skin | $0.2\,\%$ |
| fat – muscle | $1\,\%$ |
| soft tissue – bone | $40\,\%$ |
| air – skin | $99.9\,\%$ |
| Substance | Speed of sound |
|---|---|
| air | $343\,\text{m/s}$ |
| fat | $1450\,\text{m/s}$ |
| soft tissue, average | $1540\,\text{m/s}$ |
| muscle | $1580\,\text{m/s}$ |
| bone | about $3500\,\text{m/s}$ |
Investigate: reading the echo curve
Made-up example data
The graph shows an A-scan, that is, an echo curve, of a thigh. The probe is on the skin, and the pulse is sent at the moment $t = 0$. All the echoes are in the same direction under the skin. Echo A comes from the boundary of fat and muscle, echo B from a membrane between two muscles and echo C from the surface of the femur. The machine assumes that sound travels in tissue at a speed of $1540\,\text{m/s}$.
Read the times of echoes B and C from the graph. How thick is the layer of tissue between boundaries B and C? Answer in centimetres.
Hint 1: a guiding question
What distance does the time difference of echoes B and C correspond to? The pulse travels through the layer twice.
Hint 2: what is the connection
The time difference of the echoes is
and the thickness of the layer is $\tfrac{v \Delta t}{2}$.
Hint 3: an intermediate step
Show the solution
Given
speed of sound in tissue $v = 1540\,\text{m/s}$; from the graph the time of echo B and the time of echo C
Find
the thickness $\Delta d$ of the layer between boundaries B and C in centimetres
Principle
During the travel time of an echo the pulse goes to the boundary and back, so the depth is $d = \tfrac{vt}{2}$. The time difference of the echoes is the time the pulse needs to travel down through the layer and back, when the speed is the same all the way.
Reading the graph
The peak of echo B is on the grid line $50\,\mu\text{s}$ and the peak of echo C on the grid line $90\,\mu\text{s}$. The time difference is
Formula
so $\Delta d = \dfrac{v \Delta t}{2}$
Conversions
$40\,\mu\text{s} = 40 \cdot 10^{-6}\,\text{s}$; finally $1\,\text{m} = 100\,\text{cm}$.
Substitution
Answer
$3.1\,\text{cm}$
Two significant figures, because the time difference $40\,\mu\text{s}$ read from the graph is known to about a microsecond.
Reasonableness
The boundaries are at depths $d_\text{B} = 3.85\,\text{cm}$ and $d_\text{C} = 6.93\,\text{cm}$, and their difference is the same $3.08\,\text{cm}$. A layer of muscle a few centimetres thick on top of the femur is plausible in an adult.
Investigate: three seismograph stations
Values calculated by a model
An earthquake sends out two waves that travel through the interior of the Earth: the P wave and the S wave. The graph shows the seismograms of three stations from the same earthquake. The stations are at different distances from the earthquake, and the time is counted from the moment of the earthquake. The values have been calculated from a model in which the waves travel straight and at constant speeds.
Read the difference $\Delta t$ in the arrival times of the P wave and the S wave at each station.
How does it depend on the distance of the station? How many kilometres of distance correspond to one second of time difference? Answer in the unit km/s.
Hint 1: a guiding question
From each row read the start times of P and S and calculate their difference. Compare the differences with the distances.
Hint 2: what is the connection
The time differences are $10\,\text{s}$, $20\,\text{s}$ and $30\,\text{s}$. When the distance doubles, the time difference also doubles.
Hint 3: an intermediate step
$\tfrac{84\,\text{km}}{10\,\text{s}}$
Show the solution
Given
the distances of the stations $84\,\text{km}$, $168\,\text{km}$ and $252\,\text{km}$; from the graph the arrival times of the P wave and the S wave
Find
the distance per second of time difference, $\tfrac{d}{\Delta t}$, in the unit km/s
Principle
Both waves travel the same distance at a constant speed, so the travel time of each is directly proportional to the distance. Then the difference of the travel times is also directly proportional to the distance, and the ratio $\tfrac{d}{\Delta t}$ is the same at all stations.
Reading the graph
At the top station P starts at $14\,\text{s}$ and S at $24\,\text{s}$, so $\Delta t = 10\,\text{s}$. At the middle one $28\,\text{s}$ and $48\,\text{s}$, $\Delta t = 20\,\text{s}$. At the bottom one $42\,\text{s}$ and $72\,\text{s}$, $\Delta t = 30\,\text{s}$. The time difference grows in the same proportion as the distance.
Formula
$d = k\,\Delta t$, from which $k = \dfrac{d}{\Delta t}$
Conversions
The distances are in kilometres and the times in seconds, so no conversions are needed.
Substitution
Answer
$8.4\,\text{km/s}$
Two significant figures, because the arrival times are read from the graph to about a second.
Reasonableness
The number is not the speed of either wave: it is greater than the speed of the P wave $\tfrac{84\,\text{km}}{14\,\text{s}} = 6.0\,\text{km/s}$, because the time difference is smaller than either travel time. All three stations give the same ratio, so the time difference is directly proportional to the distance.
Theory summary
Key idea
An echo tells the depth
Ultrasound is sound with a frequency above $20\,\text{kHz}$, that is, too high for a person to hear. In medical imaging frequencies of about $2\text{–}15\,\text{MHz}$ are used. The probe sends a short ultrasound pulse into the body and then listens for echoes. When the pulse meets the boundary of two tissues, part of it is reflected back and the rest continues deeper (chapter A wave meets an obstacle or a gap). So every boundary gives an echo of its own. The machine measures how long it takes for each echo to return. Because the pulse travels to the boundary and back, during the travel time of the echo it covers twice the depth of the boundary. The machine assumes that sound travels in soft tissue at an average speed of $1540\,\text{m/s}$, and calculates the depth of every echo.
