This stems from a riddle I read in a magazine perhaps 20 years ago so I apologise for the imprecise recollection.

A dog that can run infinitely fast is placed on an infinitely large flat surface and an alarm clock is tied to his tail. The dog has been trained to double the speed he is running when he hears the bell go off.

So this dogs sets out running at, let's say 5 m/s and every 10 seconds, the alarm goes off. The question is how fast is he running after two minutes.

Also can someone find the actual riddle? I have not been able to.

I have cloaked the rest of the question in case you want to answer that one first.

So the tricky part of this is of course that the dog supposedly stops doubling his speed after he doubles past the speed of sound, outrunning the alarm My question is, would he not intercept previous sound waves and start doubling again?

Thanks

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What happens is you overtake one of these previous sound-waves? The frequencies will be shifted up, so at some point the dog can't hear those anymore at some point? –  Bernhard Jan 27 '12 at 14:38
And they would be fainter as well. –  queueoverflow Jan 27 '12 at 17:02
The sound waves will travel through his body so he will always hear the alarm very quickly. –  Philip Gibbs - inactive Jan 27 '12 at 19:06
Assume the alarm is on a string that does not carry the sound –  mplungjan Jan 27 '12 at 19:13
Does the alarm give off light as well? –  Dimensio1n0 Jun 8 '13 at 4:31

There are only a non-infinite number of waves that escaped the dog. So he will double a couple more times, and then he will reach his final speed.

I wrote a small Python simulation for this. The output of the program is also on the same gist page.

To run it, download dog.py and call it with python dog.py, assuming you have a python interpreter on your machine.

It starts off with 6 waves catching the dog, and then the dogs catching the waves.

So I think there indeed is one final speed, the program suggests 20480 m/s. This is only true, if the dog can hear the faint and differently pitched sounds.

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Yes, that would be my thought, but that would be faster than the riddle-writer expected as answer in my opinion Can whoever voted this down tell us why? –  mplungjan Jan 27 '12 at 15:04
The Python program looks cool. Can you run it on github, or do you have to download it, compile and run? Could you just post the output yourself? I mean, you went through all the effort to write that after all. –  Alan Rominger Jan 27 '12 at 18:13
You have to download the file and call it with python dog.py, assuming you have a python interpreter on your computer. –  queueoverflow Jan 27 '12 at 18:30
@mplungjan The original post didn't include the Python script. So now it is improved. Adding figures would improve the post even further. –  Bernhard Jan 27 '12 at 18:32
That was fun :) –  mplungjan Jan 27 '12 at 19:16

I wanted to just raise a comment, but dont have the reputation to do so... I didnt get the same result as the python simulation, so ill just detail my thoughts below.

Some assumptions (do deal with riddle as probably intended, as opposed to a real world problem):

• Instantaneous acceleration to new speed when we hear an alarm.
• An alarm is the wave-front of the sound emission from a point source on the dogs tail.
• Alarms start behind the dog (ears are behind tail).
• The alarm propagates through air (earth-like atmosphere).
• The alarm does not register through the dog (reverberation through dog body not felt).
• Once an alarm reaches us, causing us to increase speed, this alarm will be considered in front of us (to avoid an acceleration dependent race condition on the wave that pushes us over speed of wave propagation).

We'll start at 5 m/s. So long as we're slower than speed of sound(330 m/s), the wave will reach/pass us, and we will double speed. The wave that causes us to break speed of sound, will be the last wave to pass us (last wave ahead of us that we may overtake).

So basically, we will double once ever 10 seconds: 5,10,20,40,80,160,320,640... Each comma above is a wave that passed us, causing us to double speed. There are 7 waves in front of us when we're traveling at 640 m/s. We may potentially overtake all these. This means doubling our initial speed 14 times. 5*(2^14) = 81920 m/s. This max is achieved if all waves are overtaken in the 2 mins.

The first wave will be the furthest away, emitted at position 50m, and traveling as far as 36350m by the 2 min mark (50 + 330*(120-10)).

The dog breaks speed of sound at the 70sec mark, located at 6350m. From there, if it never speeds up again in the final 50 seconds, it will reach 38350m by the 2 min mark (6350 + 640*(120-70)).

Since 38350 > 36350, it will overtake all 7 waves in front of it (thereby going well past 38350m) and reach the cap of 81920 m/s.

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Thanks for your input –  mplungjan Aug 1 '14 at 21:28