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Homework Help: Doppler effect

  1. Apr 12, 2005 #1
    A stationary light source S wit ha natural frequency Fo is viewed in a mirror M by a stationary observer O. The mirror moves away from the observer wit ha velocty of Vrel << c

    a) what frequency of light is recorded by a detector attached to the moving mirror

    because Vrel << c classical may be used
    let F1 be this observed frequency observed
    then [tex] f_{1} = f_{0} (1 - \frac{v_{rel}}{c}) [/tex]
    is this correct??

    b) what frequency in terms of fo will the stationary observer measure for the light reflected off the mirror?
    the mirror will now emit the f1 from above wouldn't it ??
    sine this mirror is moving away wouldnt the doppler shift be [tex] f_{2} = f_{1} \frac{c}{c+v_{rel}} [/tex]
    which would be [tex] f_{2} = f_{0} (1 - \frac{v_{rel}}{c}) \frac{c}{c+v_{rel}} = f_{0} \frac{c-v}{c+v} [/tex]

    but i got the second part wrong! Whats wrong with it??

    Also when asked for the lowest average speed of atom at some temperature T given some molar mass M
    which formula should be used??
    is it [tex] v_{rms} = \sqrt{\frac{3RT}{M}} [/tex] or [tex]v_{avg} = \sqrt{\frac{8RT}{\pi M}} [/tex]
    Last edited: Apr 12, 2005
  2. jcsd
  3. Apr 12, 2005 #2
    Surely if v_rel << c, the second term in that bracket would go to 0?
  4. Apr 12, 2005 #3
    are you talking about part a) or b)? I did get the first part correct by the way
  5. Apr 12, 2005 #4
    Either! Assuming v_rel is << c in both cases..
  6. Apr 12, 2005 #5

    Can you find a polynomial expression that approximates (c-v)/(c+v) when v<<c?
  7. Apr 12, 2005 #6
    k first of all the first one isnt wrong because i wasnt marked wrong thae fact that v<<c doesnt mean that the result in null so get off that!
    Of course my approximation is lousy but im trying to answer my prof's question properly according to him, at least

    [tex] \frac{c-v}{c+v} = \frac{1-\frac{v}{c}}{1+\frac{v}{c}} = \frac{1-\beta}{1+\beta} = 1 - \beta + \frac{\beta^2}{2} + ... [/tex]

    something like that? Doesnt that give the same answer as a) though??
    Last edited: Apr 12, 2005
  8. Apr 12, 2005 #7

    Try this. Divide the numerator and denominator of (c-v)/(c+v) by c. Then define x = v/c. The new denominator will be 1+x. Can you find a power series for 1/(1+x)?
  9. Apr 12, 2005 #8

    I can't keep up with you!

    You're very close to the right answer, but you're guessing on the power series. Figure it out. Write 1/(1+x) as (1+x)^-1, then it's easy to see all the derivatives: -(1+x)^-2, 2(1+x)^-3....
  10. Apr 12, 2005 #9
    what you're saying is put v/c = x whihc would give

    [tex] \frac{1 + x}{1 - x} [/tex] whiuch is clearly not [tex] \frac{1}{1-x} [/tex] and thus u cannt ot expand it out like the latter
  11. Apr 12, 2005 #10
    and power series for
    [tex] \frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + ... [/tex] for abs (x) < 1 where abs means absolute value
  12. Apr 12, 2005 #11

    But (1-x)/(1+x) = (1-x)*1/(1+x). And 1-x is already a power series. So just get the series for 1/1+x and you'll see the answer.
  13. Apr 12, 2005 #12
    are you sure that can be done??
  14. Apr 12, 2005 #13

    A power series for 1/(1+x)? Why not?

    y = (1+x)^-1 >> y(0) = 1
    y' = -(1+x)^-2 >> y'(0) =-1
    y'' = 2(1+x)^-3 >> y''(0) =2


    So, y = 1/(1+x) = 1 - x + x^2....

    So, what's 1/(1+x) when x << 1?
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