Wind's effect on the Doppler Effect... EK 1001 Physics... Help!

[MedGrl@2022]

I am having some issues understanding wind's effect on the Doppler Effect. I thought I understood it but apparently I did not. Please help me if you can. I think I am sort of understanding the concepts but I would like to get it down solid for MCAT. Thank you for all your help!

These are EK 1001 Physics problems

749. If the source of a sound wave and the observer are stationary, and there is a steady wind blowing from the observer to the source, how will the Doppler Effect change the observed frequency?

A. The observed frequency will be greater.
B. The observed frequency will be smaller.
C. There will be no Doppler Effect.
D. The change in the frequency cannot be predicted.

So I originally thought that the answer was B because I thought the wind would blow the sound wave away which would affect the frequency of the sound wave (because I thought it would decrease the frequency). According to EK:
“C is correct. The wind is equivalent to giving both the source and the receiver a velocity equal in magnitude and in the opposite direction to the wind. Their relative velocity remains the same. In this case, the relative velocity is zero, and there is no Doppler effect.”

Is wind not affecting the sound wave at all and only the velocity of the observer and source? Or is it just affecting the velocity of the sound wave and not the frequency and therefore not having effect on changing the pitch of the sound wave (and therefore not having any Doppler effect)?




750. 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?

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.

I thought D because once again I thought wind would decrease the frequency and therefore minimize the Doppler Effect.

According to EK: “A is correct. Most problems (probably all problems) on the MCAT concerning the Doppler effect can be solved with the formula v/c =(change in)f/fs =(change in)wavelength/wavelength(s). This is an approximation that is good only when the relative velocity v of the source and observer is much less than the velocity of the wave c. This equation cannot account for movement of the wave medium, like wind. For such problems, we must use the full equation: fo = fs[(c ± v0)/ (c ± vs)]. In this problem, the observer moves toward the source, increasing the observed frequency. The wind can be replaced by giving both the source and the observer
an extra velocity in the opposite direction to the wind. This increases the Doppler effect in this case. The ratio of their velocities plus sound velocity is the amount by which the frequency is increased. If this ratio was 350/340 without the wind, with a 10 m/s wind the ratio becomes 340/330; a greater ratio; a greater increase. (Note: The numbers are hypothetical to illustrate the problem. The velocity of sound has been chosen to be 340 m/s.)”

Thus, EK is saying that wind is changing the medium which makes sense. Since it is changing the medium it should be changing the velocity and amplitude of the wave, but not the frequency, right? Because it’s effect should be similar to a wave refracting into a second medium. Why does wind “give both the source and the observer an extra velocity in the opposite direction to the wind”?

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[slz1900]

Hard question, but the answers are not contradictory. The equation is f = (v+v(observer))/(v+v(source))f(emitted). In the first question you have to look at the reference frame from wind's perspective. It is moving in the same direction relative to the observer and the source, so its velocity is added to both the top and the bottom of the equation. Your fraction becomes (v+20)/(v+20) Adding the same thing to the numerator and the denominator doesn't change the fraction, so no net effect.

On the second question you go through the exact same process. This time the source is moving towards the observer, so your fraction is (v+5)/(v) without wind. Now let's add the wind in. Both your source and your observer are moving relative to the wind, so you get (v+25)/(v+20). I just chose random numbers for the source and the wind velocity, but if you compute this with the speed of sound, you'll see the fraction is bigger in the second case than the first, so frequency increases.

Also, I think you are confusing yourself with refraction, medium changes, etc. Take the questions more at face value.

 

[MedGrl@2022]

Hard question, but the answers are not contradictory. The equation is f = (v+v(observer))/(v+v(source))f(emitted). In the first question you have to look at the reference frame from wind's perspective. It is moving in the same direction relative to the observer and the source, so its velocity is added to both the top and the bottom of the equation. Your fraction becomes (v+20)/(v+20) Adding the same thing to the numerator and the denominator doesn't change the fraction, so no net effect.

On the second question you go through the exact same process. This time the source is moving towards the observer, so your fraction is (v+5)/(v) without wind. Now let's add the wind in. Both your source and your observer are moving relative to the wind, so you get (v+25)/(v+20). I just chose random numbers for the source and the wind velocity, but if you compute this with the speed of sound, you'll see the fraction is bigger in the second case than the first, so frequency increases.

Also, I think you are confusing yourself with refraction, medium changes, etc. Take the questions more at face value.

Why are we choosing wind as the reference frame? Also, EK says "This equation cannot account for movement of the wave medium, like wind." But I shouldn't think about it changing the medium, right?

 

[slz1900]

You have to choose something as your reference frame. Whether it is wind or a spot 20 miles to the left of where the source and observer are won't really make a difference, but you want to choose a reference frame that makes the math easy. If you choose to view the wind relative to the observer and source you can just add the velocity of the wind to the top and bottom so it is a little easier.

EK's explanations kind of suck. I think they mean the equation doesn't directly account for the movement of air, but that's neither here nor there. The medium itself isn't changing, but its movement does affect the frequency of sound. Medium properties really don't have anything to do with the question. If the source and observer were immersed in a tub of palladium you would still figure it out the same way (but sound velocity would change). In this case the medium isn't changing.

 

[MedGrl@2022]

You have to choose something as your reference frame. Whether it is wind or a spot 20 miles to the left of where the source and observer are won't really make a difference, but you want to choose a reference frame that makes the math easy. If you choose to view the wind relative to the observer and source you can just add the velocity of the wind to the top and bottom so it is a little easier.

EK's explanations kind of suck. I think they mean the equation doesn't directly account for the movement of air, but that's neither here nor there. The medium itself isn't changing, but its movement does affect the frequency of sound. Medium properties really don't have anything to do with the question. If the source and observer were immersed in a tub of palladium you would still figure it out the same way (but sound velocity would change). In this case the medium isn't changing.

Okay I guess I am having issues discerning how to use the formula then.

In number 750 The observer is moving toward the source so we added a hypothetical 10m/s to the top of the equation because the observer is getting closer to the source. The wind subtracts a hypothetical 10m/s from both the observer and the source because they are moving 10m/s in the opposite direction of the wind. Is this right? Thus, we come up with the ratios which help us solve the problem and figure out that wind would increase the frequency and Doppler effect.

In a similar 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