Doppler's effect for two trains

  • Thread starter physgirl
  • Start date
  • Tags
    Trains
In summary, the two trains approaching each other in a relative sense, with the first train moving at 28 m/s and blowing a horn at a frequency of 125 Hz and the second train hearing a frequency of 132 Hz, leads to a calculation of the second train's speed to be 48.78 m/s. However, the mistake of assuming that the observer is stationary and being approached by a moving source leads to a different calculation of 42.7 m/s. This is incorrect and the correct calculation involves finding the Doppler shifted frequency of the sound in air twice, first as the sound from the first train and then as the sound from the air itself, which results in the correct answer of 48.78 m/s.
  • #1
physgirl
99
0
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? :confused:
 
Physics news on Phys.org
  • #2
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.
 
  • #3
I equation is easier to rearrange if you say.

[tex]f' = [(v+v_o)/(v-v_s)]f. [/tex]
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.
 
  • #4
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)
 
  • #5
physgirl said:
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.
 
  • #6
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...
 
  • #7
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.
 
Last edited:
  • #8
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 :-/
 
  • #9
physgirl said:
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?
 
  • #10
nope... the unit of the answer they want is in m/s :(
 
  • #11
physgirl said:
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.
 

What is the Doppler's effect for two trains?

The Doppler's effect for two trains is a phenomenon where the perceived frequency of sound or light waves changes when an observer is moving relative to the source of the waves. This effect can be observed when two trains are moving towards each other or away from each other.

How does the Doppler's effect for two trains work?

The Doppler's effect for two trains works by changing the wavelength of the sound or light waves depending on the relative motion between the observer and the source. When the source is moving towards the observer, the waves are compressed, resulting in a higher frequency and a higher pitch. Conversely, when the source is moving away from the observer, the waves are stretched, resulting in a lower frequency and a lower pitch.

What factors affect the perceived frequency in the Doppler's effect for two trains?

The perceived frequency in the Doppler's effect for two trains is affected by the relative velocity between the two trains, the speed of sound, and the frequency of the source. If the trains are moving towards each other at a higher velocity, the change in frequency will be more noticeable. Additionally, the speed of sound and the frequency of the source also play a role in determining the perceived frequency.

How does the Doppler's effect for two trains relate to the speed of sound?

The Doppler's effect for two trains is directly related to the speed of sound. As the trains move towards each other, the perceived frequency increases due to the compression of sound waves, but this effect is limited by the speed of sound. Once the trains reach the speed of sound, the effect becomes more pronounced, resulting in a sonic boom. However, if the trains are moving away from each other, the perceived frequency will decrease until it reaches the speed of sound, where it becomes inaudible.

Can the Doppler's effect for two trains be observed with other types of waves?

Yes, the Doppler's effect can be observed with other types of waves, such as electromagnetic waves. This effect can be observed with visible light, resulting in a shift in the perceived color of an object depending on its relative motion with the observer. It can also be observed with radio waves, which is utilized in the Doppler radar system to detect the speed and direction of moving objects.

Similar threads

  • Introductory Physics Homework Help
Replies
20
Views
873
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
904
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
8
Views
179
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Back
Top