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The following physics responses are provided by a student and, as such, should be considered in that context. While every effort is made for accuracy, absolute correctness cannot be fully guaranteed.
| Question | |
C | 14 |
I | |
Question A bat, flying at 5.00 m/s toward a wall, emits a chirp at 50.0 kHz. If the wall reflects this sound pulse, what is the frequency of the echo received by the bat? (vsound = 340 m/s) | |
Answer 51.5 kHz | |
Notes
| User | Added | Note | e | d |
|---|---|---|---|---|
| KillGorack | 2023-07-19 17:27:11 | The reason why the calculation needs to be done twice is because there are two distinct parts to this problem involving the Doppler effect: when the bat emits the sound, and when the bat hears the reflected sound. When the bat emits the sound towards the wall: The bat is moving toward the wall and emits a sound. In this case, because the sound source (the bat) is moving towards the observer (the wall), the frequency increases. This is the first application of the Doppler effect. When the bat hears the reflected sound: The sound reflects off the wall and returns to the bat. Now, the sound source is the wall and the observer is the bat. Again, because the observer (the bat) is moving towards the source (the wall), the frequency increases. This is the second application of the Doppler effect. Each stage involves a relative motion between the sound source and observer, and each contributes to an increase in the perceived frequency of the sound. That's why we need to apply the Doppler effect twice in this case. |
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