A commuter train blows its horn as it passes a passenger platformat a constant s

Question

A commuter train blows its horn as it passes a passenger platformat a constant speed of 30.0 m/s. Thetrain horn sounds at a frequency of 340 Hz when the train is at rest. (a) What is the frequency observed by a personon the platform as the train approaches?
1 Hz

(b) What is the frequency observed by a person on the platform asthe train recedes from him?
2 Hz

(c) What wavelength does the observer find in each case?
3 m
4 m


Will rate. Thanks
(a) What is the frequency observed by a personon the platform as the train approaches?
1 Hz

(b) What is the frequency observed by a person on the platform asthe train recedes from him?
2 Hz

(c) What wavelength does the observer find in each case?
3 m
4 m


Will rate. Thanks

Explanation / Answer

When the person is standing stationary on the platform and thetrain is approaching the frequency observed will be higher than theactual frequency and as the train recedes (passes the person) thefrequency will be lower than the actual frequency. Theequation we will be using isfo=fs(1/1+or-vs/v) wherefo is frequency observed, fs is frequency ofthe source, vs is the speed of the source, and v is thespeed of sound (343 m/s). The + is used when the frequency isreceding and the - is used when the source is approaching. Our knowns from the problem are:
vs=30.0m/s fs=340Hz vsound=343m/s (this is a constant you shouldmemorize)

So.... a) fo=fs(1/1-vs/v) plug in the numbers and you will get fo=373 Hz. I will let you do the math to save the room on here. b) fo=fs(1/1+vs/v) again plug in the numbers and you get fo=315 Hz. This seems reasonable according to the logic I appliedearlier about approaching frequency being higher than actualfrequency and receding frequency being lower than actual frequencywhich is what we got with these two answers. c) Now all you have to do is use the equationv=f(lambda) (sorry my computer won't let me add the lambdacharacter.) You manipulate the equation algebraically to solve for lambdausing the frequency from a) and there is your answer. Youshould get lambda = 8.04e-2. d) do the same thing as c) but use the frequency from b) andyou are all done. You should getlambda=9.52e-2.
Hope this not only gets you the answers you need but a betterunderstanding of why. Good luck!

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