Moving vehicles and Doppler Effect

AI Thread Summary
The discussion revolves around a physics problem involving the Doppler Effect and sound intensity from a moving fire engine. The calculations for the frequency heard by passengers in both a moving car and a stationary van were presented, yielding results of 516.5 Hz and 566 Hz, respectively. The intensity level heard by the van's passengers was calculated to be 88 dB, based on the intensity and decibel formulas. There was a debate about whether the power received by the van is influenced by the frequency of the sound, with some confusion regarding the relationship between intensity, power, and energy. The conversation highlights the complexities of sound propagation and the assumptions made about the vehicles' states during the calculations.
Nishikino Maki
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Homework Statement


Here is the problem: http://faculty.kfupm.edu.sa/PHYS/kuhaili/doppler_problem.htm

{Mentor's edit: Here's the text copied from the url:
A fire engine moving to the right at 40 m/s sounds its horn ( frequency 500 Hz ) at the two vehicles shown in the figure. The car is moving to the right at 30 m/s, while the van is at rest.

(a) What frequency is heard by the passengers in the car?
(b) What is the frequency as heard by the passengers in the van?
(c) When the fire engine is 200 m away from the car and 250 m from the van, the passengers in the car hear a sound intensity of 90 dB. At that moment, what intensity level is heard by the passengers in the van?
}

Homework Equations


Doppler Effect:
f' = f\frac{v±v_o}{v∓v_s}

Intensity:
I = \frac{P}{4\pi r^2}

Sound level:
\beta = 10 \log \frac{I}{10^{-12}}

The Attempt at a Solution


I actually got answers for the problem, however, this was an even problem and I could not check my answers anywhere.

Part a:
f'=500(\frac{343 - 30}{343 - 40})
This turned out to be 516.5 Hz

Part b:
f'=500(\frac{1}{1 - \frac{40}{343}})
This was 566 Hz

Part c:
For this part I assumed that everything was standing still, and just used intensity and decibel formulas.
90=10 \log \frac{I}{10^{-12}}
I = 10^{-3}
P = I*4\pi 200^2
I_2 = \frac{P}{4\pi 250^2}
\beta = 10\log \frac{I_2}{10^{-12}}
\beta = 88 dB
 
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There's some sort of problem with the link to the question. It can't be accessed.
 
The website seems to show up for me.

Here is a screenshot
 

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For part c, isn't the power received also dependent on received frequency?
 
Is it?

I used the second equation and got power by multiplying intensity and the area the sound wave went over.
 
Nishikino Maki said:
Is it?

I used the second equation and got power by multiplying intensity and the area the sound wave went over.
Each wave carries a fixed total energy, E. An area A parallel to the wavefront at distance x gets energy ##\frac{A E}{4\pi x^2}## from each wave. If the wave frequency is ##\nu## then the power is ##\frac{A E\nu}{4\pi x^2}## , no?
 
My book gives the formula Intensity = \frac{Power}{4\pi r^2}. I think you are using energy as power and power as intensity?
 
Nishikino Maki said:
My book gives the formula Intensity = \frac{Power}{4\pi r^2}. I think you are using energy as power and power as intensity?
I believe what I wrote is consistent with the equation you quote from your book.

Let me try to put my thinking another way. Suppose there is a sound source at one end of a tube, so there's no spreading out. The sound intensity is the same all along the tube. A given volume of air is carrying, at any instant, a certain quantity of sound energy. If you sit at some point in the tube and wait for the sound to come to you, it comes at the speed of sound, c. That determines the power.
Specifically, if the tube cross-section is A and the energy density per unit volume is p then the power you receive is Apc. If instead you move towards the source, you get more of the energy in each unit of time, so more power, Ap(v+c).
 
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I understand your tube example, but I don't really see how it applies here. The question says "At that moment", so I assumed that the cars were basically standing still. Is this a wrong assumption?
 
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Nishikino Maki said:
I understand your tube example, but I don't really see how it applies here. The question says "At that moment", so I assumed that the cars were basically standing still. Is this a wrong assumption?
At a given moment, you can still have velocity, momentum, kinetic energy, acceleration...
 
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