Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition

12-1: Characteristics of Sound
12-2: Intensity of Sound; Decibels
12-3: Loudness
12-4: Sources of Sound: Strings and Air Columns
12-5: Quality of Sound, Superposition
12-6: Interference; Beats
12-7: Doppler Effect
12-8: Shock Waves; Sonic Booms

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 59
Q

As a bat flies toward a wall at a speed of 6.0 m/s, the bat emits an ultrasonic sound wave with frequency 30.0 kHz. What frequency does the bat hear in the reflected wave?

A
31.1 kHz31.1 \textrm{ kHz}
Giancoli 7th Edition, Chapter 12, Problem 59 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. So, this bat is flying towards this wall and the bat is emitting a frequency of frequency f and the ball or the wall does not receive that frequency though because the bat is moving towards the wall. And so the wall is like a stationary observer with a sound source moving towards it. And so the frequency that the wall perceives you could say is f prime, and that's gonna be the frequency emitted by the bat divided by 1 minus the speed of the bat divided by the speed of sound. And the wall will also reflect this frequency f prime. And in that case the bat will not perceive that frequency f prime however because the bat can then be considered an observer moving towards a stationary sound source. The wall is like a stationary sound source reflecting the sound frequency, f prime. And the bat is moving towards that sound source. So, f double prime, the frequency perceived by the bat as the frequency reflected from the wall, that will be 1 plus the speed of the bat divided by the speed of sound times the frequency reflected by the wall. This frequency reflected by the wall is this which we can then substitute in here. And we get that the frequency perceived by the bat reflected from the wall is gonna be 1 plus speed of the bat over the speed of sound times f, originally emitted by the bat divided by 1 minus the speed of the bat divided by speed of sound. So, that's 1 plus 6 meters per second divided by 343 meters per second times thirty kilohertz divided by 1 minus 6 meters per second divided by 343. And that gives about 31.1 kilohertz.

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