# 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.

... subs in water - sonar doppler-shift illustrating the double-shift.

https://www.physicsforums.com/archive/index.php/t-335219.html

Not following your reasoning - what is the relationship between the beat frequency and the doppler shifted frequency?
We need to see how you are getting these results in order to help you properly.

Last edited: Feb 2, 2012