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,
⇒ ![f = f' (v-vo)/(v-vs)](https://tex.z-dn.net/?f=f%20%3D%20f%27%20%28v-vo%29%2F%28v-vs%29)
Where
= observed frequency,
= actual frequency,
= speed of sound,
= speed of observer,
= speed of source.
One car approach stationary listener (
=
)
∴
=
×![(340)/(340-vs)](https://tex.z-dn.net/?f=%28340%29%2F%28340-vs%29)
One car parked, so (
) and ![vo=0](https://tex.z-dn.net/?f=vo%3D0)
∴
=
From beats,
∴
no. of beats.
Where
= frequency of approaching car,
= frequency of stationary car.
∴
×
= ![9](https://tex.z-dn.net/?f=9)
While solving above equation,
∴
5.35 m/s.
Therefore, the speed of approaching police car is 5.35m/s.