EK Physics 1001 Question 750

[sillyjoe]

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This question (750) as well as 749 and 751 are killing me. I literally cannot understand the explanation. Any help would be MUCH appreciated!!!!

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 decreases the frequency.
D. The wind minimizes the Doppler decreases the frequency.
A is correct. 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 oftheir 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.)

I have looked up previous threads on this problem and still cannot seem to understand the answer.

 

[t5Nitro]

That explanation is too long. Here it is simply:

Consider when both the sound source and the observer are stationary. The sound source is emitting sound waves that, for hypothetical purposes, let's say every 10 seconds a new sound wave reaches the observers ears.

Now consider when the sound source is stationary but the observer is moving towards the source. The sound source is still emitting sound waves at the same frequency of 10 waves per second, but the observer is walking towards the source so the perceived frequency is higher. If you want to think of it really simply, instead of waiting a full 10 seconds for the observer to run into another sound wave, s/he will "walk into it" before 10 seconds elapses.

Now consider when the sound source is stationary but the observer is moving away from the source. Again, the sound source is emitting at our hypothetical 10 waves per second, but now the observer is walking away from the wave. In simple terms, the observer is "walking away" from the emitted waves. So as one wave is emitted, it will take longer than 10 seconds to reach the observer since s/he is "walking away" from it. Therefore, it takes longer to reach the observer and the perceived frequency decreases.

The same scenarios apply for a stationary observer and a moving sound source. If the source is moving toward a stationary observer, the observer perceives higher frequency. If the moving source is moving away from the observer, the perceived frequency is less.

These are the concepts you will want to master, and you'll be fine. Just remember, any time the source and observer are approaching one another, the perceived frequency will be higher. Any time the source and observer are moving away from each other, the perceived frequency will be lower (relative to when the source and observer are both stationary).

Hope this helps.

Edit: The part where they say "The wind can be replaced by giving both the source and the observer an extra velocity in the opposite direction to the wind." is poor wording. The wind could give the observer an increased velocity because it is blowing in the direction s/he is walking, thereby increasing the numerator in the equation thus increasing the observer frequency. I like the approach detailed above, and I think you will only need to know it on that level of simplicity.

 

[sillyjoe]

I appreciate the response, but how does wind play into it?

 

[orangetea]

I was wondering about this too but the way I reasoned was the wind was an additional velocity term on the numerator hence why the doppler effect is larger.
I think it's centered around the concept that sound travels in mediums and one of them is air. Like if you have even been to a concert on a windy day you notice that the sound is a lot louder.. The same thing is applied here. It's kind of counterintuitive though because the wind is blowing from the observer to the source so you would think that the sound waves are being pushed away. That's what gets me. But the math makes sense so that's just how I reasoned.

 

[t5Nitro]

I appreciate the response, but how does wind play into it?

Check out my edit. I forgot to add it. I think the answer explanation is set up poorly. Orangetea also provided the answer.

 

[sillyjoe]

I was wondering about this too but the way I reasoned was the wind was an additional velocity term on the numerator hence why the doppler effect is larger.
I think it's centered around the concept that sound travels in mediums and one of them is air. Like if you have even been to a concert on a windy day you notice that the sound is a lot louder.. The same thing is applied here. It's kind of counterintuitive though because the wind is blowing from the observer to the source so you would think that the sound waves are being pushed away. That's what gets me. But the math makes sense so that's just how I reasoned.

Some EK questions really stump me. There was another one about the speed of an interstellar gas which conceptually killed me. #760

 

[t5Nitro]

I will put in some numbers to hopefully make their answer a bit better demonstrated.

c is 340 m/s in their equation. c + v_o is (340 m/s + 10 m/s) : Let's say the 10 m/s is the velocity of the observer without wind. Divide that by (340 m/s + 0 m/s) : 0 m/s because the source is not moving. So it's (340 m/s + 10 m/s)/(340 m/s + 0 m/s) = number.

Now the wind can add velocity to the observer moving in that direction. So it could be (340 m/s + 15 m/s)/(340 m/s + 0 m/s) = bigger number. (The sound source is still stationary. That bad boy is 0 m/s still).

Overall, if you manipulate the equation, you will see the numerator increases so your perceived frequency increases.

 

[sillyjoe]

I will put in some numbers to hopefully make their answer a bit better demonstrated.

c is 340 m/s in their equation. c + v_o is (340 m/s + 10 m/s) : Let's say the 10 m/s is the velocity of the observer without wind. Divide that by (340 m/s + 0 m/s) : 0 m/s because the source is not moving. So it's (340 m/s + 10 m/s)/(340 m/s + 0 m/s) = number.

Now the wind can add velocity to the observer moving in that direction. So it could be (340 m/s + 15 m/s)/(340 m/s + 0 m/s) = bigger number. (The sound source is still stationary. That bad boy is 0 m/s still).

Overall, if you manipulate the equation, you will see the numerator increases so your perceived frequency increases.

Thanks for explaining it.

 

[NextStepTutor_1]

Also, as orangetea mentioned, you can think of the wind as "pushing" the sound waves away from the detector. However, what the really does is compress the sound waves even more, making them have a higher frequency.

 

[Lunasly]

Doesn't this essentially just mean that the wind is giving you a boost. That is, you move towards the source at a faster rate with the help of the wind?