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Uncertainty of time?

  1. Sep 6, 2004 #1
    Time standards are now based on atomic clocks. A promising second standard is based on pulsars, which are rotating neutron stars (highly compact stars consisting only of neutrons). Some rotate at a rate that is highly stable, sending out a radio beacon that sweeps briefly across Earth once with each rotation, like a lighthouse beacon. Suppose a pulsar rotates once every 1.424 806 448 872 75 2 ms, where the trailing 2 indicates the uncertainty in the last decimal place (it does not mean 2 ms).
    (a) How many times does the pulsar rotate in 21.0 days?
    The answer is 1.27e9
    (b) How much time does the pulsar take to rotate 1.0 x 10^6 times? (Give your answer to at least 4 decimal places.)
    The answer is 1424.8064 seconds
    (c) What is the associated uncertainty of this time?

    For this problem I am unsure of how the uncertainty is calculated. I understand parts a and b but not C. Please help.
    Last edited: Sep 6, 2004
  2. jcsd
  3. Sep 6, 2004 #2
    I think your answer to b) is off.

    It's rotating once every 1.424 806 448 872 75 2 ms. so for it to rotate 1.000 time plus 0.0106 times, you're definately not going to be in the 1400 second range.

    That's about 20 minutes to go a little more than it was going in beyond less than a second. Recalculate that and you should get a better start for finding the uncertainty.
  4. Sep 6, 2004 #3
    This answer is the right answer, but it should say: How much time does the pulsar take to rotate 1.0 X 10^6 times?
  5. Sep 6, 2004 #4


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    Use the error propogation formula:
    [tex]s = \sqrt{ \left( \frac{\delta u}{\delta x} \right)^2_{y,z}s_x^2 + \left( \frac{\delta u}{\delta y} \right)^2_{x,z}s_y^2 + \left( \frac{\delta u}{\delta z} \right)^2_{x,y}s_z^2 }[/tex]
    This is the 3 dimensional [x,y,z] form of the equation. In this case you only have a single dimension, so, it simplifies to
    [tex]s = \sqrt{ \left( nt \right)^2 s_t^2 }[/tex]
    where s is the total uncertainty, n is number of cycles, t is cycle time and s sub t is uncertainty per cycle.
  6. Sep 6, 2004 #5
    I get some incredibly odd answer. I think this problem has a simple way that it should be solved and I just don't know what it is. :grumpy:
  7. Sep 8, 2004 #6
    last minute hope lol can anyone help?
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