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bearhunter [10]
3 years ago
13

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 9.27 percent. How fast is the ambulance driving?
Physics
1 answer:
dlinn [17]3 years ago
5 0

To develop this problem we will apply the considerations made through the concept of Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. At first the source is moving towards the observer. Than the perceived frequency at first

F_1 = F \frac{{343}}{(343-V)}

Where F is the actual frequency and v is the velocity of the ambulance

Now the source is moving away from the observer.

F_2 = F\frac{343}{(343+V)}

We are also so told the perceived frequency decreases by 11.9%

F_2 = F_1 - 9.27\% \text{ of } F_1

F_2 = F_1-0.0927F_1

F_2 = 0.9073F_1

Equating,

F\frac{343}{(343+V)}= 0.9073(F\frac{343}{(343-V)})

\frac{1}{(343+V)}= 0.9073\frac{1}{(343-V)}

0.9073(343+V) = 343-V

(0.9073)(343)+(0.9073)V = 343-V

V+0.9073V = 343-(0.9073)(343)

Solving for V,

V = 16.67 m/s

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