How can the Doppler effect be used as a motion detector?

[huffy]

Homework Statement

An audio transmitter and receiver are mounted side by side as shown in the figure. The transmitter emits sound of frequency v. A distant flat plate approaches the instrument at speed V_target and the receiver detects sound waves of frequency v' reflected from the target. The difference in frequency between the emitted and detected frequencies can be used as a motion detector.

A) Show that v'/v=v+V_target / v-V_target

B) In many cases V[/SUB]target[/SUB] <<v . For such cases show that V'/v=1+2(v_target /v)

FIGURE: http://imgur.com/bLXnm2L

Homework Equations

Doppler effect

The Attempt at a Solution

Ive already accomplished A by realizing there were two doppler shifts and that the frequency gets larger as it goes to the object, than even larger than that when it bounces back to the receiver. Part B is where I'm stuck because I am not sure what i need to assume to create that formula.

 

[haruspex]

Part B is where I'm stuck because I am not sure what i need to assume to create that formula.

Do you know how to expand (1+x)^a as a power series in x?

 

[huffy]

No, i haven't learned about power series yet, how does that relate to this?

 

[haruspex]

No, i haven't learned about power series yet, how does that relate to this?

For this question, you need to know that, for small x, (1+x)^a can be approximated as 1+ax+O(x²), where O(x²) means terms that grow smaller "like x²" as x gets smaller. I.e. you can just approximate here as 1+ax.

 

[huffy]

is x just a term for frequency in this situation? I am not sure I am following your variables.

 

[haruspex]

is x just a term for frequency in this situation? I am not sure I am following your variables.

x is anything at all, as long as it is small compared with 1.
Your initial post is very confusing because you seem to have written v for frequency and for velocity of sound in air. Maybe it's just a confusing font. Allow me to rewrite it more clearly:

The transmitter emits sound of frequency $\nu$

A) Show that $\frac{\nu'}{\nu}=\frac{V_{sound}+V_{target}}{V_{sound}-V_{target}}$

B) In many cases V_target<<V_sound

You are given V_target<<V_sound. So what quantity << 1?

 

[huffy]

so your variable x would be less than 1, or in the re formatted variable, Vsound would be <<1

 

[haruspex]

your variable x would be less than 1

Yes.

Vsound would be <<1

No, V_target<<V_sound. From that, find something that must therefore be <<1.

 

[huffy]

Perhaps the frequency would be less than 1. Would i be correct in assuming that?

 

[haruspex]

Perhaps the frequency would be less than 1. Would i be correct in assuming that?

No.

 

[huffy]

OH. The Vtarget has to be <<1

 

[haruspex]

OH. The Vtarget has to be <<1

No.
Forget the actual questionfor a moment. If you have two quantities y and z, and you are told y<<z, what quantity can you construct from y and z which is therefore << 1? (You understand what << means, right?)

 

[huffy]

<< means significantly less than. So id have to make a third quantity that is z-y?

 

[haruspex]

<< means significantly less than

Yes.

So id have to make a third quantity that is z-y

No. E.g. y=1 and z=100 would satisfy y<<z, but z-y would be 99. And I should have said that you want the absolute magnitude to be <<1, so swapping z and y won't do it either.
Think about ratios.

 

[huffy]

So what if we made a quantity that was 1/y. that would guarantee y<<Z if we just want it to be <<1

 

[huffy]

So what if we made a quantity that was 1/y. that would guarantee y<<Z if we just want it to be <<1

whoops, that wouldn't work either because if y was 1/3 than it would be 1/(1/3) which is >>1. ill keep thinking.

 

[huffy]

nvm I am confused again