# Doppler Effect Formula Question

1. Jun 12, 2013

### tompalatine

I am a high school teacher and I just came across this issue when I began teaching at a new school. I am positive that if I am stationary and a 500 Hz siren comes towards me at 30 m/s, then the observed frequency will increase. I am 99% positive that if the 500 Hz siren is stationary and I move towards it at 30 m/s, the observed frequency will be the exact same.

At my new school, we use the formula:

Frequency observed = frequency [343 / (343 +/- v)]
Obviously the 'v' was the relative velocity of the two objects and adding/subtracting depended on their motion. This made 1st paragraph I described above work out correctly. Whether I move towards a siren or the siren moved towards me, then the observed frequency was the same.

At my old school, we used the formula seen on Wikipedia:

Frequency observed = frequency [(343+/-Vr) / (343 +/- Vs)]

This is what confused me. Using the example above, if a 500 Hz siren comes towards (a stationary) me you would get:

Frequency observed = 500 [(343 + 0) / (343 - 30)] = 547.92 Hz

If I move at the same speed towards a stationary 500 Hz siren, you get:

Frequency observed = 500 [(343 + 30) / (343 + 0)] = 543.73 Hz

How can those be different? The relative velocity between the two is the same. Anyone know what I am doing wrong? Which way is the correct way?

2. Jun 12, 2013

### Bobbywhy

3. Jun 12, 2013

### nasu

For waves propagating in a medium, the two values are not supposed to be the same.
The medium breaks the symmetry. The relative speed source-observer is the same but the speed of the medium relative to the observer is not. So the speed of the observer relative to the medium is different in the two cases.

4. Jun 12, 2013

### tompalatine

So when I assumed that a siren moving at 30 m/s towards a stationary listener would create the same observed frequency as the listener moving at 30 m/s moving towards a stationary siren....I was wrong?

5. Jun 13, 2013

### nasu

I am afraid so.

The formula used at your new school is a special case of the general one, for stationary observer.
Most of the examples on the web are for this special case.

The "old school" formula is more general. I am surprised that you did not realize the asymmetry between source and observer, if you say you used this general formula for some time.

6. Jun 15, 2013

### tompalatine

Can someone explain why there is an asymmetry in a little bit more detail?