Doppler Shift for Traffic Light Colors

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To determine how fast a motorist must travel for a yellow traffic light (595 nm) to appear green (550 nm) due to the Doppler shift, the correct equation is crucial. The initial attempt used an incorrect formulation, leading to an erroneous speed calculation of 2.35x10^7 m/s. Another participant suggested using the equation f_observed = f * sqrt((c + v)/(c - v)), which proved to be effective. The discussion highlighted the importance of using the right equations in physics, especially when dealing with relativistic effects. Correct application of the Doppler effect equations is essential for accurate results in such scenarios.
Cheezay
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Homework Statement


How fast would a motorist have to be traveling for a yellow (l = 595 nm) traffic light to appear green (l = 550 nm) because of the Doppler shift?


Homework Equations



v=[(c)(f0/fs)2-c] / [(f0/fs)2+1]

The Attempt at a Solution


v= Speed of motorist
c= Speed of light
f0= Observed wavelength (green, (5.5x10^-7 m)
fs= Source wavelength (yellow, (5.95x10^-7 m)

First of all, am I using the correct equation? Because I'm almost certain my work is correct, since after plugging in the above known numbers, i get 2.35x10^7 m/s for an answer, but this answer is not correct. Any help would be greatly appreciated!
 
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Cheezay said:
First of all, am I using the correct equation?
Looks OK. (Except for some reason you use f to represent wavelength. f usually stands for frequency.)

Because I'm almost certain my work is correct, since after plugging in the above known numbers, i get 2.35x10^7 m/s for an answer, but this answer is not correct.
That answer looks good to me.
 
oooh i love doppler shift =]

ive always used F(observed)= [c/(c+v)]*F(actual)
 
Thanks RoryP. I don't know what was wrong with the equation that i posted, but i tried yours out and it worked!
 
RoryP said:
ive always used F(observed)= [c/(c+v)]*F(actual)
That equation isn't quite right; it should be:

f_{obs} = f \sqrt{\frac{c + v}{c - v}}
 
No worries Cheezay, yeah I've never seen the equation you started with, but then again I've only been doing physics for 2 years now so i might bump into it soon!

Yeah i just checked my notes from 6th form and the equation i used is for c>>v, so don't konw if that makes any difference =]
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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