acetylmandarin

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Never would I have thought that wind has any effect on sound waves, but...
"The source of a sound wave is stationary. The observer is moving toward the source. There is a steady wind blowing from the observer to the source. How does the wind change the observed frequency?"

The answer is "The wind magnifies the doppler effect and increases the frequency."

I used the equation, Frequency of listener = freq. of source * (v sound + v listener / v sound )

So I guess the wind velocity would add to the numerator, to add to the component of the fraction that shows the velocity of the listener towards the source, and therefore you would be multiplying by a larger number and have a larger observed frequency?
 

theonlytycrane

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Something like that. Maybe the wind is pushing the observer to move faster!
 

TheLongGame

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You need to change your frame of reference from yourself to the source of the sound wave. Sound waves propagate outwards from the source at a set frequency. In stationary air, the frequency remains unchanged, however, sound wave's relative velocity is reduced by the difference between the speed of sound in stationary air and the wind speed. By reducing the relative speed of sound you essentially compress the wave and therefore increase its frequency/shorten its wave length relative to a stationary frame of reference. I kind of like to think of it as sound running on a treadmill!

Compare this to the red shift observed when looking at light from distant stars. It's the exact same thing, just in reverse. (And of course, we all know c is a constant, but the speed of sound is not)

Does this make sense?


Sent from my iPhone using SDN mobile
 
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Disregard this post, @sazerac explains this way better than me below..


@TheLongGame can you clear up what you're saying? It's not quite clear to me.

The question is really asking: Why does wind speed influence the Doppler shift when the source or observer is moving, but not when they are both stationary?

Here are two previous SDN attempts at explaining this question:
http://forums.studentdoctor.net/threads/winds-effect-on-the-doppler-effect-ek-1001-physics-help.982631/
http://forums.studentdoctor.net/threads/wind-and-doppler-effect-how-does-relative-velocity-of-wind-come-into-play.654148/

Honestly, I can't make sense of any explanation I've read. In fact, my own attempts have found the opposite answer. Here they are:

Attempt 1: assuming wind speed is simply added to the speed of sound (the speed of sound in the medium changes):

fo = fs (c + vo) / (c + vs) ; c = c + vw (wind speed)
fo = fs (c + vw + vo) / (c + vw + vs) ; velocity of observer is positive because moving towards source, velocity of source is 0
fo = fs (c + vw + vo) / (c + vw) ; compare this to no wind
fs (c + vw + vo) / (c + vw) < fs (c + vo) / (c)
(With wind is less than (<) without)

When we add wind speed to the numerator and denominator, the denominator is more greatly affected, so our observed frequency is actually lower. This is the opposite of what the EK answer is telling us.

However, it can be used to show that wind speed only influences the Doppler shift with a moving observer or source.
fo = fs (c + vw + vo) / (c + vw + vs) ; vo = 0 and vs = 0
fo = fs (c + vw) / (c + vw) ; (c + vw) / (c + vw) = 1
fo = fs
Frequency is unchanged with a stationary observer and source.

Attempt 2: changing the frame of reference so that wind speed is 0, and both observer and source are moving.

fo = fs (c + vo) / (c + vs) ; vo = vo + vw, vs = vs + vw ; vo is positive and vs is positive
fo = fs (c + vo + vw) / (c + vs + vw) ; vs = 0
fo = fs (c + vo + vw) / (c + vw)
Here we get the same equation as in my first attempt, and thus the same implications.

Unless someone can point out some flaw here, or find a better explanation, I think this question might be slightly off.

And for what it's worth, it seems like this question is from EK 1001 Physics for the old MCAT. Honestly, I wouldn't really bother with it for the new one. There is no real big-picture or MCAT-think way to solve it, nor is there a quick look at some equations to lead you to the answer.
 
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sazerac

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@TheLongGame can you clear up what you're saying? It's not quite clear to me.

The question is really asking: Why does wind speed influence the Doppler shift when the source or observer is moving, but not when they are both stationary?

Here are two previous SDN attempts at explaining this question:
http://forums.studentdoctor.net/threads/winds-effect-on-the-doppler-effect-ek-1001-physics-help.982631/
http://forums.studentdoctor.net/threads/wind-and-doppler-effect-how-does-relative-velocity-of-wind-come-into-play.654148/

Honestly, I can't make sense of any explanation I've read. In fact, my own attempts have found the opposite answer. Here they are:

Attempt 1: assuming wind speed is simply added to the speed of sound (the speed of sound in the medium changes):

fo = fs (c + vo) / (c + vs) ; c = c + vw (wind speed)
fo = fs (c + vw + vo) / (c + vw + vs) ; velocity of observer is positive because moving towards source, velocity of source is 0
fo = fs (c + vw + vo) / (c + vw) ; compare this to no wind
fs (c + vw + vo) / (c + vw) < fs (c + vo) / (c)
(With wind is less than (<) without)

When we add wind speed to the numerator and denominator, the denominator is more greatly affected, so our observed frequency is actually lower. This is the opposite of what the EK answer is telling us.

However, it can be used to show that wind speed only influences the Doppler shift with a moving observer or source.
fo = fs (c + vw + vo) / (c + vw + vs) ; vo = 0 and vs = 0
fo = fs (c + vw) / (c + vw) ; (c + vw) / (c + vw) = 1
fo = fs
Frequency is unchanged with a stationary observer and source.