When the probe is tilted or it has a row of small transmitters, pulses are sent in many directions. From the depths and strengths of the echoes a two-dimensional greyscale image, the B-scan, is formed. The simplest display is the A-scan, the echo curve: its horizontal axis shows the time from sending the pulse and its vertical axis the strength of the echo. A peak in the echo curve means a boundary, and the position of the peak tells how deep the boundary is.
Why gel is needed
How large a part of the wave is reflected depends on how different the substances at the boundary are as far as the propagation of sound is concerned. Soft tissues resemble each other, so only about one per cent is reflected from their boundaries. Therefore most of the pulse continues deeper and produces echoes also from deeper boundaries. From the boundary of bone and soft tissue about $40\,\%$ is reflected, and bone also strongly attenuates sound, so behind a bone there is a shadow that gives no echoes. From the boundary of air and skin about $99.9\,\%$ is reflected. If even a thin layer of air is left between the probe and the skin, almost the whole pulse is reflected back already at the skin. Gel displaces the air: from the boundary of gel and skin only about $0.2\,\%$ is reflected.
Frequency: a sharp image or an image that reaches deep
The wavelength is obtained from the equation $v = f\lambda$ (chapter How waves form and travel). At the frequency $5\,\text{MHz}$ the wavelength of ultrasound in soft tissue is about $0.3\,\text{mm}$. The smallest details that can be distinguished are roughly a few wavelengths in size, so a high frequency and a short wavelength give a sharper image. On the other hand, tissue turns part of the energy of the sound wave into internal energy, and this attenuation grows approximately in direct proportion to the frequency. A high-frequency pulse therefore weakens quickly, and its echo does not return from deep down strongly enough. Organs deep down, such as the heart and the liver, are imaged with a low frequency, and structures near the surface, such as tendons and the thyroid gland, with a high one.
| $f$ | $\lambda$ | Depth | Target |
|---|---|---|---|
| $2.5\,\text{MHz}$ | $0.62\,\text{mm}$ | $20\,\text{cm}$ | heart, abdomen |
| $5\,\text{MHz}$ | $0.31\,\text{mm}$ | $10\,\text{cm}$ | pregnancy |
| $10\,\text{MHz}$ | $0.15\,\text{mm}$ | $5\,\text{cm}$ | thyroid gland |
| $15\,\text{MHz}$ | $0.10\,\text{mm}$ | $3\,\text{cm}$ | tendons, skin |
In the table $f$ is the frequency, $\lambda$ the wavelength in tissue and the depth the typical imaging depth. The imaging depths are only indicative: they depend on the machine and the tissue. Ultrasound can also be used to measure the flow speed of blood. The frequency of the echo reflected from moving blood cells is slightly different from that of the transmitted pulse, and from the change in frequency the machine calculates the speed. This is the Doppler effect (chapter Sound as a wave: pitch, timbre and the Doppler effect).
Echolocation in nature and technology (extra)
Bats and dolphins locate their prey on the same principle. A bat sends out short ultrasonic calls whose frequency, depending on the species, is about $20$–$100\,\text{kHz}$, and works out the distance from the delay of the echoes. At the frequency $50\,\text{kHz}$ the wavelength in air is about $7\,\text{mm}$. From an object much smaller than the wavelength only a very weak echo returns, because the wave bends around it (chapter A wave meets an obstacle or a gap). That is why bats that hunt insects use high frequencies. The sonar of a ship measures the depth of the water in the same way and detects shoals of fish; sound travels in water at about $1400\text{–}1500\,\text{m/s}$, depending on the temperature.
Earthquake waves
In an earthquake the bedrock breaks, and earthquake waves set out from the point of fracture. Two kinds of waves travel through the interior of the Earth. The P wave is a longitudinal wave in which the rock is compressed and rarefied in the direction of propagation. It is the fastest and arrives at a station first. The S wave is a transverse wave in which the rock moves sideways. It is slower than the P wave. In the upper part of the crust the speed of the P wave is typically about $6\,\text{km/s}$ and that of the S wave about $3.5\,\text{km/s}$. Surface waves, which travel along the surface, are slower still, but their amplitude is the greatest, and they cause most of the damage.
A transverse wave needs a medium that resists a change of shape: when a layer of rock is displaced sideways, the neighbouring layers pull it back. A liquid does not resist a change of shape in the same way, so the S wave does not travel in a liquid. The P wave also travels in a liquid, like sound in water. When earthquakes were detected on different sides of the globe, it was noticed that S waves are not detected on the half opposite the earthquake, when the angular distance, the angle between the earthquake and the station measured at the centre of the Earth, is more than about $103^\circ$. A shadow zone is left for S waves because the outer core of the Earth is liquid.
Tsunami (extra)
An earthquake on the sea floor can suddenly raise or lower a large mass of water, and a tsunami sets out from it. Its wavelength is tens or hundreds of kilometres, much greater than the depth of the ocean. Then the water moves from the bottom all the way to the surface, and the speed of the wave depends only on the depth of the water: in the deep ocean a tsunami travels at the speed of an aeroplane. When the wave comes into shallow water near the shore, it slows down, its wavelength shortens and its energy is concentrated in a smaller amount of water. That is why the height of the wave grows many times over near the coast.
Distance from the travel time of a wave
When a pulse travels to a boundary and back along the same path at a constant speed, the distance of the boundary is
$$d = \frac{v t}{2}$$
where $d$ is the distance of the boundary (m), $v$ the speed of sound in the medium (m/s) and $t$ the travel time of the echo, that is, the time from sending the pulse to the return of the echo (s). In an ultrasound machine $v = 1540\,\text{m/s}$. The formula also holds for the echoes of a bat and a sonar, when the speed is the speed of sound in air or water. An A-scan is read by taking the time of each peak from the time axis and converting it to a depth with this formula. The thickness of a layer between two boundaries is obtained from the time difference of the echoes with the same formula.
