Understanding the Doppler Effect with Moving Sources and Observers

In summary, the conversation discusses the observed frequency when both the source and observer are in motion. The equations for the observed frequency are provided for when either the source or observer is still, but the product of these equations does not seem to match the actual equation. The conversation then suggests transforming into a frame where one of them is still, but this does not work because the medium will be in motion. It is proposed to consider a stationary observer between the source and observer and determine the frequency they would hear, and then use this frequency to calculate the frequency the other receiver would hear.
  • #1
windowofhope
5
0

Homework Statement


Can't figure how to prove when both source and observer are moving that the observed frequency is the product of the two independent cases below..

Homework Equations


We know individual cases when source is moving and observer is still, the observed frequency is equal to fs/(1-(vs/c)*cos(theta_s)) where fs is the source frequency and theta_s is the angle between the source velocity and observer

Conversely when observer is moving, the observed frequency is fs*(1+(vo/c)*cos(theta_o))

The Attempt at a Solution


When you multiply both out, you get fo2=fs2*(c+vocos(theta_o))/(c-vscos(theta_s)) but they shouldn't be squared??
 
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  • #2
Transform into a frame where one of them is still. What do you get?
 
  • #3
well of course either times
(1+(vo/c)*cos(theta_o))

or divided by
(1-(vs/c)*cos(theta_s))

I just can't comprehend how the product of the two shifts leads to the real equation..
 
  • #4
Khashishi said:
Transform into a frame where one of them is still. What do you get?
That doesn't work, since then the medium will be moving. The speeds relative to the medium are important.
@windowofhope , consider a stationary observer between them. What frequency will the observer hear? If the observer were to emit a tone at that frequency, what frequency would the other receiver hear?
 
  • #5
I assumed it was in vacuum, since the equations listed above have no "n".
 
  • #6
haruspex said:
That doesn't work, since then the medium will be moving. The speeds relative to the medium are important.
@windowofhope , consider a stationary observer between them. What frequency will the observer hear? If the observer were to emit a tone at that frequency, what frequency would the other receiver hear?
what youre alluding to isn't really what I'm getting at since swapping fo and fs wouldn't solve it either. The math is clearly right - I'm just confused as to how you can simplify fo^2=fs^2 *((1+(vo/c)*cos(theta_o)))/((1-(vs/c)*cos(theta_s)))
 
  • #7
windowofhope said:
I'm just confused as to how you can simplify fo^2=fs^2 *((1+(vo/c)*cos(theta_o)))/((1-(vs/c)*cos(theta_s)))
I'm confused as to how you get
##f_o^2=f_s^2 \frac{(1+\frac{v_o}{c}\cos(\theta_o))}{(1-\frac{v_s}{c}\cos(\theta_s))}##.
Following my own hint in post #4 I get a rather simpler result.
 
  • #8
haruspex said:
I'm confused as to how you get
##f_o^2=f_s^2 \frac{(1+\frac{v_o}{c}\cos(\theta_o))}{(1-\frac{v_s}{c}\cos(\theta_s))}##.
Following my own hint in post #4 I get a rather simpler result.
Honestly I'm not following what you're suggesting... :-/

I got the squares simply by taking each individual formula and multiplying them together.. It then becomes clear that you get the proper term in parentheses (1+vocoso/c)/(1-vscoss/c) but the fo and fs are squared as a result of that as well...
 
  • #9
windowofhope said:
I got the squares simply by taking each individual formula and multiplying them together...
ok, but unfortunately that's not a good guess.
Try following my reasoning.
Put a stationary observer O on the straight line between source S and receiver R. What frequency does O hear? Call that frequency f'.
Now forget S and consider O generating a frequency f'. What, according to the Doppler equations, is the frequency R will hear?
 

FAQ: Understanding the Doppler Effect with Moving Sources and Observers

What is the Doppler effect?

The Doppler effect is a phenomenon in which the frequency and wavelength of a wave appears to change when the source of the wave and the observer are in relative motion.

How does the Doppler effect affect sound?

The Doppler effect causes the frequency of sound waves to change when the source of the sound and the listener are in motion relative to each other. This results in a perceived change in pitch of the sound.

What is the difference between the Doppler effect for sound and light?

The Doppler effect for sound is based on the relative motion of the source and the observer, while the Doppler effect for light is based on the relative motion of the source and the observer as well as the speed of light. This results in a change in frequency and wavelength for both sound and light, but the effect is more noticeable for sound.

How is the Doppler effect used in everyday life?

The Doppler effect is used in various applications, such as radar and sonar systems, to determine the speed and direction of moving objects. It is also used in medical ultrasound to create images of internal organs and structures.

Can the Doppler effect be observed in space?

Yes, the Doppler effect can be observed in space. Astronomers use it to determine the speed and direction of celestial objects, such as stars and galaxies, based on the shift in their light spectra. It is also used to detect exoplanets by observing the slight changes in the star's light caused by the planet's orbit.

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