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Problem with calculating train speed using echo...
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evanesce...@gmail.com
Sep 29, 2012, 4:23:06 PM9/29/12
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Hi all,
From the book "ABC of Relativity" by B. Russell:
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Let us suppose that the shot is fired from the guard's-van, and the echo comes from a screen on the engine. We will suppose the distance from the guard's-van to the engine to be the distance that sound can travel in a second (about one-fifth of a mile), and the speed of the train to be one-twelfth of the speed of sound (about sixty miles an hour). We now have an experiment which can be performed by the people in the train. If the train were at rest, the guard would hear the echo in two seconds; as it is, it will take two and 2/143 seconds. From this difference, knowing velocity of sound, one can calculate the velocity of the train, even if it is a foggy night so that the banks are invisible.
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I can't figure out how they got 2/143 secs of extra time. Sure after 1sec the train would move 1/60 of the mile further, but while echo returns, guard's van (trailing wagon) would also move closer, so that would cancel each other out giving 2seconds as well. What am I missing, how do they get 2/143 difference?
Thanks for your time...
From the book "ABC of Relativity" by B. Russell:
------
Let us suppose that the shot is fired from the guard's-van, and the echo comes from a screen on the engine. We will suppose the distance from the guard's-van to the engine to be the distance that sound can travel in a second (about one-fifth of a mile), and the speed of the train to be one-twelfth of the speed of sound (about sixty miles an hour). We now have an experiment which can be performed by the people in the train. If the train were at rest, the guard would hear the echo in two seconds; as it is, it will take two and 2/143 seconds. From this difference, knowing velocity of sound, one can calculate the velocity of the train, even if it is a foggy night so that the banks are invisible.
------
I can't figure out how they got 2/143 secs of extra time. Sure after 1sec the train would move 1/60 of the mile further, but while echo returns, guard's van (trailing wagon) would also move closer, so that would cancel each other out giving 2seconds as well. What am I missing, how do they get 2/143 difference?
Thanks for your time...
Mike Terry
Sep 29, 2012, 7:29:05 PM9/29/12
to
<evanesce...@gmail.com> wrote in message
news:5615feb3-426f-4171...@googlegroups.com...
It's true that the time taken for the sound to reach the guard's van is more
than 1 second, and the time for it to return is less than 1 second, but they
don't cancel out! You need to do some accurate calculations.
Or if you just want to see that the effects don't cancel out, think through
the exagerated situation where the train travels at say 99% speed of sound.
Anyway, to do the calculation you need to work in the frame of the moving
train. What is the (relative) speed of the sound wave going from gun to the
guard's van, and what is the distance (in the trains's frame)? So you get
the time for the forwards part of the journey, and a similar calculation for
the return part...
(Note the calculations are based on the fact that the sound is travelling
through a medium (the air) through which the train is moving.)
Regards,
Mike.
news:5615feb3-426f-4171...@googlegroups.com...
than 1 second, and the time for it to return is less than 1 second, but they
don't cancel out! You need to do some accurate calculations.
Or if you just want to see that the effects don't cancel out, think through
the exagerated situation where the train travels at say 99% speed of sound.
Anyway, to do the calculation you need to work in the frame of the moving
train. What is the (relative) speed of the sound wave going from gun to the
guard's van, and what is the distance (in the trains's frame)? So you get
the time for the forwards part of the journey, and a similar calculation for
the return part...
(Note the calculations are based on the fact that the sound is travelling
through a medium (the air) through which the train is moving.)
Regards,
Mike.
evanesce...@gmail.com
Sep 30, 2012, 2:51:46 PM9/30/12
to
On Sunday, September 30, 2012 1:29:05 AM UTC+2, Mike Terry wrote:
> It's true that the time taken for the sound to reach the guard's van is more
> than 1 second, and the time for it to return is less than 1 second, but they
> don't cancel out! You need to do some accurate calculations.
Thanks Mike for the reply!
> It's true that the time taken for the sound to reach the guard's van is more
> than 1 second, and the time for it to return is less than 1 second, but they
> don't cancel out! You need to do some accurate calculations.
Unfortunately I can't get my head around how to accurately perform this calculation to get the same result. Thinking about it I can see that there is a small discrepancy.
I'd like to learn how to calculate this or similar problems. I'm not sure if it has anything to do with Galilean transformations or how to apply them to this problem.
If step-by-step solution isn't too much of a hassle to some mathematician or physicist in here, I'd be really thankful :o)
Thanks for your time...
Mike Trainor
Sep 30, 2012, 3:28:10 PM9/30/12
to
the sound signal travels are the same: the length of the train.
Let it be L.
The time taken for the 'up' journey, 'with' the train is
t1 = L/(vs - vt)
where vs and vt are, resp'ly, the speeds of the sound and
the train. Similary, the time for the retrun journey, 'against'
the train is
t2 = L/(vs +- vt)
Hence the total time, T, is
T = L( 1/((vs - vt) + 1/(vs - vt))
= (L/vs) * (1/(1 - vt/vs) + 1/(1 + vt/vs))
Now use L/vs = 1 and vt/vs = 1/12.
