Doppler's effect for two trains

[physgirl]

A train on one track moves in the same direction as a second train on the adjacent track. The first train, which is ahead of the second train and moves with a speed of 28 m/s, blows a horn whose frequency is 125 Hz. If the frequency heard on the second train is 132 Hz, what is its speed?

so I tried using the equation:

f' = [(1+vo/v)/(1+vs/v)]f

(both pluses taking into account the direction of both the observer and the source)
where:
f'=frequency heard=132 Hz
f=actual frequency=125 Hz
vo=velocity of observer=unknown
vs=velocity of source=28 m/s
v=343 m/s

so I plugged in the numbers and got 48.78 m/s. The website won't take this as the answer and I'm completely clueless as to what I did wrong... any help?

 

[OlderDan]

I suspect you have a sign mistake, but maybe you did something else. Please do the algebra to solve your equation for the thing you are looking for, vo.

 

[Max Eilerson]

I equation is easier to rearrange if you say.

f' = [(v+v_o)/(v-v_s)]f.
vs is positive because the trains are moving towards each other, also your answer should be quoted to 2.s.f since vs is only given to 2.s.f.

 

[physgirl]

I double and triple checked my numbers, I did it numerous ways by first solving the equation for what I need and plugging in numbers. The website is not so picky about sig figs and it usually takes it. If not, it tells me if my answer is within 10% of what it really is, but in this case, it did not tell me that. And I think my signs are correct, they should both be positive... I multiple checked that fact also, so correct me if I'm wrong... (and the trains aren't moving towards each other if I understood the question correctly... they're both going in the same direction)

 

[OlderDan]

I double and triple checked my numbers, I did it numerous ways by first solving the equation for what I need and plugging in numbers. The website is not so picky about sig figs and it usually takes it. If not, it tells me if my answer is within 10% of what it really is, but in this case, it did not tell me that. And I think my signs are correct, they should both be positive... I multiple checked that fact also, so correct me if I'm wrong... (and the trains aren't moving towards each other if I understood the question correctly... they're both going in the same direction)

The trains are approaching in a relative sense. The rear train is definitely moving through air in the direction of the source. I suggest you break the problem down into two steps. Find the frequency of the sound to a stationary observer between the trains (lower than source) and treat that lower frequency sound as if it came from a stationary source being approached by the second train.

 

[physgirl]

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...

 

[OlderDan]

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.

 

[physgirl]

ahhhh no... website didn't take 42.7 either. It told me it's completely off (as in, not within 10% of the correct answer). This is weird :-/

 

[OlderDan]

ahhhh no... website didn't take 42.7 either. It told me it's completely off (as in, not within 10% of the correct answer). This is weird :-/

I don't know what to tell you. The only thing I can think of is that the data was given in some other units. Speed in km/hr by any chance?

 

[physgirl]

nope... the unit of the answer they want is in m/s :(

 

[OlderDan]

nope... the unit of the answer they want is in m/s :(

I meant the units of the 28 initial velocity might have been 28 km/hr instead of 28 m/s. Just hunting for some reason they don't like your answer.