wind and doppler effect...how does relative velocity of wind come into play?

[Transformers]

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Case 1:
Lets say the observer is moving towards a stationary sound source at a velocity of 10 m/s. The source is emitting 100Hz.

1.) How does the Doppler effect account for a 10 m/s wind blowing
a.) FROM the source to the observer
b.) FROM the observer to the source.

I think the concept has to do with relative velocities but I'm not sure how the mathematical calculations go.

Case 2:
Lets say the observer is moving towards a receding sound source (both moving at 10 m/s and lets also say that the wind is blowing at 10 m/s from the observer to the source. The source is emitting 100Hz sound. Whats the perceived frequency by the observer?


Case 3: (EK PROBLEM that threw me off)...If the source of a sound wave and the observer are stationary and there is a steady wind blowing fromt he observer to the source, how will the Doppler effect change the observed frequency?

The answer is that there is no doppler shift...but I'm not sure why.

thakns

 

[RogueUnicorn]

Case 1:
Lets say the observer is moving towards a stationary sound source at a velocity of 10 m/s. The source is emitting 100Hz.

1.) How does the Doppler effect account for a 10 m/s wind blowing
a.) FROM the source to the observer
b.) FROM the observer to the source.

I think the concept has to do with relative velocities but I'm not sure how the mathematical calculations go.

Case 2:
Lets say the observer is moving towards a receding sound source (both moving at 10 m/s and lets also say that the wind is blowing at 10 m/s from the observer to the source. The source is emitting 100Hz sound. Whats the perceived frequency by the observer?


Case 3: (EK PROBLEM that threw me off)...If the source of a sound wave and the observer are stationary and there is a steady wind blowing fromt he observer to the source, how will the Doppler effect change the observed frequency?

The answer is that there is no doppler shift...but I'm not sure why.

thakns

i would think you add the windspeed to the soundspeed if the wind is blowing TOWARD the observer.. but i guess that's not the case. that kind of makes sense, as the wind itself probably doesn't compress the waves

 

[thebillsfan]

add the windspeed to the soundspeed if the wind is blowing TOWARD the observer

bleargh, in this situation can you not use the "shortened" version of the doppler effect equation provided by EK? you have to use the full version? also, i'm assuming the windspeed would subtract from the sund speed if the wind blowed from observer to source?

 

[RogueUnicorn]

bleargh, in this situation can you not use the "shortened" version of the doppler effect equation provided by EK? you have to use the full version? also, i'm assuming the windspeed would subtract from the sund speed if the wind blowed from observer to source?

i don't know what this version you speak of, but i have a feeling it is only relevant to electromagnetic waves.

 

[thebillsfan]

sorry, it's delta(wavelength)/original wavelength=relative speed/speed of sound

 

[RogueUnicorn]

Case 1:
Lets say the observer is moving towards a stationary sound source at a velocity of 10 m/s. The source is emitting 100Hz.

1.) How does the Doppler effect account for a 10 m/s wind blowing
a.) FROM the source to the observer
b.) FROM the observer to the source.

I think the concept has to do with relative velocities but I'm not sure how the mathematical calculations go.

Case 2:
Lets say the observer is moving towards a receding sound source (both moving at 10 m/s and lets also say that the wind is blowing at 10 m/s from the observer to the source. The source is emitting 100Hz sound. Whats the perceived frequency by the observer?


Case 3: (EK PROBLEM that threw me off)...If the source of a sound wave and the observer are stationary and there is a steady wind blowing fromt he observer to the source, how will the Doppler effect change the observed frequency?

The answer is that there is no doppler shift...but I'm not sure why.

thakns

ah i have the answer - to get at this, you get to answer this question:

if you are moving at 10m/s, and you hear a car honking, and this car is also moving in the same direction at 10m/s, what's the perceived shift?

 

[thebillsfan]

i was under the impression that it didnt only relate to EM waves?

 

[RogueUnicorn]

i was under the impression that it didnt only relate to EM waves?

sorry, i was thinking of a different shortened/simplified equation. the one you posted is probably fine, although i'm going to see how they derived it..

 

[thebillsfan]

there'd be no perceived shift--but how does that relate to the wind?

 

[RogueUnicorn]

if the wind is carrying sound by an increased speed of 10, let's say, what's relative speed of sound at the source vs. the observer?

 

[thebillsfan]

this is where i get tripped up. i would say the RELATIVE SPEED between the source and observer are the same, but that the speed of sound itself is faster. so in the question
(v+/- source)/(v+/-obs), the v part would change, but since source and obs are both going at 10 mph their relative speed would not change. but this would mean there's no doppler effect...