A seismograph station does not know when the earthquake happened, but it measures the difference $\Delta t$ in the arrival times of the P wave and the S wave. When the waves cover the same distance $d$ at constant speeds, the S wave falls behind the P wave the more, the longer the distance is:
$$\begin{aligned} \Delta t &= \frac{d}{v_\text{S}} - \frac{d}{v_\text{P}} \\[6pt] d &= \frac{v_\text{P} v_\text{S}}{v_\text{P} - v_\text{S}}\,\Delta t \end{aligned}$$
where $d$ is the distance of the station from the point where the earthquake started (km), $v_\text{P}$ and $v_\text{S}$ the speeds of the P wave and the S wave (km/s) and $\Delta t$ the difference between the arrival times of the S wave and the P wave (s). When $v_\text{P} = 6.0\,\text{km/s}$ and $v_\text{S} = 3.5\,\text{km/s}$, every second of time difference corresponds to a distance of $8.4\,\text{km}$. When the distance is known from three stations, the location of the earthquake is at the intersection of the circles drawn around them.
When the wavelength is much greater than the depth of the water, the speed of a shallow-water wave, such as a tsunami, is
$$v = \sqrt{g h}$$
where $v$ is the speed of the wave (m/s), $g = 9.81\,\text{m/s}^2$ the acceleration of free fall and $h$ the depth of the water (m). The equation is given here without derivation. At a depth of four kilometres the speed is about $198\,\text{m/s}$, or about $710\,\text{km/h}$, but in shallow water $10\,\text{m}$ deep only about $9.9\,\text{m/s}$.
Sources of the values
The speed of sound $1540\,\text{m/s}$ is the standard assumption of ultrasound machines for soft tissue (for example Hoskins, Martin and Thrush: Diagnostic Ultrasound, 3rd edition, 2019). The speeds of waves in the crust are approximations from the Earth speed model IASP91 (Kennett and Engdahl 1991), in which the values for the uppermost crust are $5.8\,\text{km/s}$ and $3.36\,\text{km/s}$. The percentages of the reflected fractions have been calculated from the typical acoustic properties of the tissues, and they are approximations.
Common misconception
The most common calculation error is to forget that the echo travels there and back. If an ultrasound echo returns after $65\,\mu\text{s}$, the product $v t$ is $10\,\text{cm}$, but the boundary is only at a depth of $5.0\,\text{cm}$. Another error is to think that the highest frequency is always best, because it gives the sharpest image. High-frequency ultrasound is attenuated quickly, so deep structures are imaged with a lower frequency and sharpness has to be compromised. A third error is to think of ultrasound as radiation like X-rays. Ultrasound is mechanical wave motion that needs a medium, and it does not ionise tissue.
Summary
- Ultrasound is sound with a frequency above $20\,\text{kHz}$; medical imaging uses frequencies of about $2\text{–}15\,\text{MHz}$.
- Part of the pulse is reflected from every boundary. The depth of a boundary is obtained from the travel time of the echo, $d = \tfrac{vt}{2}$, when $v = 1540\,\text{m/s}$.
- Gel displaces the air between the probe and the skin; from the boundary of air and skin almost the whole pulse would be reflected.
- A high frequency gives a short wavelength and a sharp image, but it is attenuated quickly; deep structures are imaged with a low frequency.
- The P wave is longitudinal and the fastest, the S wave transverse and slower. The S wave does not travel in a liquid, which shows that the outer core is liquid.
- The distance of an earthquake is obtained from the difference in the arrival times of the S wave and the P wave: $d = \tfrac{v_\text{P} v_\text{S}}{v_\text{P} - v_\text{S}}\Delta t$.
- The speed of a tsunami depends on the depth of the water, $v = \sqrt{gh}$; near the coast the wave slows down and grows tall.
Return to your predictions
Compare your predictions with the echo curve, the table of boundaries and the Theory summary. Then return to the opening question: how does the machine work out the depth from an echo, and what would happen without gel?
Would you change your prediction?
Apply
Worked example
Problem
A seismograph station records an earthquake. The clock of the station showed $12\,\text{h}\ 04\,\text{min}\ 18\,\text{s}$ when the P wave arrived and $12\,\text{h}\ 04\,\text{min}\ 43\,\text{s}$ when the S wave arrived. The speed of the P wave in the crust is $6.0\,\text{km/s}$ and that of the S wave $3.5\,\text{km/s}$. How far from the station did the earthquake occur? Assume that the waves travel along the same straight path at constant speeds.
Step 1/7
Given and find. $v_\text{P} = 6.0\,\text{km/s}$, $v_\text{S} = 3.5\,\text{km/s}$, arrival times $12\,\text{h}\ 04\,\text{min}\ 18\,\text{s}$ and $12\,\text{h}\ 04\,\text{min}\ 43\,\text{s}$. We are asked for the distance $d$ to the point where the earthquake started.
Step 2/7
Principle. Each wave covers the same distance $d$ at a constant speed, so the travel time is the distance divided by the speed. The moment of the earthquake is not known, but the difference $\Delta t$ in the arrival times is the difference of the travel times of the S wave and the P wave.
Step 3/7
Formula solved. Take $d$ as a common factor and solve:
$$\begin{aligned} \Delta t &= d\left(\frac{1}{v_\text{S}} - \frac{1}{v_\text{P}}\right) \\[6pt] d &= \frac{v_\text{P} v_\text{S}}{v_\text{P} - v_\text{S}}\,\Delta t \end{aligned}$$
Step 4/7
Unit conversions. The hours and minutes are the same, so the difference in arrival times is $43\,\text{s} - 18\,\text{s} = 25\,\text{s}$. The speeds are in km/s and the time in seconds, so the distance comes out directly in kilometres.
Step 5/7
Substitution. $$\begin{aligned} d &= \frac{v_\text{P} v_\text{S}}{v_\text{P} - v_\text{S}}\,\Delta t \\ &= \frac{6.0\,\tfrac{\text{km}}{\text{s}} \cdot 3.5\,\tfrac{\text{km}}{\text{s}}}{6.0\,\tfrac{\text{km}}{\text{s}} - 3.5\,\tfrac{\text{km}}{\text{s}}} \cdot 25\,\text{s} \\ &= 8.4\,\tfrac{\text{km}}{\text{s}} \cdot 25\,\text{s} \\ &= 210\,\text{km} \end{aligned}$$
Step 6/7
Result. The earthquake occurred about $210\,\text{km}$ from the station. The result is given with two significant figures, because the speeds $6.0\,\text{km/s}$ and $3.5\,\text{km/s}$ are known to two figures.
