Does Sound Intensity Change with Frequency Shift in Doppler Effect?

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In the discussion about the Doppler Effect, a car with a 1000Hz siren moving away at 15m/s produces a frequency of 958Hz as heard directly and 1046Hz when reflected off a wall. The stationary observer experiences a lower frequency due to the source moving away, which is calculated using the Doppler equation. The question of which sound is louder arises, with the direct sound being identified as the quieter 958Hz wave. The concept of beats is mentioned, suggesting that both frequencies could be perceived together. Ultimately, the calculation confirms that the frequency decreases as the source moves away from the observer.
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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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