Respuesta :
Answer:
The speed of approaching police car is 5.35 m/s.
Explanation:
Given :
Police car actual frequency = 570 Hz
No of beats heard by stationary observer = 9
The speed of sound = 340 m/s
Beats is nothing but the difference of the frequency that heard by the observer.
According to the doppler equation,
⇒ [tex]f = f' (v-vo)/(v-vs)[/tex]
Where [tex]f[/tex]= observed frequency, [tex]f'[/tex] = actual frequency, [tex]v[/tex] = speed of sound,
[tex]vo[/tex] = speed of observer, [tex]vs[/tex] = speed of source.
One car approach stationary listener ([tex]vo\\[/tex] = [tex]0[/tex] )
∴ [tex]f1[/tex] = [tex]570[/tex]×[tex](340)/(340-vs)[/tex]
One car parked, so ( [tex]vs = 0[/tex]) and [tex]vo=0[/tex]
∴ [tex]f2[/tex] = [tex]570[/tex]
From beats,
∴ [tex]f1 -f2=\\[/tex] no. of beats.
Where [tex]f1[/tex] = frequency of approaching car, [tex]f2[/tex] = frequency of stationary car.
∴ [tex]570[/tex]×[tex]340/(340-vs)[/tex] [tex]- 570[/tex] = [tex]9[/tex]
While solving above equation,
∴ [tex]vs =[/tex] 5.35 m/s.
Therefore, the speed of approaching police car is 5.35m/s.
