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(a) How far apart are two layers of tissue that produce echoes having round-trip times (used to measure distances) that differ by $0.750 \mu \mathrm{s}$ ? (b) What minimum frequency must the ultrasound have to see detail this small?

a) 5.78 $\times 10^{-4}$ m b) 2.67 $\times 10^6$ Hz

Physics 101 Mechanics

Chapter 17

Physics of Hearing

Sound and Hearing

Cornell University

Hope College

University of Sheffield

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okay, It's on this question. We want to get how far two layers of tissue are If they have different echo, let go round trip times by 0.750 microseconds. So let's write that down Delta Time 0.750 And then our goal is to get a Delta X. He calls some question mark and, um, we want this is for a tissue. So I assume it's for fat tissues and we can use the speed properties of fat. Um, and so we can see from, like, a confirmation that when they say tissue, they mean, um, fat tissue. Oh, I concede here that they say it's 15 40. Is that even in this table that's so annoying that it's not in this table of 17.8, but it's like, buried in the text, So Okay, well, my answer to the last problem that I did is slightly off. That's okay. Okay, So that's the velocity on. And so, in general to calculate the time, um, that it takes time, is CMI involves city is distance over time. So then time is equal to distance over velocity, so the velocity is the same for both So then we know the Delta T is going to be the distance to the back layer two by two by velocity minus the distance to the top layer divided by velocity. And so that's, um, won over v times Delta eggs. So then we know that Delta eggs is equal to V. Dr T. But actually, let's be careful, um, with this in factors of two, because it is echoing so the time that it takes. So it's gonna be 1/2 the time it takes to get through, um, the time that it takes to actually get to the tissue and back. So, um so let me think how to to redo this. So this is not the time between, um, receiving the issue, the signals. Um, so it actually has to travel ah to twice the distance to acts to and then twice the distance to x one. So then what? Weaken begin. Put a to here. And so now we're prepared to solve for a Delta X. So Delta X, um tightly Delta X when you should divide both sides by two. So what we want to do is do 15 40 times, 0.7 50 it's 10 to the minus six and we wonder by about two. Then with that, I got 5.7 a times tend to the minus four meters and we want the minimum frequency. And this question is sort of cut off for me. So let me go to the buck and minimum frequency to see a detail so small. So we want this to be the wavelength of, um of the wave. And then we can relate that to frequency by frequency is equal to the velocity and the material divided by the wavelength so we can take this 15 40 divided by 5.78 times 10 to the minus four lunge 2.67 times 10 to the six hertz.

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