Doppler Effect Question

[HopefulOncoDoc]

Hey guys I need some help with this doppler effection question. It's in reference to one of the trials in the passage..

Trial III: A jet flies in a straight line at a constant speed of 200 m/s above the radar dish as the radar dish emits a uniform EM signal with a wavelength that increases uniformly from 150 to 250m. The wavelength is 200 m as the jet passes directly over the dish.

In trial III, how does the frequency of the reflected wave from the jet differ from the frequency emitted by the dish while the jet is moving towards the dish?

A. The reflected frequency is greater than the emitted frequency by a factor of 0.00006%.
B. The reflected frequency is greater than the emitted frequency by a factor of 0.006%.
C. The reflected frequency is greater than the emitted frequency by a factor of 0.6%.
D. The reflected frequency is less than the emitted frequency by a factor of 0.006%.

I know D is out since the frequency has to increase. I'm just a little confused on how to calculate the magnitude of the increase. According to the solution, it is essentially the speed of the source divided by the speed of the wave x 100% making choice A correct. But I'm confused as to how this answer is derived from the doppler equation of Fl = (V + Vl / V - Vs)Fs

Thanks a lot everyone.

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[Geekchick921]

Well, we know if neither the source nor the observer were moving, the observed frequency would be the same as the real (V/V = 1). Since the
observer" is moving towards the source, if we just want to know how much the frequency increases, we can just take (V0/V). 2.0x 10^2 / 3.0x10^8 = 6x10^-7. x100% = .00006%.

 

[addo]

it is .000006%
(200/3*10^8)*100=6.6 *10^-5

on the mcat, you will not usually use the full equation. You usually use
v/c=delta F/F=delta wavelength/wavelength
So you could also use wavelength as follows:
(200m/s/(3*10^8m/s))*200m=1.33*10^-4
1.33*10^-4m/200m=6.6*10^-7*100=6.6*10^-5
same result that you got as above

 

[HopefulOncoDoc]

Well, we know if neither the source nor the observer were moving, the observed frequency would be the same as the real (V/V = 1). Since the
observer" is moving towards the source, if we just want to know how much the frequency increases, we can just take (V0/V). 2.0x 10^2 / 3.0x10^8 = 6x10^-6.

Are those percents in the original question? And if so, are you sure you have the right number of 0's. I think it should just be .000006, not .000006%.

Isn't the observer (Dish) stationary so its actually V/V - Vs ? Please correct if I'm wrong. Thanks for the help btw!

 

[HopefulOncoDoc]

it is .000006%
(200/3*10^8)*100=6.6 *10^-5

on the mcat, you will not usually use the full equation. You usually use
v/c=delta F/F=delta wavelength/wavelength
So you could also use wavelength as follows:
(200m/s/(3*10^8m/s))*200m=1.33*10^-4
1.33*10^-4m/200m=6.6*10^-7*100=6.6*10^-5
same result that you got as above

addo thanks a lot! I'll remember that equation since utilizing it makes much more sense to me.

 

[Geekchick921]

Isn't the observer (Dish) stationary so its actually V/V - Vs ? Please correct if I'm wrong. Thanks for the help btw!

That is a very good point. I think you're probably right. Either way, you will arrive at the right answer.

(3.0x10^8)/(3.0x10^8 - 200) = 1.00000066 x 100% = 100.00006%

 

[plsfoldthx]

it is .000006%
(200/3*10^8)*100=6.6 *10^-5

on the mcat, you will not usually use the full equation. You usually use
v/c=delta F/F=delta wavelength/wavelength
So you could also use wavelength as follows:
(200m/s/(3*10^8m/s))*200m=1.33*10^-4
1.33*10^-4m/200m=6.6*10^-7*100=6.6*10^-5
same result that you got as above

These are different values.

 

[plsfoldthx]

Ok guys, I hate waves.

Can someone explain to me how the equation df/fs = v/c = dwavelength/wavelength source is being used here?

Isn't the question asking how does the frequency of the reflected wave from the jet (@ wavelength 200m/s) differ from the frequency of the reflected wave from the source (@ wavelength 150m/s)??

Isn't delta wavelength 50 and source wavelength 150?

 

[Geekchick921]

Ok guys, I hate waves.

Can someone explain to me how the equation df/fs = v/c = dwavelength/wavelength source is being used here?

Isn't the question asking how does the frequency of the reflected wave from the jet (@ wavelength 200m/s) differ from the frequency of the reflected wave from the source (@ wavelength 150m/s)??

