Problem 12.61 A stationary police car emits a sound of frequency 1205 Hz that bo
ID: 1468528 • Letter: P
Question
Problem 12.61
A stationary police car emits a sound of frequency 1205 Hz that bounces off of a car on the highway and returns with a frequency of 1265 Hz . The police car is right next to the highway, so the moving car is traveling directly toward or away from it.
Part A
How fast was the moving car going?
16.13
SubmitMy AnswersGive Up
Incorrect; Try Again; 4 attempts remaining
Part B
Was the moving car moving towards or away from the police car?
It's was moving towards the police car.
SubmitMy AnswersGive Up
Correct
Part C
What frequency would the police car have received if it had been traveling toward the other car at 16.0 m/s ?
1324.54
SubmitMy AnswersGive Up
Incorrect; Try Again; 4 attempts remaining
Problem 12.61
A stationary police car emits a sound of frequency 1205 Hz that bounces off of a car on the highway and returns with a frequency of 1265 Hz . The police car is right next to the highway, so the moving car is traveling directly toward or away from it.
Part A
How fast was the moving car going?
v =16.13
m/sSubmitMy AnswersGive Up
Incorrect; Try Again; 4 attempts remaining
Part B
Was the moving car moving towards or away from the police car?
It's was moving towards the police car.
SubmitMy AnswersGive Up
Correct
Part C
What frequency would the police car have received if it had been traveling toward the other car at 16.0 m/s ?
f =1324.54
HzSubmitMy AnswersGive Up
Incorrect; Try Again; 4 attempts remaining
Explanation / Answer
Part A:
Here the frequency that is received by the police car is greater than the actual frequency, so the car should be moving towards the police car. The apparent frequency that is received by the police car is given by the following formula:
f' = f(V+Vc)/(V-Vc) ......................(i)
Here in the question, f is given to be 1205 Hz and f' is 1265 Hz. Assuming the velocity of the sound to be 343 m/s, we get from the above equation:
1265/1205 = (343+Vc)/(343-Vc)
Vc = 343*60/2470 = 8.33 m/s
Part B:As above said, It's was moving towards the police car.
Part C:
f = 1205 Hz, V = 343 m/s and Vc = 16 m/s, f' = ?
as discussed in part A, f' can be found by putting the values in equation (i):
f' = f(V+Vc)/(V-Vc)
f' = 1205(343+16)/(343-16) = 1322.92 Hz.