Step 7/7
Reasonableness. Let us check the travel times: the P wave covers $210\,\text{km} : 6.0\,\text{km/s} = 35\,\text{s}$ and the S wave $210\,\text{km} : 3.5\,\text{km/s} = 60\,\text{s}$. Their difference is $25\,\text{s}$, as measured. So the earthquake occurred $35\,\text{s}$ before the arrival of the P wave, that is, when the clock of the station showed $12\,\text{h}\ 03\,\text{min}\ 43\,\text{s}$.
Apply to a new situation
An earthquake on the sea floor creates a tsunami $2000\,\text{km}$ from the coast. The depth of the ocean along the path of the wave is on average $4.0\,\text{km}$.
How many hours do the people on the coast have to evacuate when the alarm is given immediately after the earthquake? Assume that the tsunami travels the whole distance at the speed $v = \sqrt{gh}$. Answer in hours.
Hint 1: a guiding question
What determines the speed of a tsunami? First calculate the speed and then the time in which the distance is covered.
Hint 2: what is the connection
Convert the depth to metres: $h = 4000\,\text{m}$. The speed is $v = \sqrt{gh}$ and the time $t = \tfrac{s}{v}$.
Hint 3: an intermediate step
$t = \tfrac{2\,000\,000\,\text{m}}{198\,\text{m/s}}$, and divide the seconds by $3600$.
Show the solution
Given
distance $s = 2000\,\text{km}$, depth $h = 4.0\,\text{km}$, $g = 9.81\,\text{m/s}^2$
Find
the travel time $t$ of the tsunami in hours
Principle
The wavelength of a tsunami is much greater than the depth of the ocean, so it travels as a shallow-water wave at the speed $v = \sqrt{gh}$. When the depth is assumed to be the same all the way, the speed is constant and the travel time is the distance divided by the speed.
Formula
$v = \sqrt{gh}$ and $s = vt$, so $t = \dfrac{s}{\sqrt{gh}}$
Conversions
and finally $1\,\text{h} = 3600\,\text{s}$.
Substitution
Answer
$2.8\,\text{h}$
Two significant figures like the depth $4.0\,\text{km}$.
Reasonableness
The speed $198\,\text{m/s}$ is about $710\,\text{km/h}$, of the order of a jet aircraft, and covering $2000\,\text{km}$ at this speed takes just under three hours: $\tfrac{2000\,\text{km}}{710\,\text{km/h}} \approx 2.8\,\text{h}$. In reality the wave slows down in the shallow water near the coast, so the warning time is slightly longer.
Find the error
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- A bat hunts moths and other insects about a centimetre in size. Why is the frequency of its calls tens of kilohertz and not, for example, $1\,\text{kHz}$?
Answer
From an object much smaller than the wavelength only a very weak echo returns, because the wave bends around it. At the frequency $50\,\text{kHz}$ the wavelength in air is $\tfrac{343\,\text{m/s}}{50\,000\,\text{Hz}} \approx 7\,\text{mm}$, which is of the order of the size of an insect, but at the frequency $1\,\text{kHz}$ it is about $34\,\text{cm}$, tens of times greater, so only a very weak echo would return from an insect.
- Why can a tsunami that is only about a metre high in the ocean rise to many metres high at the coast?
Answer
The speed of a tsunami is $v = \sqrt{gh}$, so in the shallow water near the shore the wave slows down. The back of the wave catches up with the front, the wavelength shortens and the energy is concentrated in a smaller amount of water, so the height of the wave grows.
1. The core of the Earth
2. The length of an eye
Before cataract surgery the length of the eye is measured with an A-scan. The echo from the retina returns $31.0\,\mu\text{s}$ later than the echo from the surface of the cornea. Sound travels in the eye at an average speed of $1550\,\text{m/s}$.
How long is the eye from the cornea to the retina? Answer in centimetres.
Hint 1: a guiding question
The pulse travels through the eye to the retina and back.
Hint 2: what is the connection
$d = \tfrac{v \Delta t}{2}$, when $\Delta t = 31.0 \cdot 10^{-6}\,\text{s}$.
Hint 3: an intermediate step
$\tfrac{1550\,\text{m/s} \cdot 31.0 \cdot 10^{-6}\,\text{s}}{2}$
Show the solution
Given
time difference of the echoes $\Delta t = 31.0\,\mu\text{s}$, speed of sound in the eye $v = 1550\,\text{m/s}$
Find
the length of the eye $d$ in centimetres
Principle
In the time difference of the echoes the pulse travels from the cornea to the retina and back, so it covers the length of the eye twice. The speed is assumed to be the same all the way.
Formula
$2d = v \Delta t$, so $d = \dfrac{v \Delta t}{2}$
Conversions
$31.0\,\mu\text{s} = 31.0 \cdot 10^{-6}\,\text{s}$; finally $1\,\text{m} = 100\,\text{cm}$.
Substitution
Answer
$2.40\,\text{cm}$
Three significant figures like the time difference $31.0\,\mu\text{s}$.
Reasonableness
The eyeball of an adult is about $2.4\,\text{cm}$ in diameter, so the result is plausible. Had the return trip been forgotten, the eye would have been almost $5\,\text{cm}$ long, which is clearly too much.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
3. Explain the choice of frequency
The heart is usually imaged with a probe of about $3\,\text{MHz}$, but a tendon of the wrist with a probe of about $12\,\text{MHz}$. Explain why the frequencies are chosen like this. Use the concepts wavelength and attenuation.
