# 8.1: Ultrasound

**At Grade**Created by: CK-12

## Benefits of Ultrasonography

- Non–invasive: the probe does not breach the tissue and reliable imaging can be recorded without surgery.
- Inexpensive and routinely available in North America.
- Provides a clearer picture of soft tissues that do not image well using ionizing radiation.
- Provides flow rates for blood using Doppler technique.
- Provides immediate images which can be used to guide other procedures or surgeries.
- Can be repeated reasonably often.
- Can be used on possibly pregnant women and on fetuses.

Ultrasound imaging provides a view of the human body that is not accessible by other means. While more energetic electromagnetic beams like \begin{align*}X-\end{align*}

The ultrasound image can be produced in real time so that the image can be used to guide surgical procedures. The equipment is routinely available in North America for a nominal cost. Advanced ultrasound machines can use the Doppler Effect to determine speed of blood flow in arteries and veins. By analyzing the blood velocity, physicians can locate aneurysms and blood clots.

Risks and Shortcomings of Ultrasonography

- No known risks, but major medical organizations such as the World Health Organization have discouraged the popular practice of imaging fetuses to determine sex or to take “home movies.”
- Very limited ability to image structures with bone or air. See the section on impedance.
- Very long period (>30 minutes) of ultrasound associated with damage in small rodents.

Ultrasound has been generally recognized as a safe procedure when used for medically significant imaging. The recent popularity of making home movies of the developing fetus has been discouraged by major medical organizations. There is some danger of thermal heating of the tissue by long exposures at high power. A recent study found that prolonged exposures of over thirty minutes to developing rat fetuses produced some genetic damage. This is an area of ongoing research.

## More Risks

One of the risks of any diagnostic device is that the energy beam is thermalized by the body; that is, the incoming energy heats the tissue. Because homeostasis (maintaining the same temperature) is the hallmark of mammals, changing the temperature of target tissue is a problem.

As a worst case scenario, assume that the probe delivers \begin{align*}10 \;\mathrm{W}\end{align*} of energy to a cylinder of tissue \begin{align*}10 \;\mathrm{cm}\end{align*} deep and \begin{align*}5 \;\mathrm{cm}\end{align*} in diameter for 15 minutes. Also assume that there is no blood flow to the affected area so that the heat stays in the cylinder and that the technician does not move the probe for the entire 15 minutes.

The energy delivered to the tissue is \begin{align*} 10 \;\mathrm{W} \times 60 \ \;\mathrm{s/min} \times 15 \;\mathrm{min} = 9,000 \;\mathrm{J} \end{align*}.

The change in temperature of the tissue can be compued by assuming that the energy of the ultrasound is converted into heat:

\begin{align*} Q & = mc\ \Delta T \\ Q & = \rho\ V\ c\ \Delta T \\ \Delta T & = \frac{Q} {\rho V c} \\ \Delta T & = \frac{9,000\ \text{J}}{\left(1025 \ kg/m^3 \right) \cdot \left(1.96 \times 10^{-4} \ \text xm^3 \right) \cdot \left(4050 \ \text{J/kg} \cdot \text{K}\right)}, \end{align*}

where \begin{align*}Q\end{align*} is the change in the internal energy associated with a temperature rise \begin{align*}\Delta T \end{align*}, \begin{align*}m\end{align*} is the mass of the tissue, \begin{align*}c\end{align*} is the heat capacity, and \begin{align*}T\end{align*} is the temperature. Substituting for the mass \begin{align*} m= \rho V \end{align*}, where \begin{align*}\$ \rho \$\end{align*} is the density and \begin{align*}V\end{align*} is the volume, yields the second equation. Solving for the change in temperature gives the third equation. The final temperature change is about \begin{align*}11 \ \;\mathrm{K}\end{align*}, which is appreciable. Clearly, there is the possibility of changing the temperature of the target tissue under extreme circumstances.

