Frequency/Speed of sound

[BooGTS]

Problem goes as follows:

A person standing close to a railroad crossing hears the whistle of an approaching train. He notes that the pitch of the whistle drops as the train passes by and moves away from the crossing. The frequency of the distant approaching whistle is 546 Hz; it drops to 469 Hz after the train is well past the crossing. What is the speed of the train? Use 340 m/s for the speed of sound in air.

Hint: Calculate the ratio of frequency of the whistle before and after the crossing. That ratio does not include the frequency of the train at rest.

The best I've been able to come up with is:

546 Hz=507.5 Hz (340 m/s/(340 m/s-X))
1.076=(340/(340-X))
1.076(340-X) = 340
365.8 - 1.076X = 340
-1.076X = - 25.8
X=23.97 m/s

Any suggestions?

[nrqed]

Problem goes as follows:

A person standing close to a railroad crossing hears the whistle of an approaching train. He notes that the pitch of the whistle drops as the train passes by and moves away from the crossing. The frequency of the distant approaching whistle is 546 Hz; it drops to 469 Hz after the train is well past the crossing. What is the speed of the train? Use 340 m/s for the speed of sound in air.

Hint: Calculate the ratio of frequency of the whistle before and after the crossing. That ratio does not include the frequency of the train at rest.

The best I've been able to come up with is:

546 Hz=507.5 Hz (340 m/s/(340 m/s-X))
1.076=(340/(340-X))
1.076(340-X) = 340
365.8 - 1.076X = 340
-1.076X = - 25.8
X=23.97 m/s

Any suggestions?

You cannot assume that the frequency of the whistle is midway between the two frequencies given to you! It is not!
You must treat the frequency of the whistle as being an unknown. Write the equation for the Doppler effect when the source is moving toward the observer (and the frequency observed is 546 Hz) and then the equation when it is moving away (with the frequency observed 469 Hz). That will give you two equations with two unknowns (the speed of the train and the frequency of the whistle.) If you take the ratio of the two equations the frequency of the whistle will cancel out.

[BooGTS]

Thanks nrqed,

Thats basically what the hint said, but it helped. Here is the finished problem:

((340/340-Vs)*((340+Vs/340)) = 1.164

(115600+340Vs)/(115600-340Vs) = 1.164

115600+340Vs = 1.164 (115600-340Vs)

115600+735.76Vs = 134558.4

735.76Vs = 18958.4

=25.77 m/s


I have one more problem that I could use some help on, I'll bump it to the top ;)