Model answer
The wavelength is $\lambda = \tfrac{v}{f}$, so the wavelength of $12\,\text{MHz}$ ultrasound in tissue is about $0.13\,\text{mm}$ and that of $3\,\text{MHz}$ about $0.5\,\text{mm}$. The smallest details that can be distinguished are a few wavelengths in size, so a higher frequency gives a sharper image. However, attenuation grows approximately in direct proportion to the frequency, so a $12\,\text{MHz}$ pulse weakens quickly and its echo does not return from deeper than a few centimetres. A tendon is right under the skin, so a small depth is enough for imaging it and a high frequency can be used. The heart extends to a depth of over $10\,\text{cm}$, so a low frequency is used that reaches that far, and sharpness has to be compromised.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Exercises
Level 1 · basic practice 5 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 17-1Core
In monitoring a pregnancy an ultrasound machine sends a pulse from the skin of the abdomen. The echo reflected from the surface of the fetus's head returns to the probe after $91.0\,\mu\text{s}$. Sound travels in tissue at a speed of $1540\,\text{m/s}$.
How deep below the surface of the skin is the head of the fetus? Give the answer in centimetres.
Answer
$7.01\,\text{cm}$
Show the solution
Given
travel time of the echo $t = 91.0\,\mu\text{s}$, speed of sound in tissue $v = 1540\,\text{m/s}$
Find
the depth $d$ of the surface of the fetus's head in centimetres
Principle
The pulse travels from the skin to the surface of the fetus's head and back as an echo, so during the travel time it covers twice the depth. The speed is assumed to be the same all the way, as the machine assumes.
Formula
$2d = v t$, so $d = \dfrac{v t}{2}$
Conversions
$91.0\,\mu\text{s} = 91.0 \cdot 10^{-6}\,\text{s}$; finally $1\,\text{m} = 100\,\text{cm}$.
Substitution
$d$$= \dfrac{1540\,\text{m/s} \cdot 91.0 \cdot 10^{-6}\,\text{s}}{2}$$= 0.070\,07\,\text{m}$$= 7.007\,\text{cm}$Answer
$7.01\,\text{cm}$
Three significant figures like the travel time $91.0\,\mu\text{s}$.
Reasonableness
In round numbers sound travels in tissue about $1.5\,\text{mm}$ in a microsecond, so in $90\,\mu\text{s}$ about $14\,\text{cm}$ there and back and $7\,\text{cm}$ in one direction. A fetus a few centimetres under the abdominal wall is plausibly at this depth.
- Exercise 17-2Core
The arteries of the neck are imaged with a probe whose frequency is $7.50\,\text{MHz}$. Sound travels in tissue at a speed of $1540\,\text{m/s}$.
What is the wavelength of the ultrasound in tissue? Give the answer in millimetres.
Answer
$0.205\,\text{mm}$
Show the solution
Given
frequency $f = 7.50\,\text{MHz}$, speed of sound in tissue $v = 1540\,\text{m/s}$
Find
the wavelength $\lambda$ in millimetres
Principle
The speed of a wave is the product of the frequency and the wavelength. The frequency is determined by the probe, and the speed by the tissue.
Formula
$v = f\lambda$, so $\lambda = \dfrac{v}{f}$
Conversions
$7.50\,\text{MHz} = 7.50 \cdot 10^{6}\,\text{Hz}$; finally $1\,\text{m} = 1000\,\text{mm}$.
Substitution
$\lambda$$= \dfrac{1540\,\text{m/s}}{7.50 \cdot 10^{6}\,\text{Hz}}$$= 2.053 \cdot 10^{-4}\,\text{m}$$= 0.2053\,\text{mm}$Answer
$0.205\,\text{mm}$
Three significant figures like the frequency $7.50\,\text{MHz}$.
Reasonableness
In the table of this chapter the wavelength at $5\,\text{MHz}$ is $0.31\,\text{mm}$ and at $10\,\text{MHz}$ it is $0.15\,\text{mm}$. The result is between these, like the frequency. A wavelength of less than a millimetre is enough to distinguish the wall of an artery.
- Exercise 17-3
The echo sounder of a boat sends a sound pulse straight down into a lake. The echo from the bottom returns after $32.0\,\text{ms}$. Sound travels in lake water at a speed of $1480\,\text{m/s}$.
How deep is the lake at this spot? Give the answer in metres.
Answer
$23.7\,\text{m}$
Show the solution
Given
travel time of the echo $t = 32.0\,\text{ms}$, speed of sound in water $v = 1480\,\text{m/s}$
Find
the depth of the lake $d$ in metres
Principle
The pulse travels from the sonar attached near the surface to the bottom and back as an echo, so during the travel time it covers twice the depth. The speed of sound is assumed to be the same in the whole column of water.
Formula
$2d = v t$, so $d = \dfrac{v t}{2}$
Conversions
$32.0\,\text{ms}$$= 32.0 \cdot 10^{-3}\,\text{s}$$= 0.0320\,\text{s}$Substitution
$d$$= \dfrac{1480\,\text{m/s} \cdot 0.0320\,\text{s}}{2}$$= 23.68\,\text{m}$Answer
$23.7\,\text{m}$
Three significant figures like the travel time $32.0\,\text{ms}$.
Reasonableness
In round numbers sound travels in water about $1.5\,\text{m}$ in a millisecond, so in $32\,\text{ms}$ about $48\,\text{m}$ there and back and $24\,\text{m}$ in one direction. A depth of a couple of dozen metres is common in a Finnish lake.
- Exercise 17-4
A bat detects a moth: the echo returns $15\,\text{ms}$ after the call. Sound travels in air at a speed of $343\,\text{m/s}$.
How far away is the moth? Assume that the bat and the moth are stationary during this time. Give the answer in metres.
Answer
$2.6\,\text{m}$
Show the solution
Given
travel time of the echo $t = 15\,\text{ms}$, speed of sound in air $v = 343\,\text{m/s}$; the bat and the moth stationary
Find
the distance $d$ to the moth in metres
Principle
The call travels from the bat to the moth and back as an echo, so during the travel time the sound covers twice the distance. Because neither of them moves, the distance is the same in both directions.
