Question

You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind and you perceive the frequency as 1310 Hz. You are relieved that he is in pursuit of a different speed when he continues past you, but now you perceive the frequency as 1240 Hz. What is the frequency of the sirenin the police car

Answer 1

1270.44 Hz

Step-by-step explanation:

`v_(L)` = velocity of the our car = 35.0 m/s

`v_(P)` = velocity of the police car = ?

`v_(S)` = velocity of the sound = 343 m/s

`f_(app)` = frequency observed as police car approach = 1310 Hz

`f_(rec)` = frequency observed as police car go away = 1240 Hz

`f` = actual frequency of police siren

Frequency observed as police car approach is given as

`f_(app)= ((v_(s)-v_(L))f)/(v_(s) -v_(P) )`

inserting the values

`1310 = ((343 - 35)f)/(343 -v_(P) )` eq-1

Frequency observed as police car goes away is given as

`f_(rec)= ((v_(s) + v_(L))f)/(v_(s) + v_(P) )`

inserting the values

`1240 = ((343 + 35)f)/(343 + v_(P) )` eq-2

Dividing eq-1 by eq-2

`(1310)/(1240) = \left ( (343 - 35)/(343 - v_(P) ) \right )((343 + v_(P)))/(343 + 35 )\\`

`v_(P)` = 44.3 m/s

Using eq-1

`1310 = ((343 - 35)f)/(343 - 44.3 )`

f = 1270.44 Hz