Calculating Doppler Effect: Car Speed and Frequency Analysis

In summary, the conversation discusses a problem involving a car approaching a reflecting wall and the resulting sound frequencies heard by a stationary observer. By using equations for moving observer and moving source, the speed of the car can be determined by combining the two sound frequencies. The frequency of the car horn is 800 Hz, and the frequency of the sound reflected from the wall is 863 Hz. The correct answers to the multiple choice questions are 1. D - 54.2 m/s, 2. B - 800 Hz, and 3. A - 931 Hz.
  • #1
sonutulsiani
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Homework Statement



A car is approaching a reflecting wall. A stationary observer behind the car hears a sound of frequency 745 Hz from the car horn and a sound of frequency 863 Hz from the wall.

1. How fast is the car traveling?

A. 25.2 m/s
B. 33.4 m/s
C. 18.9 m/s
D. 54.2 m/s

2. The frequency of the car horn is

A. 690 Hz
B. 800 Hz
C. 804 Hz
D. 875 Hz

3. The frequency the car driver hears reflected from the wall is

A. 931 Hz
B. 804 Hz
C. 800 Hz
D. 926 Hz

I got 1 as 54.2 m/s but I am not sure about it. I don't know how to do the rest. Please help me out.

Homework Equations





The Attempt at a Solution

 
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  • #2
(1)
+ Observer - Carhorn: in this case, observer is fixed and source is moving => f = f0 * v / (v + vS)
+ Observer - wall: observer receives a sound hitting the wall from car => f = f0 * v / (v - vS)

(2) You have frequency in 2 situations, it will be easily to find f0.

(3) Car driver hears a sound reflected from the wall. The original frequency is the same that is incident upon the wall. In this case, you can consider driver as observer, the wall as source.
 
  • #3
I didn't really understand what you just wrote..

1 is like we need to find the speed. What is f0, v and vs?
And why are there 2 equations, we just need one answer.
 
  • #4
Sry for this unconvinience.

f is a frequency received by observer. v is, in this problem, speed of sound, and vS is speed of source.

(1)
+ observer - carhorn: because a car is driving away from observer and generates a sound, we can consider that car is source. Thus, eq for this is f1 = f0 * v / (v + vS), with f1 is sound frequency received by observer (745hz).

+ observer - wall: a wall reflects a sound from car. Original sound frequency is f0. But car is moving towards wall, therefore, wall will reflect a sound with frequency f2 instead of f0. f2 = f0 * v / (v - vS).

Because you don't have f0, you should combine these 2 eqs to find vS (in this case, vS is speed of car).

Hope this help.
 
  • #5
Oh okay! I finally understood it, I got 1 as A - 25.2 m/s
2 as B - 800 HZ
And what will 3 be? 800 as well?
 
  • #6
No. In last problem, car is not source but an observer. The source is a wall and you have a sound frequency reflected from a wall. Plug it into formula for moving observer case.
 
  • #7
So it is 800*343/(343-25.2) = 863 ? But it's not there in the options.
 
  • #8
No. You are using formula for moving source case. For moving observer, it should be f = f0 * (v +/- vO) / v (+ in the case observer is going towards source).
 
  • #9
I got the 3rd 1 as 926. But then you said f0 is 863, what is f0? Isn't it the frequency of the car horn?
 
  • #10
OH NO ! You have confused yourself : Here you have taken f2

ApexOfDE said:
Sry for this unconvinience.

f is a frequency received by observer. v is, in this problem, speed of sound, and vS is speed of source.

(1)
+ observer - carhorn: because a car is driving away from observer and generates a sound, we can consider that car is source. Thus, eq for this is f1 = f0 * v / (v + vS), with f1 is sound frequency received by observer (745hz).

+ observer - wall: a wall reflects a sound from car. Original sound frequency is f0. But car is moving towards wall, therefore, wall will reflect a sound with frequency f2 instead of f0. f2 = f0 * v / (v - vS).

Because you don't have f0, you should combine these 2 eqs to find vS (in this case, vS is speed of car).

Hope this help.


But here you have taken f2 as f0

ApexOfDE said:
No. You are using formula for moving source case. For moving observer, it should be f = f0 * (v +/- vO) / v (+ in the case observer is going towards source).
 
  • #11
If car is not moving, f0 will be 800hz. However, car is going towards wall, so wall will receive a higher sound frequency and it will reflect this sound.
 
  • #12
So the 2nd answer will be 804 Hz?
 
  • #13
2B is right...

For 3rd problem, original f is 800hz, but source is moving, therefore, observer in wall will receive higher f. Wall receives this f and it reflect this f.
 
  • #14
So that's what I get. 863 * (343+25.2)/343=926 Hz isn't it right?
 
  • #15
D is correct. I never say you are wrong in 3rd :(
 
  • #16
Ha ha, I said that long time back! Anyway, I understood it well! Thanks a lot for helping me :) Peace
 

FAQ: Calculating Doppler Effect: Car Speed and Frequency Analysis

1. What is the Doppler Effect?

The Doppler Effect is a change in the frequency and wavelength of a wave (such as sound or light) in relation to an observer who is moving relative to the wave source.

2. How does the Doppler Effect work?

The Doppler Effect works by altering the perceived frequency of a wave based on the relative motion between the source of the wave and the observer. If the source is moving towards the observer, the frequency will appear higher, and if the source is moving away from the observer, the frequency will appear lower.

3. What causes the Doppler Effect?

The Doppler Effect is caused by the relative motion between the source of a wave and the observer. This change in frequency is due to the compression or stretching of the wave as the source moves towards or away from the observer.

4. How is the Doppler Effect used in everyday life?

The Doppler Effect is used in everyday life in various ways. For example, it is used in weather forecasting to track the movement of storms, in medical imaging to measure blood flow, and in radar technology to detect the speed and direction of objects.

5. How is the Doppler Effect related to the color of stars?

The Doppler Effect is related to the color of stars because the motion of a star towards or away from Earth can cause a shift in the wavelengths of light it emits. This shift can be observed as a change in the star's color, which can provide information about its motion and distance from Earth.

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