# Doppler Effect.

1. Jan 23, 2008

### _Mayday_

I have a few questions concerning the Doppler Effect, I know what it is in terms of a moving object, and how speed effects fequency etc. but I'm more interested in the explantions. My actual knowledge of the two explanations is reasonably vague.

The one I am interested in is the compression of the sound waves being emitted by say for example a train, for the people that it is approaching they will hear a high pitch, and people below a lower pitch. Can anyone expand on this idea, and possibly explain it a bit more if necessary. I've heard people talking about compression of the waves?

I have been told that there are two formulas for the Doppler effect, one a simpler one and the other a more complex one. I think the simpler one is $$v = f \lambda$$, this is useful in that it illistrates that if v remains constant ie. the speed of sound around 320m/s and the wave frequency increases, then the wavelength must decrease.

I don't know of a more complex one, but is there one that would take more factors into account? I don't seem to be able to find it, I would have thought it would have taken into account distance aswell? Any ideas?

Thank you, any help would be great

2. Jan 23, 2008

### Staff: Mentor

Sound is the effect of compression waves traveling in some medium - solid, liquid or gas.

The relative speeds of the source and the receiver affect the frequency, but the speed of sound is a property of the medium.

The source (vibration) simply imparts momentum to the atoms/molecules. The rate of change of momentum is determined by the force, and the force per unit area is the pressure. The rate at which momentum is transfered depends upon the surface velocity of the vibrating membrane or solid with respect to the medium, so the translational velocity of the vibrating source affects the surface velocity relative to the medium. Similarly the velocity of the receiver affects the rate at which the compression waves in the medium interact with the receiving surface.

I'm not sure about simple and complex equations. The complexity depends on whether the source and/or receiver are moving or not.

3. Jan 23, 2008

### _Mayday_

Sorry, I should have been a bit more specific. I understand that sounds is the effect of the compression waves. But take the example of a train, as it tavels faster does the sound waves in front of the train become even more compressed casuing an increase in frequency, but behind the waves are more spread out resulting in a lower frequency.

For more complex equation, the reciever is staionary, and the source is moving like a train going past a station.

Thanks for the help so far :)

4. Jan 23, 2008

### _Mayday_

I think I've found one for when the source is moving towards the observer. Can anyone check this and see if it's ok? If it is correct could someone take me through how you would get to that?

$$\Delta f = \frac {v} {(c-v)} f$$

5. Jan 23, 2008

### Mentz114

6. Jan 24, 2008

### _Mayday_

Is this equation correct for when the source is moving towards the stationary observer?

$$\Delta f = \frac {v} {(c-v)} f$$

And this one is the equation for when the source is moving away from the observer?

$$\Delta f = \frac {v} {(c+v)} f$$

Thanks, the link didn't mention either of these formulas.

7. Jan 24, 2008

### Mentz114

Do some algebra. The source gives -

receding source -

f' = f.v/(v+v_source)

approacing source -

f' = f.v/(v-v_source)

work out f'-f = delta f

8. Jan 26, 2008

### greeniguana00

If you want a way to think about why this must happen, you can think about it this way:

In a given time period, the train has traveled a certain distance determined by its speed. Let's assume this train produced only two clicks, one at the starting point and one at the ending point. If you had been standing right at the ending point (which wouldn't be such a good idea) and the train hadn't moved at all during the time period, you would have heard two clicks right at the beginning and end of the time period. If the train had been moving, you would have heard the end click right when the train reached the end point, but the beginning click would have been delayed by the speed of sound, making the clicks sound closer together.

9. Jan 28, 2008

### delta_moment

The derived equations of your general formula provide insight:

f'=f[1/(1-+vs/v)]

that is, frequency equals original_freq*1/1-+velocity_of_source/relative velocity to observer.

Necessary to be that general for cases where there's not a direct approach or reproach.

The __wavefront__ is what's perceived. Although the frequency doesn't change, since the next incurring sound waves hit the receiver of the sound sooner than they should due to velocity, the are higher. Opposite for when they are receding.

It's mainly due to the sound waves reaching the ear either sooner/later than if the source where at rest, relatively. Since lamba doesn't change, but the velocity is +or-, there's a discernible difference.