Doppler-shifted frequency problem

[hitman0097]

Homework Statement

A radar "speed gun" emits microwaves of frequency f_o=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 f_o and its detected wave at f'. These waves interfere, creating a beat pattern whose beat frequency is f_beat=f'-f_o.

a.) Show that: v≈cf_beat/2f_o,

if f_beat <<f_o. If f_beat=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 Δf_beat is given by : Δf_beat=2f_oΔv / c

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

Homework Equations

f=c/λ = f_o[(c+v/c-v)^1/2]
λ=λ_o[(c-v/c+v)^1/2]
1/f_o=λ_o/c

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

 

[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]

 

[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

 

[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.

https://www.physicsforums.com/showthread.php?t=223186
... subs in water - sonar doppler-shift illustrating the double-shift.

https://www.physicsforums.com/archive/index.php/t-335219.html
... for radar gun!

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.

 

[asteriatic]

for reaching out. It sounds like your teacher may have been referring to the Doppler effect, which is the change in frequency of a wave due to the relative motion between the source of the wave and the observer. In this problem, the radar "speed gun" is the source of the microwaves and the moving object is the observer.

To find the speed of the object, we can use the equation v=λf, where v is the speed of the object, λ is the wavelength of the waves, and f is the frequency of the waves. We can also use the Doppler effect equation, f'=f(v/c+v)/(c-v), where f' is the frequency detected by the radar apparatus, f is the emitted frequency, v is the speed of the object, and c is the speed of light.

For part a), we are given the beat frequency, fbeat, and we want to find the speed of the object, v. We can rearrange the Doppler effect equation to solve for v, giving us v=c(fbeat/2fo). Since we are given the beat frequency in Hz, we need to convert it to m/s by multiplying by the speed of light, c=3.0x10^8 m/s. We also need to convert the emitted frequency, fo, from GHz to Hz by multiplying by 10^9. Plugging in these values, we get v≈cfbeat/2fo≈3.0x10^8 m/s x 6670 Hz/(2x36.0x10^9 Hz)≈34.8 m/s.

To convert this to km/h, we can multiply by 3.6, giving us v≈125 km/h.

For part b), we need to find the difference in beat frequency, Δfbeat, when the speed of the object changes by Δv. We can use the same equation for v as before, but now we have two different speeds, v and v+Δv. Plugging these values into the equation, we get Δfbeat=2fo((v+Δv)/c)/(c-(v+Δv)) - 2fo(v/c)/(c-v). Simplifying this expression, we get Δfbeat=2foΔv/c(c-v). Again, we need to convert the emitted frequency, fo, from GHz to Hz by multiplying by 10^9. Pl