# Doppler Effect and the speed of sound

1. Apr 12, 2015

### Potatochip911

1. The problem statement, all variables and given/known data
You are moving at a speed of 35m/s and hear a siren coming from behind you and observe the frequency to be 1370 Hz. The siren goes past you and the new frequency heard is 1330 Hz. What is the speed of the siren traveling at? The speed of sound in air is 340m/s.
f1=1370 Hz
f2= 1330 Hz
D=35m/s
V=340m/s
2. Relevant equations
fl=fo*((V+D)/(V+S))

3. The attempt at a solution
Since the original frequency is not given I decided to make 2 equations with the different frequencies then just divide to eliminate the original frequency.
f1=fo*((V+D)/(V-S)) <-Since the siren is approaching
f1*(V-S)=fo*(V+D)
f2=fo*((V-D)/(V+S)) <-Since the siren is moving away
f2*(V+S)=fo*(V-D)
Dividing the 2nd equation by the first,
[f2*(V+S)]/[f1*(V-S)]=(V-D)/(V+D)
Letting C=f2/f1
C(V+S)(V+D)=(V-D)(V-S)
C(V2+SV+DV+SD)=(V2-DV-SV+SD)
CV2+CSV+CDV+CSD=V2-DV-SV+SD
CV2+CDV+DV-V2=-SV+SD-CSD-CSV
CV(V+D)+V(D-V)=S(-V+D-CD-CV)
S={V[C(V+D)+(D-V)]/[V(-1-C)+D(1-C)]}
Which gives me S=-40m/s which doesn't make any sense, I'm not sure where my mistake is but the correct answer is +40m/s.

2. Apr 12, 2015

### Staff: Mentor

Check the signs in the numerators of the equations for fi.

When the siren is approaching, you are moving away from the siren (at least relative to the air).
When the siren is moving away, you are approaching the siren (at least relative to the air).

3. Apr 12, 2015

### Potatochip911

So in f1 are both of the signs negative and in f2 both signs are positive?

4. Apr 12, 2015

### Staff: Mentor

Right.
In both cases, the velocity of the ambulance and your velocity have opposite effects, in one case you have larger numerator and denominator and in the other case both get reduced. Which one is which depends on your choice of the coordinate system.