## How Can an Ultrasound "See"?

The Rayleigh criterion gives the resolution of waves, whether the wave is light, sound, or any other kind. It states that the wavelength must be at least as small as the object in order to "see" it. When the wavelength is larger than the object, then diffraction occurs. This “smears” out the beam so that an image cannot be formed. The same process applies to computer screens. If the image to be displayed is smaller than the pixels on the screen, then the image cannot be represented. The best results for the computer screen are when the images are much, much bigger than the size of the pixels that make up the computer display.

For medical imaging, the smaller the object to be imaged means the smaller the wavelength (and the higher the frequency) of the imaging beam.

Rayleigh criterion for different wavelengths

Rayleigh criterion:

- The limit to imaging an object (if everything else is perfect) is diffraction.
- This means that the wavelength of the incoming detector beam can be no smaller than the size of the object.
- When the minimum of one peak just overlaps the maximum of the next peak, the two peaks are resolved.
- If the peaks are closer, then they cannot be told apart.

## Choosing the Best Frequency

The frequency of the ultrasound beam depends on two factors: the speed of sound in the tissue and the wavelength of the imaging beam. Note that the smaller the wavelength, the higher the frequency. And, of course, the wavelength is fixed by the size of the object to be examined.

The frequency of the required ultrasound depends on the wavelength and the speed of sound in the material:

\begin{align*}f = \frac{\nu_{material}} {\lambda}\end{align*}

where \begin{align*}v_{material}\end{align*} is the speed of sound in the substance and \begin{align*}\lambda\end{align*} is the wavelength of the sound wave.

The speed of sound in a solid or a fluid depends on the density of the material, \begin{align*}\rho\end{align*}, and the stiffness of the material.

For fluids, this is the bulk modulus, \begin{align*}B\end{align*}, while for solids this is usually Young’s modulus, \begin{align*}Y\end{align*}.

The speed of sound in human tissue varies by more than a factor of three. The speed of sound depends on the density of the tissue. While human bone is fairly dense, subcutaneous fat is much less dense than water. The overall density of a human is just about that of water. How do you know? Because the average human just barely floats in water. The more muscle and bone that a person has, the lower that person floats.

The other factor for determining the speed of sound is the stiffness of the material. Again bone is fairly stiff while subcutaneous fat is not. The stiffness is measured by the bulk modulus for fluids or Young’s modulus for solids.

\begin{align*}\nu_{sound} = \sqrt{\frac{B} {\rho}}\end{align*}

Material | Velocity \begin{align*}\;\mathrm{(m/s)}\end{align*} |
---|---|

air | \begin{align*}331\end{align*} |

human soft tissue | \begin{align*}1540\end{align*} |

human brain & amniotic fluid | \begin{align*}1541\end{align*} |

liver | \begin{align*}1549\end{align*} |

kidney | \begin{align*}1561\end{align*} |

blood | \begin{align*}1570\end{align*} |

muscle | \begin{align*}1585\end{align*} |

skull-bone | \begin{align*}4080\end{align*} |

fat | \begin{align*}1450\end{align*} |

(Taken from http://www.yale.edu/ynhti/curriculum/units/1983/7/83.07.05.x.html)

Which references Christensen, E. E., Curry,T. S., Dowdey, J. E.: *Introduction to the Physics of Diagnostic Radiology*. Philadelphia: Lea & Febeger, 2nd Edition; 1978: Chapter 25.

Most soft tissues where the ultrasound is most effective have a speed of sound that is about \begin{align*}1550 \;\mathrm{m/s}\end{align*}, which is about five times faster than the speed of sound in air. Everyone is familiar with watching a distant event like a lightning strike where the light arrives almost immediately, but the sound of the thunderclap arrives some time later (about 5 seconds for every mile away).

## Now Try This!

You can demonstrate the change in the speed of sound with the change in density by using dry coffee creamer (or hot chocolate mix) in a coffee cup. Put a couple of spoonfuls of the fatty creamer in the bottom of the cup and carefully add hot water so that the cream stays on the bottom. Strike the cup with your spoon so that the cup "rings." Now stir the creamer into the hot water while continuing to strike the side of the cup. As the density of the fluid changes, the notes will change.

