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Moving vehicles and Doppler Effect

  1. Nov 27, 2015 #1
    1. The problem statement, all variables and given/known data
    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?
    }


    2. Relevant equations
    Doppler Effect:
    [itex]f' = f\frac{v±v_o}{v∓v_s}[/itex]

    Intensity:
    [itex]I = \frac{P}{4\pi r^2}[/itex]

    Sound level:
    [itex]\beta = 10 \log \frac{I}{10^{-12}} [/itex]

    3. 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:
    [itex]f'=500(\frac{343 - 30}{343 - 40})[/itex]
    This turned out to be 516.5 Hz

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

    Part c:
    For this part I assumed that everything was standing still, and just used intensity and decibel formulas.
    [itex]90=10 \log \frac{I}{10^{-12}}[/itex]
    [itex]I = 10^{-3}[/itex]
    [itex]P = I*4\pi 200^2[/itex]
    [itex]I_2 = \frac{P}{4\pi 250^2}[/itex]
    [itex]\beta = 10\log \frac{I_2}{10^{-12}}[/itex]
    [itex]\beta = 88 dB[/itex]
     
    Last edited by a moderator: Nov 27, 2015
  2. jcsd
  3. Nov 27, 2015 #2

    SteamKing

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    There's some sort of problem with the link to the question. It can't be accessed.
     
  4. Nov 27, 2015 #3
    The website seems to show up for me.

    Here is a screenshot
     

    Attached Files:

  5. Nov 27, 2015 #4

    haruspex

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    For part c, isn't the power received also dependent on received frequency?
     
  6. Nov 27, 2015 #5
    Is it?

    I used the second equation and got power by multiplying intensity and the area the sound wave went over.
     
  7. Nov 27, 2015 #6

    haruspex

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    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?
     
  8. Nov 27, 2015 #7
    My book gives the formula [itex]Intensity = \frac{Power}{4\pi r^2}[/itex]. I think you are using energy as power and power as intensity?
     
  9. Nov 28, 2015 #8

    haruspex

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    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: Nov 28, 2015
  10. Nov 30, 2015 #9
    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?
     
  11. Nov 30, 2015 #10

    haruspex

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    At a given moment, you can still have velocity, momentum, kinetic energy, acceleration....
     
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