Question

A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequency of 1275 HzHz. The police car is right next to the highway, so the moving car is traveling directly toward or away from it. And what frequency would the police car have received if it had been traveling toward the other car at 25.0 m/s?

Answer 1

Answer:

frequency =  1475.45 Hz

Explanation:

given data

frequency f1 = 1215 Hz,

frequency f2 = 1265 Hz

police car moving vp = 25.0 m/s

solution

speed of sound u = 343 m / s

speed of the other car = v

when the police car is stationary

the frequency the other car receives is

f2 =  f1  ×  $\dfrac{u+v}{u}$      ................1

and

the frequency the police car receives is

 f2 =  f1  ×  $\dfrac{u}{u-v}$      ..................2

now from equation 1 and 2

$\frac{f2}{f1} = \dfrac{u+v}{u-v}$

 $\frac{1275}{1240} = \frac{u+v}{u-v}$

$v =\frac{1275-1240}{1275+1240}\times 343$  

v = 4.77 m/s

and

frequency the other car receives is  

f2 = f1 ×   $\dfrac{u+v}{u-vp}$       ......................3

and

the frequency the police car receives is

f2 = f1 ×  $\dfrac{u+vp}{u - v}$       .......................4

now we get

f2 = f1 ×  $\dfrac{(u+v)(u + vp)}{(u-v)(u-vp)}$      

f2 =    $1240\times \frac{(343+4.77)(343+25)}{(343-4.77)(343-25)}$        

f2 =  1475.45 Hz