## Just a Bag of Water

The first thing to notice is that the speed of sound in choosing the best frequency is very close to that of water (why?) and is about five times faster than the speed of sound in air.

So in order to image an object that is \begin{align*}1\end{align*} cm in size, the frequency of the ultrasound probe that travels through muscle should be \begin{align*}158.5 \;\mathrm{kHz}\end{align*}. In general, to image a smaller object, the frequency must be increased. Why?

The speed of sound in human tissue is about that of water, \begin{align*}1540 \;\mathrm{m/s}\end{align*}. If the size of the target is \begin{align*}1 \;\mathrm{cm} \ (0.010 \;\mathrm{m})\end{align*} then

\begin{align*}f = \frac{1540 \ \text{m/s}} {0.01\ \text{m}} = 154,000\ \text{Hz}\end{align*}

or about \begin{align*}154\;\mathrm{kHz}\end{align*}. As the size of the target decreases, the frequency increases. To image a target that is \begin{align*}1.0\;\mathrm{mm}\end{align*}, the imaging beam must have a frequency of \begin{align*}1.54\;\mathrm{MHz}\end{align*}. The normal range of diagnostic ultrasound is \begin{align*}7-9\;\mathrm{MHz}\end{align*}.

Now try some problems.

- What is the frequency required to image an object that is \begin{align*}1\;\mathrm{mm}\end{align*} in diameter?
- What is the frequency required to image an object that is \begin{align*}0.2\;\mathrm{mm}\end{align*} in diameter?
- A typical medical ultrasound is \begin{align*}9\;\mathrm{MHz}\end{align*}. What is the smallest object that can be imaged with this wave in human tissue?

## Echolocation

The process of imaging is the same as the echo-locating sonar of a submarine or a bat. The observer sends out a brief pulse of ultrasound and waits for an echo. The pulse travels out, reflects off the target and returns. The ultrasound machine uses pulses because the same device acts as both transmitter and receiver. If it continually sent out sounds, then the receiver would not hear the much softer echo over the louder transmission. The duty cycle of the ultrasound imager is the amount of time spent transmitting compared to the total time of transmitting and listening.

Reflected Wave

- The pulse travels out and returns to the transducer where it is converted to electrical signal.
- But the same device is both sender and receiver.
- Duty cycle: emit pulse, wait, and listen.
- Same procedure as SONAR.

## Wait Time

In order to be as efficient as possible the machine should send out the next pulse just after the target pulse arrives.

To calculate the wait time:

\begin{align*}t = \frac{2 \times d} {\nu_{material}} = \frac{2 \times0.05\ \text{m}} {1540 \ \text{m/s}} = 65 \ \mu \text{s}\end{align*}

Notice that there is a very short return time for the echo. Some bats do this naturally and even change the duty cycle as they close in on their prey. Essentially, as soon as the bat hears an echo, it sends out a new chirp for additional information.

The duty cycle is the amount of time that the probe is producing a pulse compared to the time that it is listening.

The size of the object that can be imaged with the transducer is a function of wavelength, therefore, the user should move to a transducer with the highest frequency and smallest wavelength.

But, sound waves that travel are subject to attenuation, i.e., gradual loss of intensity.

## Attenuation

All of this suggests that in order to image very small targets, the frequency of the ultrasound should be increased to something like \begin{align*}150\end{align*} MHz, which is certainly technically feasible. But this is where the other aspect of sound travel through a medium comes into play. Waves can lose their energy by scattering or by absorption. Together these two processes are called *beam attenuation*.

Scattering results from parts of the beam deflecting from the straight path of travel. The most familiar example is the scattering of sunlight by our atmosphere. While the Sun is very bright, the rest of the sky is lit by the scattered light of the atmosphere. Scattering is often most important in mixtures and materials that are heterogeneous. The scattering of sunlight is aided by solid dust particles in the air. The most famous example of dust scattering sunlight was the spectacular sunsets that resulted from the famous Krakatoa volcano eruption in 1883.

