# Doppler Shift/ Doppler effect HELP

Doppler Shift/ Doppler effect!! HELP

1. A police car is traveling at a constant speed of 110. km/h with a radar gun which generates and detects radiowaves. The police car and the ambulance drive directly towards each other. The radar gun produces radiowaves with a wavelength of 0.0440 m.
An ambulance is traveling at 64.0 km/h towards an intersection
The ambulance has a siren which produces sound at a frequency of 4.30 kHz. Assume the speed of sound in air is 340. m/s.

(i) What is the relative speed between the police car and the ambulance?

(ii) What doppler shift (Δf = f′–f) is detected by the radar gun after the radiowave has been reflected by the ambulance?

## Homework Equations

relative speed = v1 + v2

f obs = fs ((vw + v obs) / (vw + vs))
Vw = 340 m/s (speed of sound...)
v obs = 64km/h (17.778 m/s)
vs = 110 km/h (30.556 m/s)
fs = 4.3 kHz

(Δf = f′–f)
f = 4.3 kHz

v = fλ (I got λ = 0. 044 so what am I using it for??)

## The Attempt at a Solution

A) Easy 110 +64 = 174 km/h

B) The problem child.
I plug in with the equation above, with + because the ambulance and the police are moving towards each other, but I don't get the right answer (2,200 Hz)
I know I need to find Δf, but how do I go about to find it? Why is this equation not working, and which equation am I meant to use? Whats the radar wavelength for? How do I implement it?

Any help would be much appreciated!!

## Answers and Replies

f obs = fs ((vw + v obs) / (vw + vs))
Vw = 340 m/s (speed of sound...)
v obs = 64km/h (17.778 m/s)
vs = 110 km/h (30.556 m/s)
fs = 4.3 kHz

you have taken a wrong value of Vs. Check it.

you have taken a wrong value of Vs. Check it.

What do you mean?
I don't understand.
Vs is the source, and the ambulance is the source of the siren sound.
It is travelling towards the observer which is the police car (with the radar detector of λ wavelength)....

If the ambulance isn't the source, then what is? The police car?
Then the ambulance is the observer?
I'm so confused....

The sound wave is changing frequency twice. One the frequency observed by the observor in ambulance. Another one is the that observed b ht observor in the police car(that is radar detector). So the equation

f obs = fs ((vw + v obs) / (vw + vs))

will be used twice.

NOTE: Use the values of Vs and Vobs carefully as the observor and the source is changing and keep in mind when the value will be taken negative/positive

Okay,
so I'm meant to use the equation twice.

In that case,
f obs (police) = fs (( vw + vobs) / (vw + vs ) )

I keep fs = 4.3kHz and as both sources are still going towards each other (+)
vw = 340 m/s
V obs (detector) = v = fλ = 4,300 * 0.044 = 189.2 km/h (???) (52.56 m/s)
Vs = 64 km/ h (17.778 m/s)

f (obs police) = 4.718 kHz

f obs (ambulance) = fs (( vw + vobs) / (vw + vs ) )

As above keep vw and (+)
v/λ = fs = 110 km/h * 0.044 = 1.34 kHz (???)
Swap Vobs = 64 km/h (17.778 m/s)
vs (police) = 110 km/h (30.558 m/s)

f (obs ambulance) = 1.293 kHz

Δf = 4.718 - 1.293 = 3.42 kHz ??? (not the right answer of 2,200 Hz)...

Mind pointing out what I'm getting wrong?

I give up.
Don't get it.
Good luck to the rest!