# Doppler's effect

1. Aug 18, 2010

### mizzy

1. The problem statement, all variables and given/known data
A radar trap operates at a frequency of f = 10GHz. The speed of microwaves is 3.0x10^8m/s. The microwave beam is aimed at a car approaching at 20m/s.
a) calculate difference between observed frequency and original frequency.
b) the car becomes the moving source at frequency, fo. calculate the frequency difference between the reflected signal and fs.

2. Relevant equations

for part a, we have a fixed source and a moving observer. the equation for this is fo = fs(1 + vo/v)

for part b, the equation is fo = fs (v/v-vs), but i'm not sure what's v and vs. is v the speed of the radar? or the car?

3. The attempt at a solution

for part b, i need to look for the observed frequency of the car(moving source).

can someone clarify the speeds please?

2. Aug 18, 2010

### vela

Staff Emeritus
This is the kind of thing you can find in your textbook where they derive the formula. What does the book say?

3. Aug 19, 2010

### mizzy

usually the v is speed of sound, but in this case the medium is the microwaves (3.0x10^8m/s)

for the first part i calculated it using fo = fs (1+vo/v) where fs is the frequency emitted by the radar, vo is the car's speed and v is 3.0x10^8 and i get an answer of 10.00000067GHz.

now it asks for the difference between observed and original. I took my answer - the original frequency of the radar and i get 6.70x10^-7.

is that correct??

it's part b that i'm a bit confused since it's talking about the reflected signal. =(
can someone guide me with this please?

4. Aug 19, 2010

### vela

Staff Emeritus
Since you're dealing with light, you should be using the formula for the relativistic Doppler effect:

$$f_o = f_s \sqrt{\frac{1+v/c}{1-v/c}}$$

where fo is the observed frequency, fs is the source frequency, v is the velocity of the source, and c is the speed of light. At low speeds, like in your problem, you can use the approximate formula

$$\frac{f_o-f_s}{f_s} \cong \frac{v}{c}$$

It turns out if you solve for fo, you get the same equation you used in part (a), so your answer for that part is correct.

When the car reflects the microwaves, it becomes the source of microwaves and the radar detector is now the observer, so you essentially do the same calculation using the frequency you found in part (a) as the source frequency to find the frequency the radar detector sees. To get the final answer to part (b), you want to find the difference between the frequency that radar detector emitted and the frequency it sees.

5. Aug 20, 2010

### mizzy

There's a third part to this question: When the reflected signal is combined with the original signal, a "beat" frequency is produced. What is its frequency?

What I did was to take my answer (observed frequency) from part b - my answer from part a. Beat frequency is the difference between two sources.

Is that right?

6. Aug 20, 2010

### vela

Staff Emeritus
Right idea, but you used the wrong frequency. The original signal has the given frequency, not the frequency you found in part (a).

7. Aug 20, 2010

### mizzy

k. thanks

8. Aug 22, 2010

### mizzy

I redid my work and i think i did something wrong. I get the same answer for both a and b?????

For b, i used fo = fs (1 + v0/v) where, fo is the frequency radar sees, fs is car's frequency, v0 is the car and v is speed of light. when i calculate this i get the same answer as a.

9. Aug 22, 2010

### vela

Staff Emeritus
What value did you use for fs in part (b)?

10. Aug 22, 2010

### mizzy

10.00000067 (the answer from part a)

11. Aug 22, 2010

### vela

Staff Emeritus
OK, you're probably just running into a problem where your calculator just doesn't have the necessary precision. It's a common problem when you have small changes like this. To get around it, you have to do the calculation differently. You actually want just the shift in frequency, so with a little algebra, you get:

$$\Delta f = f_o-f_s = f_s\frac{v_o}{v}$$

12. Aug 22, 2010

### mizzy

ok wait...part a asks for the difference. That's not the answer i use, right? because i used the observed frequency calculated before that