Scattering is not as important in ultrasound as absorption. The ultrasound beam is thermalized as the sonic energy is converted into tissue heating. This problem of absorption of the sound has a large impact on the amount of echo that returns to the probe.

Scattering

- Where a portion of the wave deflects from the straight line path and is lost.
- Typically this will happen more often in materials that are heterogeneous (mixed) materials (not generally a problem here).

Absorption

- Where the wave energy is converted into thermal heating of the material. Absorption is a major issue in ultrasound imaging.

## Absorption Coefficient

The absorption process follows first order kinetics, which is familiar from radioactive decay. The intensity is measured as a function of the distance \begin{align*} x \end{align*} travelled in the tissue, i.e., \begin{align*} I = I (x)\end{align*}. The change in intensity \begin{align*}\Delta I = I(x + \Delta x) - I(x)\end{align*} is proportional to the intensity and the distance \begin{align*}\Delta x\end{align*}, i.e., \begin{align*} \Delta I = - \alpha I(x) \ \Delta x\end{align*}. Here \begin{align*} \alpha\end{align*} is the absorption coefficient in units of inverse meters. It follows that the intensity decays exponentially with distance,

\begin{align*}I (x) = I_0\ e^{-\alpha x}\end{align*}

where \begin{align*}I_0\end{align*} is the initial intensity at \begin{align*} x=0\end{align*}. The absorption coefficient \begin{align*}\alpha\end{align*} governs how rapidly the sound is absorbed. The absorption coefficient is the product of two factors, the medium and the frequency of the ultrasonic beam. The frequency matters because the transfer of energy from the ultrasound beam to the tissue is more efficient if the frequency of the ultrasound matches the frequency of a (microscopic) process in the tissue.

This is called resonance and is familiar from a child on a swing set. The parent exerts a small force each time the child is closest to him/her. If these forces are synchronized with the oscillations of the child, theses pushes "add up," resulting in a large amplitude [and thus energy] of the child.

The competition between the need for the highest possible frequency to provide a good target image and the need for the lowest possible frequency to return a good echo is usually made around \begin{align*}7-2 \;\mathrm{MHz}\end{align*}, which provides reliable imaging of objects that are \begin{align*}0.20\;\mathrm{mm}\end{align*} in size.

The absorption coefficient is the product of a coefficient \begin{align*}\gamma\end{align*} that depends on the medium and the frequency \begin{align*}f\end{align*} of the ultrasonic beam,

\begin{align*} \alpha & = 2\gamma\ f \\ I(x) & = I_0\ e^{-2 \gamma\ f\ x} \end{align*}

This means that as the frequency is increased (wavelength decreases), more of the signal is absorbed by the medium.

Some Sample Amplitude Absorption Coefficients

Tissue | \begin{align*}y_{sound}\ (\mathrm{s/m})\end{align*} |
---|---|

blood | \begin{align*}2.1 \times 10^{-6}\end{align*} |

abdomen | \begin{align*}5.9 \times 10^{-6}\end{align*} |

fat | \begin{align*}7.0 \times 10^{-6}\end{align*} |

soft tissue | \begin{align*}8.3 \times 10^{-6}\end{align*} |

muscle | \begin{align*}2.3 \times 10^{-5}\end{align*} |

bone | \begin{align*}1.6 \times 10^{-4}\end{align*} |

lung | \begin{align*}4.7 \times 10^{-4}\end{align*} |

(Taken from Irving, H. B., Physics of the Human Body. Berlin: Springer Verlag; 2007:562.)

