Problem

You're standing between two speakers that are dri…

03:02

Problem 68 Hard Difficulty

A bat flies toward a wall, emitting a steady sound of frequency 2000 $\mathrm{Hz}$ . The bat hears its own sound, plus the sound reflected by the wall. How fast should the bat fly in order to hear a
beat frequency of 10.0 $\mathrm{Hz}$ ? (Hint: Break this problem into two parts, first with the bat as the source and the wall as the listener and then with the wall as the source and the bat as the listener.)

Answer

0.858 $\mathrm{m} / \mathrm{s}$

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Physics 101 Mechanics

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Video Transcript

So here the wall first acts of the listener, then it acts as a source. So we can say that the force, the frequency of the source, someone is going to be equal to 2,000 hertz. And we can say that the frequency of the listeners have one minus the frequency of the sub of the source Sub one, eagle two ten 10 hertz. We know that the frequency of the listener soap one is greater than the frequency of the source of won. This is because the Doppler effect, the rather the bat, is moving towards the wall. So that means that the Doppler effect is goingto increase the frequency and the final frequency. So let's first, first part would be the wall receives the sound. So we can say that the frequency of the source equals the frequency of the source of one, and we can sit at the velocity of the source equals the negative velocity of the bat and the velocity of the listener equals zero. So we can say that we also out there We also know that the frequency of the listener equals frequency of the listeners of one. So at this point, we can say that the frequency of the listener equals the velocity plus the velocity of the listener divided by the velocity plus the velocity of the source. This is times F sub s. So we can say that f sub else of one will equal with the What about the minus V that times the frequency of the source? Someone so is simply just substituting our knowns into this equation. The second part, we're going to say that here, the wall, Yeah, I can hear the world acts as a source so we can say that f sub two equals f sub else of one. The velocity of the source equal zero, and the velocity of the listener now equals positive velocity of the bat. And so we can say that the frequency of the listeners some two equals the velocity plus the velocity of the bat divided by the times, the frequency of the source up, too. This is going to be equal to the velocity, plus the philosophy of the bat divided by the The frequency of the source up to this is simply this. So we can say that it'll be times the divided by the minus the bat times the frequency of the source. Someone, um, velocity. Obviously. Counsel's out, and we have that the frequency of the listener sub two equals the plus B of the bat, divided by the minus velocity of the bat times again, frequency of the source of one. We know that the frequency of the listeners up to minus the frequency of the source of one is going to be equal to the delta changing frequency. And this will be equal to velocity plus the velocity of the bat divided by the velocity minus the velocity of the bat. This will be minus one times again, the frequency of the source. Someone Now, at this point, my minus the velocity is about as this point we can say that f equals two times the velocity of the bat divided by velocity minus the velocity of the bat times the frequency of the source someone and we can say that the velocity of the bat will be equal two. The times the change and frequencies is this will be the speed of sound through air divided by two times the frequency of the source of one. Plus the change in the frequency and this is equaling the speed of sound through air 344 meters per second times the changing frequency of 10 hertz divided by two times 2,000 hertz plus 10 hertz. And we find that the velocity of the bat is going to be equal 2.858 meters per second. This is our final answer. That would be the velocity of the bat. That is the end of the solution. Thank you for watching.

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