This gives T = 288/143.
The difference with the stationary case is
288/143 - 2 = 2/143.
Now, can you physically, without calculations, grasp
why this is a positive number?
mt
>
Mike Terry
Sep 30, 2012, 4:33:33 PM9/30/12
to
<evanesce...@gmail.com> wrote in message
news:8b327177-fac3-4216...@googlegroups.com...
Yes, it's to do with Galilean transformations, but these are simple to
understand, e.g. if I'm riding a bike at 10mph and throw a ball forwards at
5mph (relative to me on the bike), the ball's speed relative to the ground
is 15mph ( = 10mph + 5mph ). Relative velocities in Galilean
transformations just add and subtract like we would intuitively expect...
Lets write V for the speed of sound in metres per second.
Speed of sound relative to stationary frame (air not moving) is V.
Speed of train relative to stationary frame (air not moving) is (1/12)V.
Speed of sound (travelling forwards) in train frame = V - (1/12)V =
(11/12)V.
Distance sound travels forwards in train frame =
distance from gun to guard's van = V (metres). [Given in problem]
So time taken for sound wave (travelling forwards) in train frame =
distance/speed = V / [(11/12)V] = 12/11 (seconds).
The calculation for the echo travelling backwards is similar:
Speed of sound (travelling backwards) in train frame = V + (1/12)V =
(13/12)V.
Distance sound travels forwards in train frame =
distance from gun to guard's van = V (metres). [Given in problem]
So time taken for sound wave (travelling backwards) in train frame =
distance/speed = V / [(13/12)V] = 12/13 (seconds).
Total time = 12/11 + 12/13 = (156 + 132)/(11*13)
= 288 / 143
= 2 + 2/143
Regards,
Mike.
> Thanks for your time.
news:8b327177-fac3-4216...@googlegroups.com...
understand, e.g. if I'm riding a bike at 10mph and throw a ball forwards at
5mph (relative to me on the bike), the ball's speed relative to the ground
is 15mph ( = 10mph + 5mph ). Relative velocities in Galilean
transformations just add and subtract like we would intuitively expect...
Lets write V for the speed of sound in metres per second.
Speed of sound relative to stationary frame (air not moving) is V.
Speed of train relative to stationary frame (air not moving) is (1/12)V.
Speed of sound (travelling forwards) in train frame = V - (1/12)V =
(11/12)V.
Distance sound travels forwards in train frame =
distance from gun to guard's van = V (metres). [Given in problem]
So time taken for sound wave (travelling forwards) in train frame =
distance/speed = V / [(11/12)V] = 12/11 (seconds).
The calculation for the echo travelling backwards is similar:
Speed of sound (travelling backwards) in train frame = V + (1/12)V =
(13/12)V.
Distance sound travels forwards in train frame =
distance from gun to guard's van = V (metres). [Given in problem]
So time taken for sound wave (travelling backwards) in train frame =
distance/speed = V / [(13/12)V] = 12/13 (seconds).
Total time = 12/11 + 12/13 = (156 + 132)/(11*13)
= 288 / 143
= 2 + 2/143
Regards,
Mike.
> Thanks for your time.
evanesce...@gmail.com
Sep 30, 2012, 5:30:39 PM9/30/12
to
Thank you very much Mike!
evanesce...@gmail.com
Sep 30, 2012, 5:31:20 PM9/30/12
to
Thank you too for chiming in... Very helpful!
Michael Stemper
Oct 5, 2012, 12:45:36 PM10/5/12
to
In article <8b327177-fac3-4216...@googlegroups.com>, evanesce...@gmail.com writes:
>On Sunday, September 30, 2012 1:29:05 AM UTC+2, Mike Terry wrote:
>> It's true that the time taken for the sound to reach the guard's van is more
>> than 1 second, and the time for it to return is less than 1 second, but they
>> don't cancel out! You need to do some accurate calculations.
>
>Thanks Mike for the reply!
>
>Unfortunately I can't get my head around how to accurately perform this calculation to get the same result. Thinking about it I can see that there is a small discrepancy.
>
>I'd like to learn how to calculate this or similar problems. I'm not sure if it has anything to do with Galilean transformations or how to apply them to this problem.
You might wish to check out this book:
>On Sunday, September 30, 2012 1:29:05 AM UTC+2, Mike Terry wrote:
>> It's true that the time taken for the sound to reach the guard's van is more
>> than 1 second, and the time for it to return is less than 1 second, but they
>> don't cancel out! You need to do some accurate calculations.
>
>Thanks Mike for the reply!
>
>Unfortunately I can't get my head around how to accurately perform this calculation to get the same result. Thinking about it I can see that there is a small discrepancy.
>
>I'd like to learn how to calculate this or similar problems. I'm not sure if it has anything to do with Galilean transformations or how to apply them to this problem.
<http://archive.org/details/einsteintheoryof032414mbp>
Lieber goes through this type of calculation in her presentation.
--
Michael F. Stemper
#include <Standard_Disclaimer>
Life's too important to take seriously.
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