## Sample Problems

- Calculate the attenuation for a \begin{align*}15\;\mathrm{MHz}\end{align*} ultrasound that penetrates average soft tissue for a distance of \begin{align*}5\;\mathrm{cm}\end{align*} and returns to the transponder. Consider the initial beam to be \begin{align*}100\end{align*}%.
- Repeat the calculation for a beam with twice the wavelength \begin{align*}(7.5\;\mathrm{MHz})\end{align*}.
- Repeat the calculation for a beam with half the distance \begin{align*}(2.5\;\mathrm{cm})\end{align*} and twice the wavelength \begin{align*}(7.5\;\mathrm{MHz})\end{align*}.

## Equipment

The probe for the ultrasound is a transducer. This is a crystal of piezoelectric material (*piezo* is Greek for pressure or squeezing). The most common example of piezoelectricity is the high school demonstration with a crystal mounted under a lever. As the lever is rocked the crystal is squeezed and electric sparks shoot across a gap beside the crystal. A somewhat similar, but not exactly the same effect, is the triboluminescence of crushing Wint-O-Green Lifesavers.

**Try It!**

Take some fresh Wint-O-Green Lifesavers into a darkened room with a mirror. Put one in your mouth and crush it. You should be rewarded with a visible blue spark as the electrons are moved out of the sugars by crushing. The Wint-O-green flavor, methyl salicylate, (oil of wintergreen) acts to convert the normally ultraviolet light of this transition into visible light. When an electric field is applied to the crystal it contracts, as the field is reversed, the crystal expands. The contraction and expansion produce a pressure sound wave. The same process works in reverse so that when the echo comes back the pressure wave produces an electrical signal. The compression high pressure will cause the crystal to produce one electric field and the low-pressure rarefaction will produce the opposite electric field.

Transducer

Transducers consist of a piezoelectric material. A varying electrical signal will cause the material to contract and expand, which produces a pressure sound wave.

The same process can work in reverse: A sound wave hitting the piezoelectric material will give rise to a varying electrical signal.

## Another Problem

The other transmission problem occurs when the ultrasound wave encounters a new medium with a different speed of sound. As the incident intensity of the beam encounters the new material it is either reflected or transmitted (refracted). The amount transmitted plus the amount reflected must be equal to the incident amount in order to conserve energy. The amount that is reflected compared to the amount that is transmitted depends on a property of the two materials called the *impedance*. The short answer is that if the impedances of the two materials match then the ultrasound is transmitted. If they don’t, the wave is reflected.

The ultrasound wave represents a beam energy that must move into the new material if transmission is to occur. But, because the materials are different, the materials have different speeds of sound. An analogy is an airplane delivering packages to an airport that then transfers those packages to trucks for home delivery. If the air freight plane delivers packages (energy) faster than the trucks can haul them away, packages pile up. In that case, the packages are sent back (re-elected).

Impedance, or complex resistance, can also be found in electricity, specifically in an LRC circuit. The inductor \begin{align*}(L)\end{align*}, resistor \begin{align*}(R)\end{align*}, and capacitor \begin{align*}(C)\end{align*} have a natural resonance where the electric current is the highest. Two circuits with matching impedances will resonate together.

In the case of ultrasound, two processes can happen when the beam changes two media: Reflection and refraction.

As the ultrasound pressure wave hits the boundary between the media, there can be no net pressure so exactly how much is reflected and how much is transmitted (refracted) depends on the impedances of tthe media.

If the impedances match then ALL of the incident intensity if transmitted. For example, if the material is the same on both sides, then the beam is transmitted and not reflected.

## More Impedance

The intensity, \begin{align*}I\end{align*}, of some pressure sound wave is defined as the power \begin{align*}P\end{align*} per unit area \begin{align*}A\end{align*}. But the power is the kinetic energy per unit time. Expressing the mass of the kinetic energy term as the product of the density, \begin{align*}\rho\end{align*}, and the volume, \begin{align*}V\end{align*}, yields the fourth equation. But the volume is simply the unit area \begin{align*}A\end{align*} times the distance that the wave travels at the speed of sound \begin{align*}c\end{align*} in time \begin{align*}t\end{align*}, i.e., \begin{align*}V = A \cdot c t\end{align*}.

