# EK Doppler approximation question

Discussion in 'MCAT Study Question Q&A' started by tncekm, Mar 15, 2008.

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1. ### tncekmMS-1

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Okay, I've got another one for you guys.

An interstellar gas circles the core of earth's galaxy. If the wavelength of the light reflecting off the gas coming toward the earth is 499 nm, and the wavelength of light reflecting off the gas moving away from the earth is 501 nm, what is the speed of the gas?

A. 4.2 x 10&#8308; m/s
B. 1.2 x 10&#8309; m/s
C. 6.0 x 10&#8309; m/s - Correct answer
D. 1.5 x 10&#185;&#185; m/s

The explanation in the back is screwing with my understanding more than anything. I can get the correct answer just by using v=c(&#916;&#955;/&#955 (where &#916;&#955;=1nm and &#955;=500nm). But, then the answer says something about the velocity of the gas needing to be divided by two because of the round trip issue? That changes my answer to something that isn't available in A<->D. That really just confused the heck out of me.

Thanks.
2. ### tncekmMS-1

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Anybody?
3. ### physics junkieSDN Two Year Member

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You're already taking the "round trip issue" into account(a horribly unintuitive name for it). I suspect the author got confused when he wrote that explanation. He was probably drawing on the same concept that when you run 1 mile further down the beach thats an extra 2 miles added to your run to try to explain why you use half the actual wavelength shift to determine the object's velocity. You, on the other hand, realized the obvious conclusion in this problem that the real lambda value of the non-moving object is 500nm and interpreted the &#916; to refer to the deviation from the wavelength for the non-moving object. The author assumes you to be too dull to reach this conclusion on your ownp). In his eyes the &#916;&#955;=2 since 501-499=2...which is the actual difference between the two &#955; values.

Get it? I have disorganized thoughts about physics but some textbooks and explanations I've heard are just flat out horrible. His is up there.
4. ### What up docFLASH

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tnc...wat number out of ek is this?

also, i really do not get ek's doppler estimation..i prefer to just memorize the eq f'"=f(v+/- Fd/v +/- fs)....but perhaps either u or junkie can explain ek's approx as i heard this was a quicker way to solve the problems...
5. ### tncekmMS-1

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The doppler approximation is: &#916;f/f = v/c (and &#916;&#955;/&#955; = v/c).

So, if a source sound wave travels at 500 hz and it is detected at 498 hz, then

&#916;f = 2hz, f = f of the sound, c = speed of sound, so you can solve for v, which is the speed of the observer.

v=(2Hz/500Hz)340m/s = (1/250)340 &#8773;1.36m/s and you know its "away from" the sound source because the observed frequency decreased.
6. ### tncekmMS-1

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Oh, that makes sense, thanks! Anyway, to make things even worse on the part of the author, in the explanation they give the &#916;&#955; = 1nm like I had, not 2nm!
7. ### What up docFLASH

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tnc, okay, so i get ek's approxmation now...and i used it for num 756 on the physics section and it worked...but, i tried the actual equation and i cant get it to work out? if youre any good with the actual equation, perhaps u can explain why its not working out?

heres what i have...

f' = f(1/(v-vs) (- minus becuz the source is moving toward, i think)

1100 = 1000 ( 1/(340-x)

x = 341.sumthing????

wat am i doing wrong please? nebody?
8. ### tncekmMS-1

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If the source is moving toward, you'll "add" because the perceived velocity increases.

E.x. if you're running at 5 m/s toward a person, and that person is running at you at 5 m/s, you're essentially approaching each other at 5+5 m/s because you close the distance between you 5 m/s and he also closes the distance between you 5 m/s.

I'll look at the actual problem later! In the meantime, maybe this will help.