# Doppler effect ambulance problem

1. Aug 5, 2006

### mikejones2000

Question: An ambulance with a siren emitting a whine at 1300 Hz overtakes and passes a cyclist pedaling a bike at 2.44 m/s. After being passed, the cyclist hears a frequency of 1280 Hz. Use 343 m/s for the speed of sound in air. How fast is the ambulance moving?

i set up the doppler equation with as follows:

1300 hz =1280 hz((343+(2.44/343))/(343+(vs/343))), I am pretty confidant the signs I choose are correct but must be making a mistake in either interpreting the data or algebra, any help is greatly appreciated..

2. Aug 5, 2006

### Mattara

If you are not sure about the positive direction, I can tell you that the positive direction is aimed towards eachother relative to the air.

3. Aug 6, 2006

### sdekivit

when both the source and the cyclist move (in this case), then the Dopplereffect is given by:

$$f_{cyclist} = f_{ambulance}\frac {(v + v_{cyclist})} {(v + v_{ambulance})}$$

but you can also say that the cyclist stands still and the ambulance passes with a relative velocity $$v_{rel}$$. Then use:

$$f_{cyclist} = f_{ambulance}\frac {v} {(v + v_{rel,amb})}$$ and then add the velocity of the cyclist.

Last edited: Aug 6, 2006