Doppler shift with microwaves i think

[Dorbo]

Doppler shift with microwaves...i think...

Homework Statement

Microwaves, which travel at the speed of light, are reflected from a distant airplane approaching the microwave source. It is found that when the reflected waves are beat against the waves radiating from the source, the beat frequency is given by 969 Hz. If the source microwaves are 150 mm in wavelength, what is the approach speed (in km/h) of the airplane?

Homework Equations

I think that the 3 relevant equations are C=\lambda \nu
and dopplers equation

F= \left(v+v_{r}\right)/\left(v+v_{s}\right)*F_{0}

F_{beat} = \left|F_1-F_2\right|

The Attempt at a Solution

My attempt at the solution is to take the speed of light, divide by .150 m, to get the frequency of my microwave. This value gives me \nu =1.998*10^9.
I then put it into the beat formula and solve for F_1= 1.997*10^9.
I would then sub into my dopplers formula, except i don't know if the source is moving towards it or if it is moving away. I am also not sure if i did any of this right because the microwave frequency is so high, that the beats have virtually no affect. Any insight would be appreciated, thank you in advance.

 

[rl.bhat]

Hi Dorbo, welcome to PH.
In the case of light, the Doppler effect is given by
Δf/f = v*cosθ/C.

 

[Dorbo]

Hi Dorbo, welcome to PH.
In the case of light, the Doppler effect is given by
Δf/f = v*cosθ/C.

I am sorry, what is \Delta f in your equation? And how do beats fit into it? From what i understand \theta = \pi because the waves are going in opposite directions. I do understand what C is though :p.

 

[rl.bhat]

I am sorry, what is \Delta f in your equation? And how do beats fit into it? From what i understand \theta = \pi because the waves are going in opposite directions. I do understand what C is though :p.

Δf is the beat frequency.
The airplane is approaching. So the source appears to be coming towards the observer. Hence θ = 0.

 

[Dorbo]

Δf is the beat frequency.
The airplane is approaching. So the source appears to be coming towards the observer. Hence θ = 0.

Ok i put in \frac{969*C}{2*10^9}, i then got a value of 145.35 m/s, i then converted it to 523.26 km/h. I then submitted this answer and got it wrong. Any ideas, and thank you for the help so far. I got the 2*10^9 by c= \lambda \nu

 

[rl.bhat]

Δf/f = v*cosθ/C.
This formula is for a source approaching the stationary observer.
In this case, when the reflector moves x meter per second with respect to the source, the image of the source in the reflector approaches the source with 2x meter per second. So try the answer 2*523.26 km/h.