Why Is Jim Hearing a Different Frequency from the Ambulance Siren?

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Homework Help Overview

The problem involves an ambulance traveling towards an observer, Jim, and the frequency of the sound from the ambulance's siren as perceived by Jim. The scenario includes a specific angle and requires the application of the Doppler effect to determine the frequency shift.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to convert the ambulance's speed from km/h to m/s and question the components of the ambulance's velocity in relation to the angle provided. There is an exploration of how to apply the Doppler shift formula given the angle of approach.

Discussion Status

Participants are actively engaging with the problem, questioning the assumptions about the direction of the ambulance's motion and how to correctly apply the Doppler effect formula. Some guidance has been offered regarding the need to consider the components of velocity in the direction of the observer.

Contextual Notes

There is a noted complexity due to the angle involved, which is not typically covered in standard examples. This has led to confusion regarding the correct application of the Doppler effect in this context.

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Homework Statement



An ambulance is traveling at 72.0 km/h towards an intersection. Jim stops his car to give way to the ambulance as shown in the diagram below. In the diagram, the angle θ = 41.0 °.
The ambulance has a siren which produces sound at a frequency of 2.1 kHz. Assume the speed of sound in air is 335.0 m/s.

[PLAIN]http://img693.imageshack.us/img693/7959/amburightjimrest.gif

What is the doppler shift (Δf = f ′ – f ) heard in the siren sound by Jim, when Jim and the ambulance are positioned as shown on the diagram above?

(correct answer: 0.0991 kHz)

Homework Equations



General Doppler-shift expression: f'=\left( \frac{v+v_o}{v-v_s} \right)f

The Attempt at a Solution



f'=\left( \frac{v+v_o}{v-v_s} \right)f= \left( \frac{335}{335-72} \right)2.1=2.62

f'-f=2.62-2.1= 0.52

Why am I not getting the right answer? Does this have something to do with the angle? All the examples in my textbook deal with situations where the observer or the source are moving directly toward one another, there are no examples with angles. Any help is appreciated.
 
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First, you have to convert km/h to m/s.
Next, does the source advance in the y direction?? does it advance in the X direction?
what is the source's velocity in y direction(if it has any..)?? what is the source's velocity in the X direction?
what should you plug in the equation?
 
Dweirdo said:
First, you have to convert km/h to m/s.
Next, does the source advance in the y direction?? does it advance in the X direction?
what is the source's velocity in y direction(if it has any..)?? what is the source's velocity in the X direction?
what should you plug in the equation?

The source is moving in the x direction but the sound that reaches the driver has both x and y component.

vx=335cos41=252.8
vy=335sin41=219.7

The resultant is:

\sqrt{252.8^2+219.7^2}=334.9

\frac{334.9}{334.9-20}2.1=1.27

then why do I still get the wrong answer?
 
Your mixing some stuff,
the velocity of the source is 20 m/s to the left, no doubt about that.
the velocity of sound in air is 335 m/s still no doubt about that.
You need to take the component of the velocity in the direction connecting the ambulance with the car.
to get that component it's 20*cos41 and you'll get the right answer..
 
Thank you! I get it now!
 

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