Speed of moth related to Doppler Effect

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Homework Help Overview

The discussion revolves around a problem involving the Doppler Effect as it relates to the speed of a moth in the context of sound waves emitted by a bat. Participants are analyzing the frequency shifts experienced by both the bat and the moth due to their relative motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss two cases of the Doppler effect, considering the sound wave's journey from the bat to the moth and back. There are questions about the calculations leading to a discrepancy between the calculated speed of the moth and the answer key. Assumptions regarding the direction of the moth's flight are also explored.

Discussion Status

The discussion is ongoing, with participants providing insights into the assumptions made in the problem. Some guidance is offered regarding the potential omission of information about the moth's flight angle, which could clarify the discrepancy in the speed calculation.

Contextual Notes

There is a noted lack of information regarding the direction of the moth's velocity, which is critical for accurately applying the Doppler effect equations. Participants are encouraged to state their assumptions when addressing the problem.

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Homework Statement
Sound wave produced by a bat has a frequency of 222 kHz and wavelength of ##1.5 x 10^{-3}## m. A stationary bat sends out a sound signal and receives the same signal reflected from a moving moth at a frequency of 230.3 kHz. Calculate the magnitude of the velocity of the moth, assuming that the velocity is constant
Relevant Equations
Doppler Effect
I treat this question as two cases of Doppler effect.

(1) When the sound wave travels from bat to moth

Speed of sound = 222 x 1.5 = 333 m/s

Frequency received by moth:
$$f_1=\frac{333+v}{333}\times 222$$

(2) When the sound wave is reflected from moth back to bat

Frequency received by bat (moth as source and bat as observer):
$$f_2=\frac{333}{333-v}\times f_1$$
$$230.3=\frac{333}{333-v}\times \frac{333+v}{333}\times 222$$

Solving this equation, I get ##v=6.1## m/s but the answer key is 12 m/s

Where is my mistake?

Thanks
 
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Both, the stationary bat and the moving moth, receive sound of the same frequency.
That common frequency is greater than the frequency of the sound emitted by the bat.
It is the relative velocity of both creatures what reduces the distances between two consecutive crests of the sound wave, regardless which one is moving.
 
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Lnewqban said:
Both, the stationary bat and the moving moth, receive sound of the same frequency.
You mean both bat and moth will detect frequency of 230.3 kHz?
 
Lnewqban said:
Both, the stationary bat and the moving moth, receive sound of the same frequency.
Not so.
 
songoku said:
the answer key is 12 m/s
Well, it doesn’t say the moth is flying directly towards the bat. Perhaps it was supposed to tell you it was flying at an angle of 60° to the direct route?
 
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haruspex said:
Well, it doesn’t say the moth is flying directly towards the bat. Perhaps it was supposed to tell you it was flying at an angle of 60° to the direct route?
How can we get this information from the question?
 
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songoku said:
How can we get this information from the question?
Assuming that's not rhetorical...

You can’t. But there’s a possibility that a 60° angle was accidentally omitted from the question - this would explain official answer.

What you can do, at the start of your answer, is to make clear any assumption(s) you need to make.

E.g. “No information is supplied about the direction of the moth’s velocity. I will assume that we need to find the magnitude of the radial component of the moth’s velocity.”
 
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Thank you very much for all the explanations Lnewqban, Gavran, haruspex, Steve4Physics
 
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