How Does the Doppler Effect Relate Detector Speed to Sound Speed?

[Doomezar]

Homework Statement

A detector initially moves at a constant velocity directly toward a stationary sound source and then (after passing it) directly from it. The emitted frequency is f. During the approach the detected frequency is f'app and during the recession it is f'rec. If the frequencies are related by (f'app-f'rec)/f=0.500, what is the ratio v(d)/v of the speed of the detector to the speed of sound?

Homework Equations

(f'app-f'rec)/f=0.500

f=frequency emmitted
f'app=f * [v+v(d)]/v
f'rec=f * [v-v(d)]/v

The Attempt at a Solution

[( f * [v+v(d)]/v)-(f * [v+v(d)]/v)]/f=0.500

f * ([v+v(d)]/v-[v+v(d)]/v)/f=0.500

[v+v(d)-v-v(d)]/v=0.500

I come up with 1/v=0.500. I guess that would be undefined but it doesn't seem right. I'm not sure where I went wrong here and I've tried a lot of other weird math techniques to not let v(d) cancel but I can't get it to work out. Any suggestions?

 

[Simon Bridge]

Homework Statement

[( f * [v+v(d)]/v)-(f * [v+v(d)]/v)]/f=0.500

Have you missed out a minus sign in the second term there?

Note: v+v(d)-v-v(d) = 0, not 1.