thainguyen
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Hello guys, I am studying A Level now. I just would like to ask a question in Physics:
The warning signal on an ambulance has a frequency of 600Hz. The speed of sound is 330m/s. The ambulance is travelling with a constant velociy of 25m/s towards an observer. Which overall change in observed frequency takes place between the times at which the ambulance is a long way behind the observer and when it is a long way in front of the observer.
A. 49Hz
B. 84 Hz
C. 91 Hz
D. 98 Hz
Well, the answer is C. How???
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Vikingninja
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Find observed frequency moving towards the observer and subtract frequency observed when moving away.
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Levisual360
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Find the wavelength of the sound,

Then find the relative velocity of the sound going to the observer ( 330 + 25 meters per second )

Now find the frequency of this sound by the equation v=f*wavelength

Then find the relative velocity of the sound when its moving away from the observer ( 330 - 25 meters per second )

Now find the frequency of this sound by the equation v=f*wavelength

Find the difference in frequencies, and you will find it to be 91hz
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Vikingninja
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(Original post by Levisual360)
Find the wavelength of the sound,

Then find the relative velocity of the sound going to the observer ( 330 + 25 meters per second )

Now find the frequency of this sound by the equation v=f*wavelength

Then find the relative velocity of the sound when its moving away from the observer ( 330 - 25 meters per second )

Now find the frequency of this sound by the equation v=f*wavelength

Find the difference in frequencies, and you will find it to be 91hz
You just need to do (v/(v-vs))*f for going towards and away and subtract.
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