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Relativistic Doppler effect 
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#1
Aug2310, 05:45 PM

PF Gold
P: 3,207

1. The problem statement, all variables and given/known data
A car is getting closer to a radar as a speed of 135 km/h. If the radar works at a [tex]2 \times 10 ^{9} Hz[/tex], what difference of frequency is observed for the radar? 2. Relevant equations [tex]\mu=\frac{\mu _0 \sqrt {1 \frac{v^2}{c^2}}}{1\frac{v}{c}}[/tex] 3. The attempt at a solution I converted 135 km/h into m/s, which gave me [tex]\frac { 1350 m }{36 s}[/tex]. Then I applied the formula and [tex]\mu  \mu _0[/tex] gave me [tex]1.024 \times 10^{13} Hz[/tex] which seems WAY too big to be realistic. Am I missing something? Is it a wrong formula?! 


#2
Aug2310, 10:13 PM

HW Helper
PF Gold
P: 3,440

The formula is correct and I agree the number is way too large. Can you show how you got it? A word of caution: v/c is too small to plug in a calculator and expect something other than μ_{0}, i.e. no frequency shift. I suggest that you try a Taylor expansion for small values of v/c.



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