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Red light to green light homework

  1. Mar 3, 2005 #1
    What speed would a vehical have to travel in order to red traffic light into a green traffic light?
    i know that green light has wavelength of 5.40x10^-7 m
    frequency of 5.56x10^14 hz

    red light has wavelength of 6.60x10^-7 m
    frequency of 4.54x10^14 hz

    so do i use velocity = wavelenght x freq

    for the 2 and then subtract the small from the large?
     
  2. jcsd
  3. Mar 3, 2005 #2

    Galileo

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    No, the speed of light is always c, it doesn't depend on the frequency.
    The only way to 'change' the frequency/wavelength is to approach or move away from the source.
    Use the doppler shift formula and plug in the numbers. Take good note of what the numbers in the formula mean, so you won't confuse signs.
     
  4. Mar 3, 2005 #3
    ok this is what i did.,

    5.56 x 10^14 = (3x10^8 / 3x10^8 x v) 4.54 x 10^14

    5.56 x 10^14 = (v) 4.54 x 10^14

    so v = 1.2ms?
     
  5. Mar 3, 2005 #4

    dextercioby

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    Nope.U need to open the book first and read the correct version of Doppler-Fizeau formulas...I think they mean the longitudinal effect...

    Daniel.
     
  6. Mar 4, 2005 #5
    is there one of you guys have grifftihs electrodynamics solution?:)please snd to my email
     
  7. Mar 4, 2005 #6

    dextercioby

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    You can search the net for solutions,but i would't trust them,if i were you...

    Daniel.
     
  8. Mar 7, 2005 #7
    Ok where could I find or could anybody give me the formula for the doppler shift when travelling at speeds close to the speed of light.
    So I need to take relativity into account.

    I needs this as soon as possible, thanks.
     
  9. Mar 7, 2005 #8

    Galileo

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    The formula you should use is not in your book?

    [tex]\frac{\lambda '}{\lambda}=\sqrt{\frac{c-v}{c+v}}[/tex]
    where [itex]\lambda '[/itex] is the Doppler-shifted wavelength and [itex]v[/itex] is the relative velocity between source and observer (positive when approaching).

    For v<<c, the approximation
    [tex]\frac{\lambda'}{\lambda}=1-\frac{v}{c}[/tex]
    will do.
     
  10. Mar 8, 2005 #9
    I havent a clue how to use this formula Galileo, because I am looking for v, and there are 2 v's in the equation.
    Would you be able to explain how I do it, or do an eg. using different values.
    I would really appreciate it. Thanks.
     
  11. Mar 8, 2005 #10

    Galileo

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    You'd have to solve the equation for v.

    Show us how far you got. Obviously, you first have to square both sides of the equation to get rid of the square root. Then bring the numerator to the other side, and collect terms with v in it.
     
  12. Mar 8, 2005 #11
    I just dont know where to start. I ve never seen this formula before.
     
  13. Mar 11, 2005 #12
    How do I calculate v [the velocity of the car needed to turn red light to green light] from this equation ?? Please can somebody show me how?
     
  14. Mar 14, 2005 #13

    Galileo

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    Elementary algebra:

    [tex]\frac{\lambda '}{\lambda}=\sqrt{\frac{c-v}{c+v}} \Rightarrow
    \frac{\lambda'^2}{\lambda^2}(c+v)=(c-v)
    \Rightarrow \frac{\lambda'^2}{\lambda^2}c+\frac{\lambda'^2}{\lambda^2}v+v=c[/tex]

    [tex]\Rightarrow (1+\frac{\lambda'^2}{\lambda^2})v=c(1-\frac{\lambda'^2}{\lambda^2}) \Rightarrow v=\frac{c(1-\frac{\lambda'^2}{\lambda^2})}{(1+\frac{\lambda'^2}{\lambda^2})}=\frac{c(\lambda^2-\lambda'^2)}{(\lambda^2+\lambda'^2)}[/tex]
     
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