Doppler Effect: Velocity of a train

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Homework Help Overview

The discussion revolves around a problem involving the Doppler Effect, specifically related to determining the speed of a train based on the observed frequencies of its whistle as it approaches and moves away from an observer. The frequencies measured are 220 Hz when approaching and 190 Hz when moving away, with the speed of sound in air given as 340 m/s.

Discussion Character

  • Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the Doppler effect equation and question the interpretation of the source and observer roles. There is an exploration of the calculations involved in determining the speed of the train, with some participants expressing confusion over the correct application of the formula.

Discussion Status

Some participants have offered guidance on the roles of the source and observer in the Doppler effect context, suggesting that the train is the source of the sound. The discussion is ongoing, with multiple interpretations being explored and no explicit consensus reached on the correct approach.

Contextual Notes

There is mention of confusion regarding the time dependence in the Doppler effect equation, and participants are clarifying the definitions of the variables involved in the calculations. The original poster's calculations differ from the textbook answer, prompting further examination of the assumptions made.

Granger
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Homework Statement


To determine the speed of a train, a student of Physics determined the frequency of the whistle in the approach to its position of observation, and measured 220 Hz. He then determined the frequency of the whistle when the train moved away, and got 190 Hz. What is the speed of train? Consider that the velocity of sound in the air is 340 ms-1

Homework Equations


[/B]
Doppler effect equation:
$$f' = f \frac{u+- v_Dt}{u+- v_St}$$

The Attempt at a Solution


What I simply did was to isolate v_D to determine it.

$$v_D=v(1-\frac{f'}{f})= 340(1+\frac{190}{200}=46.4$$

Note: I choose the minus signal in the numerator because the destiny is moving away from the source.

The answer of textbook is however:
$$v_D=v(\frac{f-f'}{f+f'})= 24.9 m/s$$

I don't get what I'm doing wrong. Can someone help me?
 
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Maybe I'm crazy but this doppler equation doesn't feel right. Why is there a time dependence? The frequency should not be changing as a function of time. Check this.

In any case, you solved for vD. Is this the train? Which object is the source of the sound, the train or the student?
 
Hey! You are right there is no time dependency (note that I didn't use it when doing the calculation). Sorry!

The way I interpreted the problem is that the student is the source and the train is the detector.
 
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Granger said:

Homework Statement


To determine the speed of a train, a student of Physics determined the frequency of the whistle in the approach to its position of observation, and measured 220 Hz. He then determined the frequency of the whistle when the train moved away, and got 190 Hz. What is the speed of train? Consider that the velocity of sound in the air is 340 ms-1

Homework Equations


Doppler effect equation:$$f' = f \frac{u+- v_Dt}{u+- v_St}$$

The Attempt at a Solution


What I simply did was to isolate v_D to determine it.$$v_D=v(1-\frac{f'}{f})= 340(1+\frac{190}{200}=46.4$$
Note: I choose the minus signal in the numerator because the destiny is moving away from the source.

The answer of textbook is however$$v_D=v(\frac{f-f'}{f+f'})= 24.9 m/s$$
I don't get what I'm doing wrong. Can someone help me?
Taking the time dependence out of your Doppler Effect Formula (I also assumed that you mean ±) gives:
##\displaystyle f' = f\, \frac{u\pm v_D}{u\pm v_S} ##​

Of course, the first thing to do is to give the definitions all of those quantities.
 
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