Isn't delta wavelength 50 and source wavelength 150?

Hold the phone! The 200 m/s is just the speed of the plane. We're talking about the radiowaves, which are moving at the speed of light. The wavelength ranges from 150 to 250, but that information isn't needed for this question. If we needed to actually calculate the frequency here, we would use it, but we don't even need to go that far.

 

[addo]

These are different values.

Oh yea, i did not notice that, I put the 0s in according to the answer choice, but 6.6*10^-5 was the actual answer that I got.
Hopefuloncodoc, are you sure you didnt slip an extra 0 in the answer choice?

 

[plsfoldthx]

Hold the phone! The 200 m/s is just the speed of the plane. We're talking about the radiowaves, which are moving at the speed of light. The wavelength ranges from 150 to 250, but that information isn't needed for this question. If we needed to actually calculate the frequency here, we would use it, but we don't even need to go that far.

Oh, I was actually referring to the 200m wavelength of the wave when the plane is directly over the source... But I think I get it now... is the question simply asking how does the frequency of the source differ from the frequency of the observer (the plane?)

Just one question though... why don't the values of delta wavelength (50m) and source wavelength (150m) correspond to the ratio of 200m/s / 3.0x10^8 when using the modified doppler effect equation?

 

[addo]

Oh, I was actually referring to the 200m wavelength of the wave when the plane is directly over the source... But I think I get it now... is the question simply asking how does the frequency of the source differ from the frequency of the observer (the plane?)

Just one question though... why don't the values of delta wavelength (50m) and source wavelength (150m) correspond to the ratio of 200m/s / 3.0x10^8 when using the modified doppler effect equation?

v is the velocity of the observer and source, and c is the velocity of the wave. Forget about the wave increasing in wavelength from 150 to 250- it is totally irrelevant. It says the wavelength is 200 m as the jet is moving towards the dish.
As calculated above, the change in the wavelength is 1.33*10^-4, which when it is divided by 200m (original wavelength) gives also gives a change of .00006%

 

[HopefulOncoDoc]

Oh yea, i did not notice that, I put the 0s in according to the answer choice, but 6.6*10^-5 was the actual answer that I got.
Hopefuloncodoc, are you sure you didnt slip an extra 0 in the answer choice?

Hey addo sorry for the typo. You're correct I accidentally typed in an extra 0. So the correct answer is 6.6*10^-5.

 

[dingyibvs]

Hey guys I need some help with this doppler effection question. It's in reference to one of the trials in the passage..

Trial III: A jet flies in a straight line at a constant speed of 200 m/s above the radar dish as the radar dish emits a uniform EM signal with a wavelength that increases uniformly from 150 to 250m. The wavelength is 200 m as the jet passes directly over the dish.

In trial III, how does the frequency of the reflected wave from the jet differ from the frequency emitted by the dish while the jet is moving towards the dish?

A. The reflected frequency is greater than the emitted frequency by a factor of 0.00006%.
B. The reflected frequency is greater than the emitted frequency by a factor of 0.006%.
C. The reflected frequency is greater than the emitted frequency by a factor of 0.6%.
D. The reflected frequency is less than the emitted frequency by a factor of 0.006%.

I know D is out since the frequency has to increase. I'm just a little confused on how to calculate the magnitude of the increase. According to the solution, it is essentially the speed of the source divided by the speed of the wave x 100% making choice A correct. But I'm confused as to how this answer is derived from the doppler equation of Fl = (V + Vl / V - Vs)Fs

Thanks a lot everyone.

1) First, the logical answer: To solve this problem, every piece of information you gave is irrelevant except the speed of the emitter(0), the speed of the observer(for a radar, it's the same as the emitter, 0), and the speed of the plane relative to the emitter/observer(200 m/s). So if an EM wave travels to the plane at 3 x 10^8 m/s, it should reflect back at the same speed. How far would the reflected wave be after one second? 3x10^8m right? No. That's because the plane would've traveled forward by 200m, so the reflected wave will actually be only 3x10^8 - 200m away. Notice the difference is 200m, which is a factor of 200/3x10^8 smaller than the expected distance, thus the frequency will increase by the same factor and the wavelength will decrease by the same factor.

