physgirl said:
I still get the same answer I got before, unless I did it wrong? for the lower frequency I got 115.566 Hz and using that number and treating as if it came from stationary source being app by second train, i still get 48.78 m/s...
I'm afraid it was I who made the mistake. My appologies.
You are right! The wavelength in the still air behind the first train is elongated to λ' = (v + v_s)/f_s. The frequency of this sound in the air is f' = v/λ' , which you calculated correctly to be 115.6Hz. The second train approaches the oncoming wave, so the wave has an effectively increased speed of (v + v_o) and an effective frequency of
f_o = (v + v_o)/λ' = f_s(v + v_o)/(v + v_s)
f_o = f_s(1 + v_o/v)/(1 + v_s/v)
which is exactly what you said it was in the first place. Solving for v_o
v_o = {(1 + v_s/v)(f_o/f_s) - 1}v = 48.78m/s
So how could anybody who had not worked as hard as you to do this problem think you made a sign mistake?
This is not the same as if the observer were stationary and being approached by a moving source producing a sound of frequency f', but it is close and it is tempting to think that this should give the same answer. We get very used to the idea that when things are moving at constant speeds only the relative velocity matters. Well in this case
it does matter who is moving and who is not moving because the sound does not go any faster in air when the source is moving than it goes when the source is stationary. If a source were moving toward the observer at speed u generating frequency f' the wavelength in the air would be λ'' = (v - u)/f' and the frequency would be
f'' = v/λ'' = vf'/(v - u) = v^2/[λ'(v - u)]
f'' = f_sv^2/[(v + v_s)(v - u)]
f'' = f_s/[(1 + v_s/v)(1 - u/v)]
Solving this for u gives
u = v{1 - (f_s/f'')/(1 + v_s/v)}
Now of course you think anyone would be crazy coming up with this equation for this problem, but that is not the thought process that leads to this result. The thought process is to find the Doppler shifted frequency of the sound in air, just like you did
f' = f_s/(1 + v_s/v) = 115.6Hz
and then find the Doppler shift again as if the air were the source approaching you instead of you moving toward the sound wave
f'' = f'/(1 - u/v)
u = v(1 - f'/f'') = 42.7m/s
This is wrong, but it is not obviously wrong, and it is close to the actual answer. I would not be surprised if this were the answer the website is waiting for you to enter, but it is not correct. You are.