\begin{align*}I &= \frac{P} {A} = \frac{E_k} {At}\\ &= \frac{0.5 \ \text{m} {\nu_{a \nu g}}^2} {At}\\ &= \frac{\rho\ A \ ct{\nu_{avg}}^2} {2 At}\\ &= \frac{\rho c {\nu_{a \nu g}}^2} {2}\\ &= \frac{1} {2} (\rho c) {\nu_{a \nu g}}^2\\ &= \frac{1} {2} Z {\nu_{a \nu g}}^2\end{align*}

Cancellation of the \begin{align*}A\end{align*} and \begin{align*}t\end{align*} terms yields an equation with the impedance, \begin{align*}Z\end{align*}, defined as the speed of sound in the material times the density of the material, \begin{align*} Z = c \ \rho\end{align*}.

The impedance, \begin{align*}Z\end{align*}, of human tissue is not that different from water, but is markedly different for air and bone. This is natural because of the differences in speeds and densities of the materials compared to water. This means an ultrasound beam will travel quite readily from water to tissue to muscle and back but will reflect off of air (lungs) or bones. The amount of reflectance can be quite remarkable as the table shows. The amount that is transmitted plummets from \begin{align*}99.8\end{align*}% for water and soft tissue to \begin{align*}0.10\end{align*}% for air and soft tissue.

Substance | Density \begin{align*}(\;\mathrm{kg/m}^3)\end{align*} | Speed \begin{align*}\;\mathrm{(m/s)}\end{align*} | Impedance \begin{align*}\;\mathrm{(Pa\ s/m)}\end{align*} |
---|---|---|---|

air | \begin{align*}1.29\end{align*} | \begin{align*}340\end{align*} | \begin{align*}439\end{align*} |

water | \begin{align*}1,000\end{align*} | \begin{align*}1,496\end{align*} | \begin{align*}1,490,000\end{align*} |

fat | \begin{align*}940\end{align*} | \begin{align*}1,476\end{align*} | \begin{align*}1,390,000\end{align*} |

muscle | \begin{align*}1,058\end{align*} | \begin{align*}1,568\end{align*} | \begin{align*}1,600,000\end{align*} |

bone | \begin{align*}1,785\end{align*} | \begin{align*}3,360\end{align*} | \begin{align*}6,000,000\end{align*} |

## Transmission

The actual intensity that is transmitted can be calculated by taking the ratio of the impedances times the incident intensity. The key point to see here is that the ultrasound "wants" some of the energy to be reflected in order to have an echo for imaging. But when all of the energy is reflected, nothing beyond that material can be directly imaged.

The most common example of this impedance mismatch is the way that sounds travel very far across water surfaces like lakes. It is not uncommon to hear conversations that are occurring a mile away as if they were in the same room. Another example of this impedance mismatch between water and air is the way that sounds are muffled when underwater. Next time you are in a pool, have someone yell at you while you are underwater. The sound reflects but does not transmit into the water. At the same time, if someone clangs on a pool ladder in the water, the sound travels quite well to your underwater ears.

\begin{align*}I_{transmitted} = \frac{Z_{transmitted}} {Z_{incident}} I_{incident}\end{align*}

- The fraction transmitted is dependent on impedance matching.
- For water/air, \begin{align*}Z/Z = 3,416\end{align*}.
- Almost all of the sound wave is reflected whether from air to water or water to air.

Interface | Reflect (%) | Transmit (%) |
---|---|---|

water/soft tissue | \begin{align*}.23\end{align*} | \begin{align*}99.77\end{align*} |

fat/muscle | \begin{align*}1.08\end{align*} | \begin{align*}98.92\end{align*} |

bone/muscle | \begin{align*}41.23\end{align*} | \begin{align*}58.77\end{align*} |

soft tissue/bone | \begin{align*}43.50\end{align*} | \begin{align*}56.50\end{align*} |

bone/fat | \begin{align*}48.91\end{align*} | \begin{align*}51.09\end{align*} |

soft tissue/lung | \begin{align*}63.64\end{align*} | \begin{align*}36.36\end{align*} |

air/muscle | \begin{align*}98.01\end{align*} | \begin{align*}1.99\end{align*} |

air/water | \begin{align*}99.89\end{align*} | \begin{align*}.11\end{align*} |

air/soft tissue | \begin{align*}99.90\end{align*} | \begin{align*}.10\end{align*} |

- Notice that just getting the ultrasound beam into the body is a problem as most of the energy is reflected at the air interface.
- A special gel is used to make good acoustic contact (match impedances) between the transducer and the body.

