How do I determine the speed of the sound source using the Doppler Effect?

The question is word-for-word:
"Standing at a crosswalk, you hear a frequency of 560 HZ from the siren of an approaching ambulance. After the ambulance passes, the observed frequency is 480 HZ. Determine the ambulance's speed from these observations."

For when the ambulance is approaching I'm using the equation F1=(V/V-Vs)F.
F1 = the observed frequency in Hertz
V= the speed of sound in meters per second (in air at 70 degrees Fahrenheit it is 343 m/s)
Vs= the speed of the source in meters per second
F= the actual frequency the actual frequency in Hertz
For when the ambulance is moving away from the observer I'm using the equation F1=(V/V+Vs)F.

The actual frequency (F) of the sound is not given.
The correct answer is 26.4 m/s, but I have no idea how to get it without the sound's actual frequency. Obviously I can calculate the actual frequency from the equations since I know the answer to the question, but I need to know how to actually do the problem. : (
Can I somehow set the two equations equal to one another or something?

Anonymous2012-11-04T08:20:58Z

Favorite Answer

When ambulance is approaching you...
F=Fo [(Vsound- Vobserver)/(Vsound-Vsource)]

560 = Fo[ 343 - 0 ] / [343 - V]
Fo = 560 / 343 * (343 - V) ------------1

WHen ambulance passes you... the formula changes...
F=Fo [(Vsound- Vobserver)/(Vsound+Vsource)]
480 = Fo[343 - 0 ] / [ 343 + V]
Fo = 480 / 343 * (343 + V) ------------- 2

Both equations have LHS equal...1 & 2
Now comes the tricky part...

560 / 343 * (343 - V) = 480 / 343 * (343 + V)
560 / 480 (343 - V) = 343 + V
400.166667-1.1666667V = 343 + V
400.166667 - 343 = V + 1.1666667V
V = 57.16667 / (1 + 1.16667) = 26.384 = 26.4m/s..

PS...Thanks to the formulas you gave and some logic I was able to do this and I know nothing about this doppler effect :]

Greywolf2012-11-04T16:33:59Z

Actually, you do know the real frequency. It is assumed that the ambulance is moving at the same speed before and after the observer, so the frequency shift is the same (just in opposite ways). Thus the actual frequency is the midpoint or mean average of the two observed frequencies, or 520 Hz.