Moving vehicles and Doppler Effect

Click For Summary
SUMMARY

The discussion focuses on the application of the Doppler Effect in a physics problem involving a fire engine moving at 40 m/s and sounding a horn at 500 Hz. The frequency perceived by a car moving at 30 m/s is calculated to be 516.5 Hz, while the frequency heard by a stationary van is 566 Hz. Additionally, the sound intensity level of 90 dB heard by the car translates to an intensity level of 88 dB for the van, demonstrating the relationship between intensity and distance in sound propagation.

PREREQUISITES
  • Doppler Effect formula: f' = f(v±v_o)/(v∓v_s)
  • Sound intensity calculation: I = P/(4πr²)
  • Decibel level calculation: β = 10 log(I/10^-12)
  • Understanding of sound wave propagation and energy transfer
NEXT STEPS
  • Study advanced applications of the Doppler Effect in various fields such as astronomy and radar technology.
  • Explore sound intensity and decibel level calculations in different mediums.
  • Learn about the effects of relative motion on sound frequency and intensity in real-world scenarios.
  • Investigate the relationship between sound wave frequency, power, and energy transfer in acoustics.
USEFUL FOR

Students studying physics, educators teaching sound wave concepts, and anyone interested in the practical applications of the Doppler Effect in real-world situations.

Nishikino Maki
Messages
17
Reaction score
0

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:
Physics news on Phys.org
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
 

Attachments

  • Screen Shot 2015-11-27 at 2.45.55 PM.png
    Screen Shot 2015-11-27 at 2.45.55 PM.png
    15.7 KB · Views: 910
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...
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
928
Replies
8
Views
1K
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K