Attempt 2: changing the frame of reference so that wind speed is 0, and both observer and source are moving.

fo = fs (c + vo) / (c + vs) ; vo = vo + vw, vs = vs + vw ; vo is positive and vs is positive
fo = fs (c + vo + vw) / (c + vs + vw) ; vs = 0
fo = fs (c + vo + vw) / (c + vw)
Here we get the same equation as in my first attempt, and thus the same implications.

Unless someone can point out some flaw here, or find a better explanation, I think this question might be slightly off.

And for what it's worth, it seems like this question is from EK 1001 Physics for the old MCAT. Honestly, I wouldn't really bother with it for the new one. There is no real big-picture or MCAT-think way to solve it, nor is there a quick look at some equations to lead you to the answer.
If the objects are moving towards or away from each other, then the perceived frequency will be altered by the speed of the waves traveling through the medium. If there is a wind, the wind will also affect the speed with which it travels through the medium (as measured over ground).

If the objects are not moving with respect to each other, then there will be no Doppler effect because there will not AND CANNOT be any change in the frequency. Why not?

This is where you get to the MCAT-worthy big picture: these pressure waves of sound are physical waves. They are only created by the source and only received by the receiver. Additional pressure waves cannot be created or destroyed. If the stationary sender is creating waves at a frequency of one per second, then the stationary receiver must receive them once a second. If he received them any slower, where did the extra waves go? If he received them any faster, who created the extra waves?

The fundamental physics concept these questions test are your fast intuition about the conservation of physical objects like pressure waves, the definition of frequency as events per second, waves traveling in a medium, and the perception of frequency as pitch. No equations are necessary to answer them, just a fast intuition.

I think these are all still fair game for the MCAT.
 
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@sazerac I'm still not seeing it. I get that there is no Doppler effect when there is no movement, but I don't see how you can predict how wind will affect the Doppler effect when there is motion.

Intuitively it seems like it makes sense that if the wind is travelling in the same direction as the sound wave the frequency will increase, but I'm just not seeing an explanation or proof that convincingly shows this.
 

sazerac

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@sazerac I'm still not seeing it. I get that there is no Doppler effect when there is no movement, but I don't see how you can predict how wind will affect the Doppler effect when there is motion.

Intuitively it seems like it makes sense that if the wind is travelling in the same direction as the sound wave the frequency will increase, but I'm just not seeing an explanation or proof that convincingly shows this.
We need a specific example to analyze. Let's take the original scenario. Person walking towards source in no wind, vs person walking towards source with lots of wind from person to source.

In the first case, the person is going to walk quickly through the sound waves between person and source. The frequency goes up (move waves per second). This is the classic Doppler effect.

Second scenario: source is making pressure waves. But the wind is blowing from person to source at 0.999 the speed of sound. The waves travel overground very very slowly to the person. The waves build up. Tons and tons of pressure waves between source and person. If the person was standing still, waves would still arrive at the original frequency. Now the person starts walking towards the source, passing through thousands if not millions of waves with each step. Frequency (number of waves observed per second) is now enormous. A million waves per second? That's way above human hearing. Therefore, the wind blowing from person to source while person walks towards source causes the Doppler effect to be magnified (vs no wind) and the frequency to increase.

Hope this helps! As you can see, a lot of the MCAT questions can be solved by making extreme assumptions and figuring out what happens. Besides F=ma, very few equations are needed.
 
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sazerac

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@sazerac I'm still not seeing it. I get that there is no Doppler effect when there is no movement, but I don't see how you can predict how wind will affect the Doppler effect when there is motion.

Intuitively it seems like it makes sense that if the wind is travelling in the same direction as the sound wave the frequency will increase, but I'm just not seeing an explanation or proof that convincingly shows this.
You seem to be trying to analyze a slightly different scenario: what if the wind was blowing from source to observer?

Okay. Let's say there is no wind. The source and observer is far enough away that there are 100 waves between them. If the observer and source are still, then the observer hears waves at the original frequency. If the observer runs towards the source and meets them in 1 second, the frequency went up by 100Hz. This is classic Doppler scenario.

Now according to you, wind blows towards observer. So now those waves travel very fast over ground. Maybe there are only 20 waves now between source and observer. If they are both standing still, the observer hears the sound at the original frequency. The wind is irrelevant. But now once again the observer runs to the source and gets there in one second. Boom, the observer caught an extra 20 waves in that second. The sound heard went up by 20Hz.

Therefore in your scenario, the frequency still goes up vs standing still, but the Doppler effect is reduced vs the no wind scenario.
 
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We need a specific example to analyze. Let's take the original scenario. Person walking towards source in no wind, vs person walking towards source with lots of wind from person to source.

In the first case, the person is going to walk quickly through the sound waves between person and source. The frequency goes up (move waves per second). This is the classic Doppler effect.

Second scenario: source is making pressure waves. But the wind is blowing from person to source at 0.999 the speed of sound. The waves travel overground very very slowly to the person. The waves build up. Tons and tons of pressure waves between source and person. If the person was standing still, waves would still arrive at the original frequency. Now the person starts walking towards the source, passing through thousands if not millions of waves with each step. Frequency (number of waves observed per second) is now enormous. A million waves per second? That's way above human hearing. Therefore, the wind blowing from person to source while person walks towards source causes the Doppler effect to be magnified (vs no wind) and the frequency to increase.

This is a great explanation! I get it now. I wish I could give you a million internet points. Thanks so much :).

Also, I did slightly misread the original question, though this misreading didn't contribute to my misunderstanding.