GR8677 #12



Alternate Solutions 
flyboy621 20101109 21:01:01  I find it easier to look at the time between successive wave crests. Let's say if the source were at rest, the time between crests would be . Then if the source is moving toward the observer at , the time between crests is . Therefore the frequency is 10 times higher, as in (E).   kicksp 20071029 10:11:11  Quick handwaving arguments save us time here. Note that the observed frequency goes to infinity as . A simple functional form for (which we could construct immediately even if we forgot the formula) is
,
yielding (E).
Go maroons!  

Comments 
flyboy621 20101109 21:01:01  I find it easier to look at the time between successive wave crests. Let's say if the source were at rest, the time between crests would be . Then if the source is moving toward the observer at , the time between crests is . Therefore the frequency is 10 times higher, as in (E).   student2008 20081011 12:48:31  ...the frequency of the wave is ....   student2008 20081011 12:44:23  The observable phase of the wave front at some point is , where and are the circular frequency and the wave vector measured in the observational reference frame. The source is moving with the constant velocity , so . Thus, . It means that in the reference frame where the source is at rest the frequency of the wave is . Since , we get eventually: .   ssp 20080905 02:46:12  Bit more handwaving...
So we take the fact that we know the frequency has to increase from above. Now we look at the speed. The shift has to large because the speed is large (speed and frequency are still proportional) and you got your answer, (E).
Little trick from GR   kicksp 20071029 10:11:11  Quick handwaving arguments save us time here. Note that the observed frequency goes to infinity as . A simple functional form for (which we could construct immediately even if we forgot the formula) is
,
yielding (E).
Go maroons!
vikhuan 20090608 02:04:49 
tyjyyty

 




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