Formula
$2d = v t$, so $d = \dfrac{v t}{2}$
Conversions
$15\,\text{ms}$$= 15 \cdot 10^{-3}\,\text{s}$$= 0.015\,\text{s}$Substitution
$d$$= \dfrac{343\,\text{m/s} \cdot 0.015\,\text{s}}{2}$$= 2.5725\,\text{m}$Answer
$2.6\,\text{m}$
Two significant figures like the travel time $15\,\text{ms}$.
Reasonableness
Sound travels in air about $0.34\,\text{m}$ in a millisecond, so in $15\,\text{ms}$ about $5\,\text{m}$ there and back. A moth a few metres away is a typical hunting distance for a bat.
- Exercise 17-5
A tsunami arrives at a continental shelf where the sea is $50\,\text{m}$ deep. The wavelength is much greater than the depth.
What is the speed of the tsunami? Give the answer in metres per second.
Answer
$22\,\text{m/s}$
Show the solution
Given
depth of the water $h = 50\,\text{m}$, $g = 9.81\,\text{m/s}^2$; wavelength much greater than the depth
Find
the speed $v$ of the tsunami in metres per second
Principle
When the wavelength is much greater than the depth of the water, the wave is a shallow-water wave, whose speed depends only on the depth and the acceleration of free fall.
Formula
$v = \sqrt{g h}$; the speed is proportional to the square root of the depth.
Conversions
The depth is in metres and $g$ in the unit $\text{m/s}^2$, so no conversions are needed: $\sqrt{\text{m/s}^2 \cdot \text{m}} = \text{m/s}$.
Substitution
$v$$= \sqrt{9.81\,\text{m/s}^2 \cdot 50\,\text{m}}$$= \sqrt{490.5\,\text{m}^2/\text{s}^2}$$= 22.15\,\text{m/s}$Answer
$22\,\text{m/s}$
Two significant figures like the depth $50\,\text{m}$.
Reasonableness
The speed is about $80\,\text{km/h}$, much less than in the ocean, where a tsunami travels at about $198\,\text{m/s}$ at a depth of four kilometres. The depth is $80$ times smaller, so the speed should be about $\sqrt{80} \approx 9$ times smaller: $198\,\text{m/s} : 9 \approx 22\,\text{m/s}$.
Level 2 · applied practice 4 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 17-6Core
The graph shows the seismogram of a seismograph station from an earthquake. The time is counted from the start of the recording, not from the moment of the earthquake. The speed of the P wave is $7.0\,\text{km/s}$ and that of the S wave $4.0\,\text{km/s}$.
How far from the station did the earthquake occur? Give the answer in kilometres.
Answer
$168\,\text{km}$
Show the solution
Given
a seismogram as a graph, $v_\text{P} = 7.0\,\text{km/s}$, $v_\text{S} = 4.0\,\text{km/s}$
Find
the distance $d$ of the station from the point where the earthquake started, in kilometres
Principle
Each wave covers the same distance at a constant speed, so the difference in arrival times is the difference of the travel times of the S wave and the P wave. The moment of the earthquake is not needed, because the time difference does not depend on the moment from which time is counted.
Reading the graph
The small oscillation, the P wave, starts on the grid line $12\,\text{s}$ and the larger oscillation, the S wave, on the grid line $30\,\text{s}$. The time difference is
$\Delta t$$= 30\,\text{s} - 12\,\text{s}$$= 18\,\text{s}$Formula
$\Delta t$$= \dfrac{d}{v_\text{S}} - \dfrac{d}{v_\text{P}}$so
$d$$= \dfrac{v_\text{P} v_\text{S}}{v_\text{P} - v_\text{S}}\,\Delta t$Conversions
The speeds are in km/s and the time in seconds, so the distance comes out in kilometres.
Substitution
$d$$= \dfrac{7.0\,\text{km/s} \cdot 4.0\,\text{km/s}}{7.0\,\text{km/s} - 4.0\,\text{km/s}} \cdot 18\,\text{s}$$= 9.33\,\text{km/s} \cdot 18\,\text{s}$$= 168\,\text{km}$Answer
$168\,\text{km}$
(the precision of the speeds $7.0$ and $4.0$ would limit the answer to two significant figures, $170\,\text{km}$; $168\,\text{km}$ is given with the intermediate result).
Reasonableness
The P wave covers $168\,\text{km}$ in $168\,\text{km} : 7.0\,\text{km/s} = 24\,\text{s}$ and the S wave in $168\,\text{km} : 4.0\,\text{km/s} = 42\,\text{s}$; the difference is $18\,\text{s}$, as read from the graph. So the recording started $12\,\text{s}$ after the earthquake.
- Exercise 17-7Core
In an image of the thyroid gland we want to distinguish structures that are $0.300\,\text{mm}$ apart. Assume that two boundaries can be distinguished when the distance between them is at least two wavelengths. Sound travels in tissue at a speed of $1540\,\text{m/s}$.
What is the lowest frequency with which this succeeds? Give the answer in megahertz.
Answer
$10.3\,\text{MHz}$
Show the solution
Given
distance between the structures $0.300\,\text{mm}$, condition for being distinguished: distance at least $2\lambda$, speed of sound $v = 1540\,\text{m/s}$
Find
the lowest frequency $f$ in megahertz
Principle
As the exercise says, assume that the structures can be distinguished when the distance between them is at least two wavelengths (the text of the chapter says more loosely "a few wavelengths"), that is, the wavelength is at most half of the distance between them. The wavelength shortens as the frequency grows, so the lowest acceptable frequency corresponds to the greatest acceptable wavelength. First solve for the wavelength, then for the frequency.
Formula
$2\lambda = 0.300\,\text{mm}$, so $\lambda = \dfrac{0.300\,\text{mm}}{2}$; $v = f\lambda$, so $f = \dfrac{v}{\lambda}$
Conversions
$\lambda$$= 0.150\,\text{mm}$$= 0.150 \cdot 10^{-3}\,\text{m}$finally $1\,\text{MHz} = 10^{6}\,\text{Hz}$.
