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A dolphin is able to tell in the dark that the ultrasound echoes received from two sharks come from two different objects only if the sharks are separated by 3.50 m, one being that much farther away than the other. (a) If the ultrasound has a frequency of 100 kHz, show this ability is not limited by its wavelength. (b) If this ability is due to the dolphin’s ability to detect the arrival times of echoes, what is the minimum time difference the dolphin can perceive?

a) $\lambda = 0.015 \mathrm { m }$b) $t = 4.55 \mathrm { mss }$

Physics 101 Mechanics

Chapter 17

Physics of Hearing

Sound and Hearing

University of Washington

University of Sheffield

McMaster University

Lectures

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The sound source of a ship…

A submarine is moving towa…

04:21

$\bullet$ The sound source…

okay. And this question, I would spot dolphins, um, and listening toe echoes so they can tell in the dark. Um, if two sharks come from two different objects on Lee, if they're separated by 3.5 meters, um, and so we know that they're ultrasound frequencies, 100 kilohertz, and we want to show that we're really not limited by wavelength. So, um, let's find maybe the smallest distance of separation that they would be able to detect. So let's write that down. The frequency is, um, 100. It was. Color hurts. Yes. Times 10 of the three hurts. So, um, basically the smallest frequency that they can supper differentiate. Uh, sorry. The smallest distance they can differentiate from what happened, be there the wavelength. So we want to find the wavelength associative with the frequency of, um, 100 times 10 of minus three hurts. So what we can do is we can say we need the velocity of the wave, so the velocity is equal to, um, Lambda Times the frequency, and we need the velocity in water. So I'm gonna look that up in the book right now. Um, high, right? Yeah. Okay, This is ultrasound to write 1500 meters per second. Yes, ultrasound. Um, so the velocity up without up here, they're really 1500 meters for a second. Going to triple check number? Um, okay. Yes, it is. So we confined Lambda by doing lambda is equal to be over off. So we take this velocity 1500 divided by, um, 100 times 10 to the three we get 30.15 me terrorists and about smaller than the wavelength. The difference distance between the sharks. So it's not limited by wavelength. Um, so I'll just write that down, not limited, because lambda, uh um is less than 3.5 meters and then we want to know, um, it's the difference in the arrivals times. So we want the minimum time difference that the dolphin can perceive. OK, so I think this deserves a picture. So basically, like, here's the dolphin, and then here shark one, and then here's shark to the dolphin gets this signal and and then it often also against this signal coming from the other shark. And then we want the time difference between these two signals. And so, um, so delta T Let's call this t the time it takes to get here a t two and the time it takes to get here a t one. So it's gonna be a t two minus two, you one. And then the time that I would take to go over this distance is, um, X two, let's call this distance from here to here X two and then this distance from here to here, x one And so time is, um, velocity. No, it's position over velocity. So that's a velocity. And, um, the water, the velocity of the wave in the water. And then, um, it has to go twice this distance like it's a two. And then this also gets it to two x one Overbey so you could see the formula to Delta acts over me. I hope this isn't in the book because I've been deriving I like every single time. Um and so we know Delta X, we can say is 3.5. Um, so we can say that Delta x 3.5, and then the velocity is, um what it is up there 1500 meters per second. So what we want to do is do. Two times 2.5, which is seven divided by 1500. And that's, um, 4.67 times 10 to the minus three seconds And small pause and check my work. Yeah, Looks good to me. So, yeah, that's how you want Teoh approach that problem.

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