- #1
Phynos
- 31
- 4
Homework Statement
Jane waits on a railroad platform while two trains approach from the same direction at equal speeds of 11 m/s. Both trains are blowing their whistles (which have the same frequency), and one train is some distance behind the other. After the first train passes Jane but before the second train passes her, she hears beats of frequency 6.00 Hz. What is the frequency of the train whistles? (Use 331 m/s as the speed of sound.)
Homework Equations
1) fb = | f1 - f2 |
where fb is the frequency of the beats.
2) f1 = f * (v/ (v-vs) )
3) f2 = f * (v/ (v+vs))
where v is the speed of sound
and vs is the speed of the source
equation 2 is for the source moving towards Jane, and equation 3 is for the source moving away from Jane.
The Attempt at a Solution
fb = | f1 - f2 |
-- Sub in equation 2 and 3 --
fb = | f * ( v / (v-vs) ) - f * ( v / (v+vs) )
-- Solve for f --
fb = | f * (2v*vs) / (v^2 - vs^2) |
-- Term in absolute value always positive since v >> vs --
f = fb [ (2v*vs) / (v^2 - vs^2 ) ]^(-1)
f = 90.2Hz
This just doesn't seem like a good answer. There's no way any train would use such a low pitch whistle, I can't even hear it on my computer at max volume. Are there any mistakes in my work? Is there perhaps a different way to solve this?
P.S. I will work on making the equations more tidy in the future, perhaps learning how to use LATEX or something along those lines. Thanks for your patience.
Last edited: