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A Doppler weather radar operates at a frequency of 3.40 GHz. The wave from this radar system reflects from an approaching weather system moving with a speed of 39.0 m/s. What is the difference in frequency between the outgoing and returning waves? Ans: 884 Hz

How do I compute this? I did

$$f = f_s (1 \pm u/c)$$

$$f = 3.4\times 10^9 (1 + \frac{39}{3 \times 10^8}) = 442 Hz$$

I suppose thats the incoming frequency? How do I know the outgoing frequency difference? Looks like its just times 2? Why?

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up vote 1 down vote accepted

The 'double doppler shift' is a subtle point but fairly obvious when you look at it clearly. The above equation describes the observed doppler shift by some other body moving relative to you. So you have only calculated the shift as 'seen' by the cloud.

The cloud will then return the waves to wards you at an identical relative velocity so that you see another doppler shift. Relative to the cloud it is stationary and you are moving towards it.

So you need to do it twice and be careful with signs...

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Hmm, at first I was wondering if its possible for the cloud originally moving toward to pass and move away making the shifts 0? Then I suppose that doesn't happen if I assume this is all happening very fast and if at the instant where the wave hit the cloud, its moving towards then as it bounces (same instance) its still moving towards? – Jiew Meng Mar 9 '13 at 1:46
consider the speed of the em wave vs the speed of the cloud... You essentially have it correct. – Nic Mar 11 '13 at 10:41

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