# Moving vehicles and Doppler Effect

• Nishikino Maki
In summary: They are just not influencing the sound wave at that moment. Intensity is simply the power divided by the area.
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.
$90=10 \log \frac{I}{10^{-12}}$
$I = 10^{-3}$
$P = I*4\pi 200^2$
$I_2 = \frac{P}{4\pi 250^2}$
$\beta = 10\log \frac{I_2}{10^{-12}}$
$\beta = 88 dB$

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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?

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...

## 1. How does the Doppler Effect affect moving vehicles?

The Doppler Effect is a phenomenon where there is a change in the frequency of a wave as the source of the wave moves relative to the observer. In the case of moving vehicles, the sound waves produced by the vehicle's engine or horn are compressed in the direction of motion and spread out in the opposite direction. This results in a change in the perceived pitch of the sound, making it higher as the vehicle approaches and lower as it moves away.

## 2. How does the speed of a vehicle affect the Doppler Effect?

The speed of a vehicle directly affects the amount of frequency change observed due to the Doppler Effect. The higher the speed of the vehicle, the greater the change in frequency. This means that a faster moving vehicle will have a more noticeable change in pitch compared to a slower moving one.

## 3. Can the Doppler Effect be observed with other types of waves besides sound?

Yes, the Doppler Effect can be observed with all types of waves, including light and water waves. In fact, the Doppler Effect was first described in relation to light waves by Austrian physicist Christian Doppler in 1842.

## 4. How is the Doppler Effect used in modern technology?

The Doppler Effect is used in various modern technologies, such as Doppler radar and sonar systems. These technologies use the change in frequency of waves to measure the speed and direction of moving objects, such as weather patterns or underwater vessels.

## 5. Can the Doppler Effect be heard in space?

No, the Doppler Effect cannot be heard in space because there is no medium for sound waves to travel through. However, it can still be observed with other types of waves, such as light waves, which are used to study the movement of stars and galaxies in space.

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