B Doppler effect/ both moving same direction

AI Thread Summary
When both the source and observer move in the same direction, the Doppler effect can be analyzed by shifting the reference frame to one where either object is at rest. If the source moves faster than the observer, the observed frequency decreases, while if the observer moves faster than the source, the observed frequency increases. The equation for frequency adjustment, f'=[(u+uo)/(u+us)]f, remains valid as long as the variables are clearly defined. Additionally, the motion of the medium, such as air, must be considered in the calculations. Understanding these dynamics allows for accurate predictions of frequency shifts in various scenarios.
QuarkDecay
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What happens if a source and the observer move at the same direction?
Always hear how only one of them doesnot move, or when they move towards each other.

For example. An ambulance moves towards direction A with Va. A biker moves towards direction A as well, with Vb.
What happens to the Doppler equation, if
(a) Va>Vb
(b) Va<Vb
 
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You consider a relative velocity between them. Imagine two objects both moving with the same velocity; you would predict no Doppler shift because they are both at rest in the center of momentum frame, for instance.
 
If you have a good idea of how Doppler Effect works when one is at rest then just shift your reference frame to one in which either of the moving objects is at rest. Then the unknown problem turns into a known situation.
 
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CrazyNinja said:
If you have a good idea of how Doppler Effect works when one is at rest then just shift your reference frame to one in which either of the moving objects is at rest. Then the unknown problem turns into a known situation.

You then have to take into consideration that the air is moving in that frame of reference.
 
pixel said:
You then have to take into consideration that the air is moving in that frame of reference.

Yes true. The velocity of sound with respect to the medium gets adjusted accordingly.
 
So is it wrong if the final frequency becomes f'=[(u+uo)/(u+us)]f, since we consider that the source is moving away from the observer, while the observer moves closer to it (when the velocity of the observer is higher than the source's)
 
QuarkDecay said:
So is it wrong if the final frequency becomes f'=[(u+uo)/(u+us)]f, since we consider that the source is moving away from the observer, while the observer moves closer to it (when the velocity of the observer is higher than the source's)

It would have been better if you had precisely mentioned what "u", "uo" and "us" are in the equation you have written. The equation is valid in all situations provided you know what "u", "uo" and "us" are in the given equation. This also matches with my previous statement about shifting reference frames. In that case the physical values of the aforementioned quantities may change, but the equation still holds.
 
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