2) Now, the mathematical answer using Fl = [(V+Vl)/(V-Vs)]Fs

So, the difference between Fl and Fs would be:
Fl - Fs

And the factor by which the frequency is increased would be:
(Fl - Fs) / Fs = Fl/Fs - Fs/Fs = Fl/Fs - 1

Now, replace Fl with Fl = [(V+Vl)/(V-Vs)]Fs, and we have:
[[(V+Vl)/(V-Vs)]Fs]/Fs - 1 = (V+Vl)/(V-Vs) - 1

Since Vs = 0(emitter/observer is stationary), we have:
(V+Vl)/V - 1

And that equals to:
V/V + Vl/V - 1 = 1 + Vl/V - 1 = Vl/V

With Vl = 200 m/s and V = 3x10^8

So, the answer, Vl/V, is equal to 200/3x10^8 = 6x10^-5%

Oh, I was actually referring to the 200m wavelength of the wave when the plane is directly over the source... But I think I get it now... is the question simply asking how does the frequency of the source differ from the frequency of the observer (the plane?)

Just one question though... why don't the values of delta wavelength (50m) and source wavelength (150m) correspond to the ratio of 200m/s / 3.0x10^8 when using the modified doppler effect equation?

The source wavelength is irrelevant in this particular problem. In any case, it is stated that the source wavelength is 200m. The 150m to 250m variation is also irrelevant to this problem. It just means that in one second, the emitted wavelength is 150m, the next second it might be 151m, and so on and so forth. Since the source wavelength is irrelevant, it doesn't matter if it varies from 150m to 250m or from 1m to 32581084m, the PERCENT change remains the same.

 

[puravida85]

it is .000006%
(200/3*10^8)*100=6.6 *10^-5

on the mcat, you will not usually use the full equation. You usually use
v/c=delta F/F=delta wavelength/wavelength
So you could also use wavelength as follows:
(200m/s/(3*10^8m/s))*200m=1.33*10^-4
1.33*10^-4m/200m=6.6*10^-7*100=6.6*10^-5
same result that you got as above

Can you explain to me where you got these equations from?
I solved this using the doppler equation, v/v-vs * 100. Which is:
(3*10^8 m/s / 3*10^8 m/s - 2.0*10^2 m/s) * 100. But this is obviously hard to do without a calculator (unless I am lacking skills). So I would love to see your break down of it. Thanks.

 

[addo]

Can you explain to me where you got these equations from?
I solved this using the doppler equation, v/v-vs * 100. Which is:
(3*10^8 m/s / 3*10^8 m/s - 2.0*10^2 m/s) * 100. But this is obviously hard to do without a calculator (unless I am lacking skills). So I would love to see your break down of it. Thanks.

as long as v is significantly smaller than c, which in this case it is by 6 orders of magnitude you can use the approximation I used- subtracting 200 from 300000000 is for all practical purposes insignificant.
this is really easy to solve using scientific notation as follows:
2*10^2/3*10^8
I find it easier to bring the powers of 10 from the denominator to the numerator, just make sure to change the sign on the exponent when you do this, and add the powers
2*10^2*10^-8/3=2/3*10-6=.66*10^-6=6.6*10^-7*10^2=6.6*10^-5%
being able to simplify, and manipulate problems to make them easier is a must. This is only a recently acquired skill and it has made everything, even in my classes much easier.
This problem was further simplified, by the fact that the change in frequency was only 1 Hz, so you did not have to find the frequency of the wave (f=c/wavelength), and then divide the change in frequency by the frequency to get the fractional change, as you would have to do using wavelength.

 

[plsfoldthx]

Can you explain to me where you got these equations from?
I solved this using the doppler equation, v/v-vs * 100. Which is:
(3*10^8 m/s / 3*10^8 m/s - 2.0*10^2 m/s) * 100. But this is obviously hard to do without a calculator (unless I am lacking skills). So I would love to see your break down of it. Thanks.

The equations are from the Exam krackers physics book

 

[puravida85]

The equations are from the Exam krackers physics book

Beautiful, checked out EK and it makes sense. Maybe addo derived it on his own? Eitherway, I appreciate it addo. Although, I think your math is more confusing. I find it easy to do

2*10^2 / 3*10^8 = ~ .6*10^-6 * 100 = 6*10^-5 = .00006%

Also, EK gives this shortcut of df/f = v/c but I think it's more effective to understand the whole idea behind the doppler effect equation because it helps one understand what exactly is going on. I will def. use EK method if the numbers are just too hard to work with otherwise understanding doppler will usually get you the answer 60% of the time all the time.