Ultrasound has many medical applications, one of which is to...

Ultrasound has many medical applications, one of which is to monitor fetal heartbeats by reflecting ultrasound off a fetus in the womb. a. Consider an object moving at speed $v_{\mathrm{o}}$ toward an at-rest source that is emitting sound waves of frequency $f_{0} .$ Show that the reflected wave (i.c., the echo) that returns to the source has a Doppler-shifted frequency $$f_{\text {echo }}=\left(\frac{v+v_{\mathrm{o}}}{v-v_{\mathrm{o}}}\right) f_{0}$$ where $v$ is the speed of sound in the medium. b. Suppose the object's speed is much less than the wave speed: $v_{\mathrm{o}} \ll v .$ Then $f_{\text {echo }} \approx f_{0},$ and a microphone that is sensitive to these frequencies will detect a beat frequency if it listens to $f_{0}$ and $f_{\text {echo }}$ simultaneously. Use the binomial approximation and other appropriate approximations to show that the beat frequency is $f_{\text {beat }} \approx\left(2 v_{0} / v\right) f_{0}$ c. The reflection of $2.40 \mathrm{MHz}$ ultrasound waves from the sur- face of a fetus's beating heart is combined with the $2.40 \mathrm{MHz}$ wave to produce a beat frequency that reaches a maximum of $65 \mathrm{~Hz}$. What is the maximum speed of the surface of the heart? The speed of ultrasound waves within the body is $1540 \mathrm{~m} / \mathrm{s}$ d. Suppose the surface of the heart moves in simple harmonic motion at 90 beats/min. What is the amplitude in $\mathrm{mm}$ of the heartbeat?

Ultrasound has many medical applications, one of which is to monitor fetal heartbeats by reflecting ultrasound off a fetus in the womb.
a. Consider an object moving at speed $v_{\mathrm{o}}$ toward an at-rest source that is emitting sound waves of frequency $f_{0} .$ Show that the reflected wave (i.c., the echo) that returns to the source has a Doppler-shifted frequency $$f_{\text {echo }}=\left(\frac{v+v_{\mathrm{o}}}{v-v_{\mathrm{o}}}\right) f_{0}$$ where $v$ is the speed of sound in the medium.
b. Suppose the object's speed is much less than the wave speed: $v_{\mathrm{o}} \ll v .$ Then $f_{\text {echo }} \approx f_{0},$ and a microphone that is sensitive to these frequencies will detect a beat frequency if it listens to $f_{0}$ and $f_{\text {echo }}$ simultaneously. Use the binomial approximation and other appropriate approximations to show that the beat frequency is $f_{\text {beat }} \approx\left(2 v_{0} / v\right) f_{0}$
c. The reflection of $2.40 \mathrm{MHz}$ ultrasound waves from the sur-
face of a fetus's beating heart is combined with the $2.40 \mathrm{MHz}$ wave to produce a beat frequency that reaches a maximum of $65 \mathrm{~Hz}$. What is the maximum speed of the surface of the heart? The speed of ultrasound waves within the body is $1540 \mathrm{~m} / \mathrm{s}$
d. Suppose the surface of the heart moves in simple harmonic motion at 90 beats/min. What is the amplitude in $\mathrm{mm}$ of the heartbeat?

Key Concepts

Simple Harmonic Motion

Simple harmonic motion describes the oscillatory movement where the restoring force is proportional to displacement. It provides a model for periodic motions such as the beating heart, allowing analysis of variables like amplitude and frequency from the observed motion.

Beat Frequency

Beat frequency is the result of interference between two waves of nearly equal frequencies. The phenomenon is exploited in ultrasound diagnostics where the small difference between the emitted and reflected frequencies produces an audible or measurable beat, indicating motion in the body.

Reflected Wave Doppler Shift

When a wave is reflected from a moving object, it undergoes a Doppler shift twice—once when the wave travels to the object and again when it returns. This double shift is central to applications such as ultrasound imaging in monitoring moving targets like a fetal heartbeat.

Doppler Effect

This is the phenomenon wherein the frequency of a wave changes relative to an observer if there is a motion between the source and the observer. It is crucial for understanding how the frequency of sound (or other waves) is altered when either the source or the observer (or both) is moving.

Binomial Approximation

This mathematical technique is used to simplify expressions that involve small quantities. In the context of the Doppler effect where the object's speed is much less than the speed of sound, the binomial approximation helps derive an approximate expression for the beat frequency.

Transcript

00:01 In part a, i'm choosing to think of it this way.

00:04 We have an object moving toward a source.

00:11 So we are given in chapter, the previous chapter, chapter 20, that when an observer is approaching a source, then the apparent frequency is v plus, v, let's see what it says here again, 0 over v times f.

01:01 But then we have a reflection off of the previous observer that is moving.

01:15 And so now the moving source, not the moving source, but the moving observer in the previous thought becomes the source for the reflection.

01:32 So we also have f plus equals f0 over 1 minus vs over v.

01:48 So this is 20 .39.

01:52 And the other one, whoops, the other one is 20 .40, i would think.

02:04 Equation 20 .40.

02:10 Okay, so let's think about what's going on here.

02:19 It's supposed to be a zero here.

02:28 So, in the second case, f, and i'm trying to go back to our question, echo, is going to be the f plus from the previous equation over 1 minus vs over v.

03:02 But the f plus is v plus v0 over v.

03:09 No, that's not right.

03:11 It's 1 plus v0.

03:13 Let's look at 40.

03:18 1 plus v0 over v.

03:20 I write that down incorrectly.

03:25 So f echo equals 1 plus v0 over vff0 over 1 minus vs is the same as v0.

03:49 I don't know why it's written differently in the book.

04:06 Okay.

04:11 So let's simplify this.

04:14 V plus v0 over v f0 over v.

04:20 F0 over v minus v0 over v.

04:31 So the frequency of the echo is v plus v0 over v minus v0 f0 f0.

04:50 That's what we wanted to prove.

04:54 Let's go to b.

04:58 Okay, so we're supposing that v -0 is much less than v.

05:14 And if that's the case, then f echo is about the same as f -sub -0.

05:24 And that makes sense, because if you add almost nothing and subtract almost nothing, then you get the same thing.

05:50 So it tells us to use the binomial approximation and other approximations to look for a beat frequency.

06:26 All right, i've thought this through, and i'm not sure where the binomial approximation comes in, but f beat equals f echo minus f sub zero.

06:45 Well, f echo is v plus v sub 0 over v minus v sub 0 times f sub 0, and then we subtract f sub 0.

07:04 That's going to give me v plus v.

07:07 Sub 0.

07:07 I'm not doing anything to the first term, but i'm going to multiply the second term times v minus v sub 0 over v minus v sub 0.

07:21 Adding those together, v minus v is zero, v sub zero minus negative v sub zero in the numerator is two v sub zero over v minus v sub zero, f sub zero.

07:39 But we said already that v minus v sub zero is almost exactly the same as v.

07:49 Because v .0 is very small.

07:53 Where'd i write that? right here...

Please give Ace some feedback