Does Sound Intensity Change with Frequency Shift in Doppler Effect?

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Homework Help Overview

The problem involves a car with a siren emitting sound at a frequency of 1000Hz while moving away from a stationary observer towards a cliff. The discussion centers on the perceived frequencies of the sound directly from the wall and the echo reflected off the wall, as well as the relative loudness of these sounds.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations for the perceived frequencies using the Doppler effect formula and question how to determine which sound is louder. Some suggest considering the concept of beats and whether both frequencies might be heard simultaneously.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have provided calculations and reasoning regarding the perceived frequencies, while others are questioning the assumptions about sound perception and loudness.

Contextual Notes

There is a focus on the individual cases of sound perception, and participants are considering the implications of the observer's stationary position and the source's movement away from them. The problem may involve assumptions about sound propagation and reflection that are not fully detailed in the original post.

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Homework Statement


A car with a 1000Hz siren moves away from you toward a cliff at 15m/s. You are stationary. What frequency of sound do you hear directly from the wall and what frequency do you hear reflected off the wall? Which sound is louder?


Homework Equations


f' = f(v +/- vo)/(v +/- vs); where v = 343m/s, vo = velocity of observer, vs = velocity of the source


The Attempt at a Solution


I found that the frequency of the sound you hear directly is 958Hz and the frequency of the echo is 1046Hz. How can I tell which one is louder?
 
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Wouldn't you be hearing both frequencies (or a mix)? What does you text say about beats?
 
I think we are supposed to look at each case individually. The answer is given to be the 958Hz wave.
 
Plug everything into the equation.

f' = f(v +/- vo)/(v +/- vs)
f' = 1000Hz((343m/s +/- (0)m/s)/(343m/s+15m/s)
f'=958Hz

Remember, since the person is stationary, their velocity is 0. Also, since it is moving away from the observer, the frequency is going to be smaller as it moves away. Therefore, you add the denominator because the object is moving away and you want to divide by a larger number, which in turn, makes your final answer smaller.
 

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