Calculating Ambulance Frequency Change: -113 Hz

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SUMMARY

The discussion centers on calculating the change in frequency of an ambulance siren as it approaches and then moves away from an observer. The ambulance emits a siren frequency of 880 Hz and travels at 22 m/s, while the speed of sound is 343 m/s. The correct change in frequency heard by the observer is -113 Hz, determined by calculating the frequency when the ambulance approaches (940 Hz) and when it recedes (827 Hz). The difference between these two frequencies confirms the observed change of -113 Hz.

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Homework Statement


:
An ambulance has a siren of frequency 880 Hz. You are sitting at a stop light as the ambulance passes you, traveling at a constant 22 m/s. If the speed of sound is 343 m/s, what is the change in the frequency that you hear as the ambulance passes you?


  • A

    -53 Hz
  • wrong-icon.png


    B

    -60 Hz


  • C

    -56 Hz
  • correct-icon.png


    D

    -113 Hz


  • E

    -116 Hz
the correct Answer is D : -113 Hz

Homework Equations


CodeCogsEqn-4.gif

WHERE FL and vL are the frequency and velocity of sound heard by the personL
fs and vs are the frequency and velocity of sound emmited by the source

The Attempt at a Solution


using the equation we get FL=880x(343+0)/(343-22) therefore FL=940.3Hz
the difference is FL-Fs=940.3-880=60.311Hz
but its not correct.
the correct answer is -113Hz why?
 
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What's the observed frequency when the ambulance is moving towards you? And then away from you?
 
Doc Al said:
WhaF frequency when the ambulance is moving towards you? And then away from you?
1) when towards me it is FL1=Fs*v/(v-vs)=880*343/(343-22)=940 Hz
when moves away from me FL2=Fs*v/(v+vs)=880*343/(343+22)=827Hz
FL2-FL1=827-940=113 Hz
Ah okay thanksssss
 

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