Moving vehicles and Doppler Effect

[Nishikino Maki]

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.
$90=10 \log \frac{I}{10^{-12}}$
$I = 10^{-3}$
$P = I*4\pi 200^2$
$I_2 = \frac{P}{4\pi 250^2}$
$\beta = 10\log \frac{I_2}{10^{-12}}$
$\beta = 88 dB$

 

[SteamKing]

There's some sort of problem with the link to the question. It can't be accessed.

 

[Nishikino Maki]

The website seems to show up for me.

Here is a screenshot

 

[haruspex]

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

 

[Nishikino Maki]

Is it?

I used the second equation and got power by multiplying intensity and the area the sound wave went over.

 

[haruspex]

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?

 

[Nishikino Maki]

My book gives the formula $Intensity = \frac{Power}{4\pi r^2}$. I think you are using energy as power and power as intensity?

 

[haruspex]

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).

 

[Nishikino Maki]

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?

 

[haruspex]

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...