 

[RogueUnicorn]

exactly. no doppler effect.

 

[thebillsfan]

wow, cant believe i got that right. so, wind only has an effect on the doppler effect if there's a relative speed. if the wind was going from observer to source (ie, slowing down the speed of sound), then the the effect of the relative speed would be greater (since youre divided it by a smaller total speed of sound). but if the wind speeds up the speed of sound in the air, then the relative speed has a less effect on the freq change.

 

[RogueUnicorn]

wow, cant believe i got that right. so, wind only has an effect on the doppler effect if there's a relative speed. if the wind was going from observer to source (ie, slowing down the speed of sound), then the the effect of the relative speed would be greater (since youre divided it by a smaller total speed of sound). but if the wind speeds up the speed of sound in the air, then the relative speed has a less effect on the freq change.

yes. but if there is no relative speed difference,t he effect is zero. you can qualitatively think of this by comparing sound to light. a car coming at you will shift sound frequencies noticeably, but certainly the color of the car doesn't seem any different as it barrels toward you.

 

[Transformers]

Yeah, but what if they are both going towards each other + wind blowing from observer to the source...if the speed of sound decreases (because its opposing the motion of sound) and thus for example say it went from 340-->330; instead of 340+10/340-10 (freq), we have 330+10/330-10 (frequncy) and since 350/330 > 340/320, the wind blowing effect would thus DECREASE the doppler shift compared to a NO WIND situation correct?

BOTTOMLINE...the wind would INCREASE the velocity of sound if its blowing AWAY from the source and the wind would DECREASE the velocity of sound if its blowing TOWARDS the source correct? Thus the v is what changes....

 

[RogueUnicorn]

you got it.

 

[Hemichordate]

This is slightly off topic, but how do you designate the +/- signs in the equation for the Doppler effect? If you are moving toward a sound that is also moving away, what would be the signs in the numerator and denominator?

 

[thebillsfan]

and slightly related--speed of sound at 0 deg is 331, believe, and 340 at room temp? will they probably tell us which of those to use in a problem?

 

[ratatat]

This is slightly off topic, but how do you designate the +/- signs in the equation for the Doppler effect? If you are moving toward a sound that is also moving away, what would be the signs in the numerator and denominator?

I believe if you're moving toward a sound that is also moving away, your velocities will have the same sign. So you can choose either both positive or both negative.

Also, if you're both moving in the same direction, subtract your velocities. If you're moving in opposite directions, add your velocities. One way I remember to do this is I think of both the source and observer standing at the same point, like say on a starting line in a race. If they're moving away from each other than they run in opposite directions, and thus you add their relative displacements will get bigger, so you ADD their velocities.

 

[MedGrl@2022]

Yeah, but what if they are both going towards each other + wind blowing from observer to the source...if the speed of sound decreases (because its opposing the motion of sound) and thus for example say it went from 340-->330; instead of 340+10/340-10 (freq), we have 330+10/330-10 (frequncy) and since 350/330 > 340/320, the wind blowing effect would thus DECREASE the doppler shift compared to a NO WIND situation correct?

BOTTOMLINE...the wind would INCREASE the velocity of sound if its blowing AWAY from the source and the wind would DECREASE the velocity of sound if its blowing TOWARDS the source correct? Thus the v is what changes....

Hi I am currently trying to figure this problem out. According to EK, the wind can be replaced by giving both the source and the observer an extra velocity in the opposite direction to the wind. Thus, wind is effecting both the source and observer in the same way. Thus the signs should be the same. For that problem is just increases the ratio.

I am having difficulty rationalizing how to mathematically factor in the wind factor. For instance in the following problem:

751. The source of a sound wave is stationary. The observer is moving toward the source. There is a steady wind blowing from the source to the observer. How does the wind change the observed frequency?
A. The wind magnifies the Doppler Effect and increases the frequency.
B. The wind minimizes the Doppler Effect and increases the frequency.
c. The wind magnifies the Doppler Effect and decreases the frequency.
D. The wind minimizes the Doppler Effect and decreases the frequency.

The answer is D because " In this case, if the same original ratio from #750 were 350/340, the ratio after considering the wind in the opposite direction would be 360/350; a lower ratio; a lower increase."

Wind is now added to both sides of the equation because the observer is moving closer to the wind? Isn't the source which is stationary moving in the opposite direction relative to the wind?

Thank you once again for all your help!

Verónica