Question

A person watches an ambulance drive by them while emitting a steady sound from its siren. When the ambulance passes the person (switching from moving towards him to moving away from him), the perceived frequency of the sound he receives decreases by 7.51 percent. How fast is the ambulance driving?

Answer 1

Answer:

 v_{s} = 27.85 m / s

Explanation:

The change of frequency by the relative movement of the bodies is explained by the Doppler effect, for this case the sender, the ambulance is moving and the receiver, the man is fixed; whereby the process equation is

         f’= f v / (v + $v_{s}$ )

Where f’ is the observed frequency, f the emitted frequency,v_{s} the speed of the source and the positive sign corresponds to the source moving away

In this case it indicates that the frequency decreases 7.51%

        f’ = f- 7.51 / 100 f = f -0.0751f

        f’ = 0.9249 f

       v +v_{s}=   v f / f ’

       v_{s} = v (f / f ’-1)

Let's calculate

       v_{s} = 343 (f / 0.9249f -1)

      v_{s} = 27.85 m / s