Substitution
$f$$= \dfrac{1540\,\text{m/s}}{0.150 \cdot 10^{-3}\,\text{m}}$$= 1.027 \cdot 10^{7}\,\text{Hz}$$= 10.27\,\text{MHz}$Answer
$10.3\,\text{MHz}$
Three significant figures like the distance $0.300\,\text{mm}$.
Reasonableness
According to the table of this chapter the wavelength at $10\,\text{MHz}$ is $0.15\,\text{mm}$, and it is used to image precisely the thyroid gland, which is only a few centimetres deep. A higher frequency would distinguish the structures even better, but it would be attenuated more quickly.
- Exercise 17-8
An ultrasound machine sends a new pulse only when the echo of the previous pulse from the deepest point being imaged has returned. The abdomen is imaged to a depth of $20\,\text{cm}$, and sound travels in tissue at a speed of $1540\,\text{m/s}$.
What is the highest rate at which pulses are sent? Give the answer in kilohertz.
Answer
$3.9\,\text{kHz}$
Show the solution
Given
greatest imaging depth $d = 20\,\text{cm}$, speed of sound $v = 1540\,\text{m/s}$
Find
the highest rate $n$ at which pulses are sent, in kilohertz
Principle
A new pulse is not sent until the echo from the deepest point has returned. First calculate how long the echo takes to travel to a depth of $20\,\text{cm}$ and back; the number of pulses that fit into one second is at most one second divided by this time.
Formula
$t = \dfrac{2d}{v}$ and $n = \dfrac{1}{t}$, so $n = \dfrac{v}{2d}$
Conversions
$d = 20\,\text{cm} = 0.20\,\text{m}$; finally $1\,\text{kHz} = 1000\,\text{s}^{-1}$.
Substitution
$t$$= \dfrac{2 \cdot 0.20\,\text{m}}{1540\,\text{m/s}}$$= 2.597 \cdot 10^{-4}\,\text{s}$$n$$= \dfrac{1}{2.597 \cdot 10^{-4}\,\text{s}}$$= 3850\,\text{s}^{-1}$$= 3.85\,\text{kHz}$Answer
$3.9\,\text{kHz}$
That is, about $3900$ pulses per second, two significant figures like the depth $20\,\text{cm}$.
Reasonableness
The echo covers the round trip of $40\,\text{cm}$ in about a quarter of a millisecond, so a few thousand pulses fit in a second. If only a depth of $5\,\text{cm}$ were imaged, four times as many pulses would fit.
- Exercise 17-9
A seismograph station detects that the S wave arrives $14.0\,\text{s}$ after the P wave. The speed of the P wave is $6.0\,\text{km/s}$ and that of the S wave $3.5\,\text{km/s}$.
How many seconds before the arrival of the P wave did the earthquake occur? Give the answer in seconds.
Answer
$20\,\text{s}$
Show the solution
Given
difference in arrival times $\Delta t = 14.0\,\text{s}$, $v_\text{P} = 6.0\,\text{km/s}$, $v_\text{S} = 3.5\,\text{km/s}$
Find
the travel time $t_\text{P}$ of the P wave in seconds, that is, the time from the earthquake to the arrival of the P wave
Principle
From the time difference we first get the distance of the station, because the time difference is the difference of the travel times of the S wave and the P wave. The travel time of the P wave is then the distance divided by the speed of the P wave.
Formula
$d$$= \dfrac{v_\text{P} v_\text{S}}{v_\text{P} - v_\text{S}}\,\Delta t$$t_\text{P} = \dfrac{d}{v_\text{P}}$
Conversions
The speeds are in km/s and the time in seconds, so the distance comes out in kilometres and the travel time in seconds.
Substitution
$d$$= \dfrac{6.0\,\text{km/s} \cdot 3.5\,\text{km/s}}{2.5\,\text{km/s}} \cdot 14.0\,\text{s}$$= 117.6\,\text{km}$$t_\text{P}$$= \dfrac{117.6\,\text{km}}{6.0\,\text{km/s}}$$= 19.6\,\text{s}$Answer
about $20\,\text{s}$, two significant figures like the speeds $6.0\,\text{km/s}$ and $3.5\,\text{km/s}$.
Reasonableness
The travel time of the S wave is $117.6\,\text{km} : 3.5\,\text{km/s} = 33.6\,\text{s}$, and $33.6\,\text{s} - 19.6\,\text{s} = 14.0\,\text{s}$, as measured. The travel time of the P wave is greater than the time difference, because the P wave also needs time for the whole distance.
Level 3 · challenging 3 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 17-10
A bat flies straight towards the wall of a barn at a speed of $10\,\text{m/s}$ and makes a call. It hears the echo from the wall $40\,\text{ms}$ later. Sound travels in air at a speed of $343\,\text{m/s}$.
How far from the wall was the bat when it made the call? Give the answer in metres.
Answer
$7.06\,\text{m}$
Show the solution
Given
speed of the bat $v_\text{bat} = 10\,\text{m/s}$ towards the wall, delay of the echo $t = 40\,\text{ms}$, speed of sound $v = 343\,\text{m/s}$
Find
the distance $d$ to the wall at the moment of the call, in metres
Principle
The sound travels to the wall a distance $d$ and returns to the bat. During the echo the bat has flown a distance $v_\text{bat} t$ towards the wall, so the return trip is $d - v_\text{bat} t$. The total distance covered by the sound is the speed of sound times the delay.
Formula
$d + (d - v_\text{bat} t) = v t$, so $2d = (v + v_\text{bat})\,t$ and $d = \dfrac{(v + v_\text{bat})\,t}{2}$
Conversions
$40\,\text{ms}$$= 40 \cdot 10^{-3}\,\text{s}$$= 0.040\,\text{s}$Substitution
$d$$= \dfrac{(343\,\text{m/s} + 10\,\text{m/s}) \cdot 0.040\,\text{s}}{2}$$= \dfrac{14.12\,\text{m}}{2}$$= 7.06\,\text{m}$Answer
$7.06\,\text{m}$
(data with two significant figures would give $7.1\,\text{m}$; the third figure is shown so that the difference from the result for a stationary bat, $6.86\,\text{m}$, can be seen).
Reasonableness
If the bat were stationary, the distance would be $343\,\text{m/s} \cdot 0.040\,\text{s} : 2 = 6.86\,\text{m}$. The motion of the bat shortens the return trip, so the wall was farther away, by $20\,\text{cm}$, half of the distance the bat flew, $10\,\text{m/s} \cdot 0.040\,\text{s} = 0.40\,\text{m}$.
- Exercise 17-11
Under the skin there is a layer of fat $4.0\,\text{cm}$ thick, in which sound travels at a speed of $1450\,\text{m/s}$, and under it muscle, in which sound travels at a speed of $1580\,\text{m/s}$. An echo returns from the lower surface of the muscle layer $80.0\,\mu\text{s}$ after the pulse is sent.
How deep below the surface of the skin is the boundary really? Give the answer in centimetres.
Answer
$5.96\,\text{cm}$
Show the solution
Given
thickness of the fat $d_\text{fat} = 4.0\,\text{cm}$, $v_\text{fat} = 1450\,\text{m/s}$, for muscle $v_\text{mus} = 1580\,\text{m/s}$, travel time of the echo $t = 80.0\,\mu\text{s}$ from the lower surface of the muscle layer
Find
the true depth $d$ of the boundary from the surface of the skin, in centimetres
Principle
The pulse first travels through the fat there and back and uses part of the travel time for it. The rest of the time is spent in the muscle there and back. In each layer the speed is constant, so the thickness of a layer is the speed times the time spent in the layer divided by two.
Formula
$t_\text{fat}$$= \dfrac{2 d_\text{fat}}{v_\text{fat}}$$d_\text{mus}$$= \dfrac{v_\text{mus}\,(t - t_\text{fat})}{2}$so
$d$$= d_\text{fat} + d_\text{mus}$$= d_\text{fat} + \dfrac{v_\text{mus}}{2}\left(t - \dfrac{2 d_\text{fat}}{v_\text{fat}}\right)$Conversions
$d_\text{fat}$$= 4.0\,\text{cm}$$= 0.040\,\text{m}$$t = 80.0 \cdot 10^{-6}\,\text{s}$; finally $1\,\text{m} = 100\,\text{cm}$.
Substitution
$t_\text{fat}$$= \dfrac{2 \cdot 0.040\,\text{m}}{1450\,\text{m/s}}$$= 55.17 \cdot 10^{-6}\,\text{s}$$d_\text{mus}$$= \dfrac{1580\,\text{m/s} \cdot 24.83 \cdot 10^{-6}\,\text{s}}{2}$$= 0.019\,62\,\text{m}$$d$$= 0.040\,\text{m} + 0.019\,62\,\text{m}$$= 0.059\,62\,\text{m}$$= 5.96\,\text{cm}$Answer
$5.96\,\text{cm}$
(the thickness of the fat $4.0\,\text{cm}$ would limit the answer to two significant figures, $6.0\,\text{cm}$; three figures are shown so that the difference from the result with a single speed can be seen).
Reasonableness
A machine that assumes the speed $1540\,\text{m/s}$ would show the depth $1540\,\text{m/s} \cdot 80.0\,\mu\text{s} : 2 = 6.16\,\text{cm}$, which is $2\,\text{mm}$ too much. If the whole distance were muscle, the result would be $1580\,\text{m/s} \cdot 80.0\,\mu\text{s} : 2 = 6.32\,\text{cm}$. The true depth is less than both of these, because sound travels more slowly in fat. The difference is still only a few per cent, so a single assumed speed is enough for most measurements.
- Exercise 17-12
In an ocean whose depth is $4000\,\text{m}$ the wavelength of a tsunami is $200\,\text{km}$. The tsunami travels to a coast where the depth of the water is $10\,\text{m}$. The period of the wave stays the same all the way.
What is the wavelength of the tsunami in the shallow water at the coast? Give the answer in kilometres.
Answer
$10\,\text{km}$
Show the solution
Given
depths $h_1 = 4000\,\text{m}$ and $h_2 = 10\,\text{m}$, wavelength $\lambda_1 = 200\,\text{km}$; period $T$ the same
Find
the wavelength $\lambda_2$ in the shallow water in kilometres
Principle
In both places the wavelength is much greater than the depth, so the speed is $v = \sqrt{gh}$. In one period the wave advances one wavelength, so $\lambda = vT$. When the period is the same, the wavelength changes in the same proportion as the speed, and $g$ cancels out.
Formula
$\dfrac{\lambda_2}{\lambda_1}$$= \dfrac{v_2 T}{v_1 T}$$= \dfrac{\sqrt{g h_2}}{\sqrt{g h_1}}$so
$\lambda_2$$= \lambda_1 \sqrt{\dfrac{h_2}{h_1}}$Conversions
The depths are in the same unit, so their ratio is dimensionless; the wavelength can be kept in kilometres.
Substitution
$\lambda_2$$= 200\,\text{km} \cdot \sqrt{\dfrac{10\,\text{m}}{4000\,\text{m}}}$$= 200\,\text{km} \cdot 0.050$$= 10.0\,\text{km}$Answer
$10\,\text{km}$
Two significant figures like the depth $10\,\text{m}$.
Reasonableness
In the ocean the speed is $\sqrt{9.81 \cdot 4000}\,\text{m/s} = 198\,\text{m/s}$, so the period is $200\,000\,\text{m} : 198\,\text{m/s} \approx 1010\,\text{s}$, or about $17\,\text{min}$. At the coast the speed is $9.9\,\text{m/s}$, and $9.9\,\text{m/s} \cdot 1010\,\text{s} \approx 10\,\text{km}$. The wavelength is still much greater than the depth, so the formula holds.