Doppler Effect: Velocity of a train

[Granger]

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?

 

[Chandler]

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 v_D. Is this the train? Which object is the source of the sound, the train or the student?

 

[Granger]

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.

 

[Chandler]

Typically, the whistle is found on the train :) try it this way. The source is the object that is doing the whistling.

https://en.wikipedia.org/wiki/Train_whistle

 

[SammyS]

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.