# Doppler-shifted frequency problem

1. Feb 1, 2012

### hitman0097

1. The problem statement, all variables and given/known data
A radar "speed gun" emits microwaves of frequency fo=36.0 GHz. When the gun is pointed at an object moving toward it at speed v, the object senses the microwaves at the Doppler-shifted frequency f. The moving object reflects these microwaves at this same frequency f. The stationary radar apparatus detects these reflected waves at a Doppler-shifted f'. The gun combines its emitted wave at fo and its detected wave at f'. These waves interfere, creating a beat pattern whose beat frequency is fbeat=f'-fo.

a.) Show that: v≈cfbeat/2fo,

if fbeat <<fo. If fbeat=6670 Hz, what is v(km/h)?

b.) If the object's speed is different by v, show that the difference in in beat frequency Δfbeat is given by : Δfbeat=2foΔv / c

c.) If the accuracy of the speed gun is to be 1 km/h, to what accuracy must the beat frequency be measured?

2. Relevant equations
f=c/λ = fo[(c+v/c-v)^1/2]
λ=λo[(c-v/c+v)^1/2]
1/foo/c

3. The attempt at a solution
I am kind of lost on this problem, I think I heard my teacher say I'd need a double doppler shift. Any help and hints are appreciated!

Thanks

Last edited: Feb 1, 2012
2. Feb 2, 2012

### hitman0097

The first Doppler shift that occurs should be in the form
[ f=36.0 G hz [(c+v)/(c-v)]^.5]

The 2nd Doppler shift would be in the form of the first Doppler shift
[ f'=(f [(c+v/c-v)]^.5)] = 36 Ghz [(c+v)/(c-v)]

fbeat can be related in terms of f not [fbeat = 36.0 Ghz [(c+v)/(c-v)] - 36.0 Ghz]

3. Feb 2, 2012

### hitman0097

Still confused to as v= c (fbeat)/ 2 fo If: fbeat<<fo. Also, I think the last fbeat is not correctly expressed or related by me. Also getting getting 27.34 m/s for part (a). Where the answer in the back of my book is 1.00 x 10^2 km/hr

4. Feb 2, 2012

### Simon Bridge

There is no difference between the moving-source and moving-observer doppler-shift expressions for light. For waves that need a medium, the wave-speed gets measured wrt the medium which is why there is a difference there.

The usual idea is that a moving car reflects the waves as it receives them - so the reflected wavelength is shifted wrt the transmitted one.
The vehicle then acts as a source for the reflected wave.