Moving vehicles and Doppler Effect

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Homework Help Overview

The discussion revolves around a problem involving the Doppler Effect, specifically related to a fire engine moving towards two vehicles while sounding its horn. The problem includes calculating the frequency heard by passengers in a moving car and a stationary van, as well as the sound intensity level experienced by the van's passengers when the fire engine is at a certain distance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the Doppler Effect formula and the calculation of sound intensity and decibel levels. There are attempts to verify the correctness of the original poster's calculations and the implications of sound frequency on power received. Questions arise regarding the assumptions made about the vehicles' motion at the moment of measurement.

Discussion Status

The discussion is active, with participants questioning the validity of certain assumptions and exploring the relationships between intensity, power, and frequency. Some participants provide insights into the physics concepts involved, while others seek clarification on the original poster's reasoning.

Contextual Notes

There is a mention of a broken link to the problem statement, which may affect participants' ability to reference the original context. Additionally, assumptions about the motion of the vehicles at the moment of measurement are being debated.

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
 
Last edited by a moderator:
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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).
 
Last edited:
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?
 
  • #10
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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