# Doppler effect car horn question

1. Jul 1, 2011

### yjk91

1. The problem statement, all variables and given/known data

A policeman with a very good ear and a good understanding of the Doppler effect stands on the shoulder of a freeway assisting a crew in a 40-mph work zone. He notices a car approaching that is honking its horn. As the car gets closer, the policeman hears the sound of the horn as a distinct B4 tone (494 Hz). The instant the car passes by, he hears the sound as a distinct A4 tone (440 Hz). He immediately jumps on his motorcycle, stops the car, and gives the motorist a speeding ticket. Explain his reasoning.

3. The attempt at a solution
f0 = f (V(sound) / (V(sound) - V(source)))
since the car is moving toward the policemen it is negative
so
494Hz = 440Hz * (343 v/s / (343 - V))
and V is 37.166 m/s...
which is obviously not..
i'm not sure where i went wrong probably doppler effect..

thank you

2. Jul 1, 2011

### Ecthelion

Well, I haven't really gone through any of the calculations in your problem... but... just off an initial glance I saw you came up with:

V = 37.166 m/s

Which is a decent answer. And you need to prove that this motorist was speeding past the limit of 40 mph... perhaps a conversion? Doing so would put the motorist's speed at ~80mph which is certainly justifiable for the policeman to pull the motorist over. Or maybe I'm misinterpreting what you're looking for here.

Last edited: Jul 1, 2011
3. Jul 1, 2011

### yjk91

i converted to mph from m/s
and i got 83 mph which still is the wrong answer
i think it's the formula

4. Jul 1, 2011

### yjk91

f(police) = v / (v + v(source)) *f(source)

440 = 494 (767 / (767 + v))

v= 94 mph

does this make sense?

5. Jul 1, 2011

### SammyS

Staff Emeritus
Actually, f(source) = 440 Hz & f(observer) = 494 Hz, so you'll get -94 mph.

6. Jul 1, 2011

### Quinzio

The correct answer is 83 mph.

7. Jul 1, 2011

### cepheid

Staff Emeritus
Yeah, I didn't look up the Doppler effect formula, but I thought about it for a few minutes and came up with the following for a stationary observer and moving source:

$$f = f_0\left[\frac{1}{1 + v/c_s}\right]$$

where cs is the sound speed, v is the source speed relative to the observer, and I use the convention that:
v > 0 if the source moves away from the observer and,
v < 0 if the source moves toward the observer.

It makes sense at least to the extent that f > f0 for v < 0 and vice versa.

Using this formula, with a sound speed of 343.2 m/s (taken from Wikipedia for dry air at 20 C) I get a result of:

v = -83.9 mi/h

Last edited: Jul 1, 2011