How Fast is the Ambulance Traveling in the Doppler Effect Problem?

In summary, the ambulance is traveling at a speed of 25.54 m/s towards the person who is traveling at 20.0 m/s. The frequency of the ambulance siren is measured to be 1.427 kHz, while the actual frequency is 1.200 kHz. As the ambulance passes the person, the measured frequency of the siren decreases to 1.288 kHz. This is due to the Doppler effect, which is calculated using the equations (v0/v) = 1-(vs/v)(f0/fs)-1 and f0=fs(343/(343+vs)).
  • #1
booj1204
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Homework Statement


You are traveling north of wayne rd at 20.0m/s, when you notice an ambulance in the distance approaching you at a high rate of speed. Using an iPhone app, you measure the siren's frequency to be 1.427 kHz. From previous experience you know that the ambulance siren emits a tone of 1.200kHz. the speed of sound in air is 343m/s.

a) How fast is the ambulance traveling south on wayne?


b) Assuming that your have stopped while the ambulance passes you, what do you measure for the frequency of the siren as the ambulance speeds away from you?


Homework Equations


a) (v0/v) = 1-(vs/v)(f0/fs)-1

b) f0=fs(1/(1+vs/v)

The Attempt at a Solution



a) (v0/v) = 1-(vs/v)(f0/fs)-1

=> (1-(20.0/343))(.1200 Hz/.1427 Hz) - 1 = .208

v0=v(0..208) = 343 m/s (0.208) = 71. 38m/s

b) f0=fs(1/(1+vs/v)

=> .1427 (1/1+20.0/343) = 0.134 Hz

I would really appreciate if someone would let me know if my solution is correct. Thanks!
 
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  • #2


I would like to clarify some points in your solution:

a) The formula you used is correct, but the value you obtained for v0 is incorrect. It should be 25.54 m/s, not 71.38 m/s. This is because you used the wrong value for v0. In this case, v0 is the speed of sound in air, which is 343 m/s. So the equation should be (343/343) = 1 - (vs/343)(1200/1427), which gives a value of 25.54 m/s for vs.

b) Your formula for b) is incorrect. The correct formula is f0=fs(343/(343+vs)), which gives a value of 1.288 kHz for the frequency of the siren as the ambulance speeds away from you.

I hope this helps clarify any misunderstandings. Keep up the good work!
 

1. What is the Doppler Effect?

The Doppler Effect is a phenomenon where the frequency of a wave, such as sound or light, appears to change when there is relative motion between the source of the wave and the observer.

2. How does the Doppler Effect work?

The Doppler Effect works by changing the wavelength of a wave when there is relative motion between the source and observer. If the source is moving towards the observer, the wavelength appears shorter and the frequency appears higher. If the source is moving away from the observer, the wavelength appears longer and the frequency appears lower.

3. How is the Doppler Effect used in real life?

The Doppler Effect is used in various practical applications, such as radar and sonar systems, where it is used to measure the velocity of objects by analyzing the frequency shift of the reflected waves. It is also used in medical ultrasound to measure the velocity of blood flow in the body.

4. What is the difference between the Doppler Effect and the Doppler Shift?

The terms "Doppler Effect" and "Doppler Shift" are often used interchangeably, but technically the Doppler Effect refers to the change in frequency of a wave due to relative motion between the source and observer, while the Doppler Shift refers specifically to the change in frequency of light waves due to relative motion between the source and observer in the context of astronomy.

5. How can I calculate the Doppler Effect?

The formula for calculating the Doppler Effect is f = f0 (v ± vs) / (v ± vo), where f is the observed frequency, f0 is the emitted frequency, v is the speed of the wave, vs is the speed of the source, and vo is the speed of the observer. The ± sign depends on whether the source is moving towards or away from the observer.

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