## Doppler Effect

The Doppler effect is the change in the frequency as heard by a listener compared to the frequency emitted by the source. As the listener moves closer to the source, the listener encounters more waves in the same time. The listener’s frequency is higher than the source. The reverse is true for the listener who moves away from the source. An analogous change occurs when the listener is stationary and the source moves.

The summary of the Doppler effect is that when the distance between the source and listener decreases, the listener hears a higher frequency. When the distance between the source and the listener increases, the listener hears a lower frequency.

\begin{align*}f_L = f_S \left (\frac{c \pm \nu_L} {c}\right )\end{align*}

- Moving Listener
- A similar argument for a listener moving away results in the same equation but with a minus sign.

- \begin{align*}f_L\end{align*} is the frequency that the listener hears.
- \begin{align*}f_s\end{align*} is the frequency of the source.
- \begin{align*}v_L\end{align*} is the velocity of the listener.

\begin{align*}f_L = f_S \left (\frac{c} {c \pm \nu_S}\right )\end{align*}

- Moving Source

- \begin{align*}f_L\end{align*} is the frequency that the listener hears.
- \begin{align*}f_s\end{align*} is the frequency of the source.
- \begin{align*}v_s\end{align*} is the velocity of the source.
- \begin{align*}c\end{align*} is the speed of sound.

\begin{align*}f_L = f_S \left (\frac{c \pm \nu_L} {c \pm \nu_S}\right )\end{align*}

- Moving Source and Listener
- Notice that if the source and the listener are moving in the same direction at the same velocity, the result is that the frequency is unchanged.

- As the listener and source close in on each other, the frequency will increase.
- As the listener and the source move away from each other, the frequency will decrease.
- Use relative motion to simplify problems by stopping the slower object.

## Doppler Demo

Now Try This!

- Either use a tuning fork on a string or a constant frequency speaker (a piezoelectric buzzer with a 9-volt battery).
- Start the buzzer or strike the tuning fork.
- Twirl around your head and listen for the Doppler shifted sounds.
- This is more effective as the speed of the sound source increases.

Did you hear it? That was the Doppler effect!

What does this have to do with Ultrasound?

## Blood Flow

Ultrasound can be used to diagnose the speed of blood flow by using the Doppler effect. The procedure is non-invasive because it doesn’t require inserting a probe into the blood vessel. If the blood is flowing at \begin{align*}2\;\mathrm{cm/s}\end{align*} in the blood vessel then the Doppler effect calculation shows that the change in the Doppler frequency is about \begin{align*}0.0025\end{align*}%, which would be incredibly difficult to measure. But the imaging system doesn’t measure the frequency directly; instead, it mixes the echo with the known original signal to produce a beat frequency (which is just the difference between the two frequencies). In this case the beat frequency is \begin{align*}178.3 \;\mathrm{Hz}\end{align*} , which is very easy to accurately measure. This is a common practice in physics to measure an unknown signal precisely by mixing it with a known frequency to produce a beat frequency.

\begin{align*} f_L & = f_S \left (\frac{c \pm \nu_L} {c \pm \nu_S}\right ) \\ f_L & = 7.000000 \ \text{MHz} \left (\frac{1570 \ \text{m/s} + 0.02 \ \text{m/s}} {1570 \ \text{m/s} - 0.02 \ \text{m/s}}\right ) \\ f_L & = 7.000178346 \ \text{MHz} \end{align*}