Doppler Shift/ Doppler effect HELP

  • Thread starter Roaku
  • Start date
  • #1
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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

  • #2
387
8
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.
 
  • #3
5
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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....
 
  • #4
387
8
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
 
  • #5
5
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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?
 
  • #6
5
0
I give up.
Don't get it.
Good luck to the rest!
 

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