Uncertainty of Time | Atomic Clocks & Pulsars

[GingerBread27]

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.

 

[falcon0311]

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 definitely 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.

 

[GingerBread27]

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?

 

[Chronos]

Use the error propogation formula:
$$s = \sqrt{ \left( \frac{\delta u}{\delta x} \right)^2_{y,z}s_x^2 + \left( \frac{\delta u}{\delta y} \right)^2_{x,z}s_y^2 + \left( \frac{\delta u}{\delta z} \right)^2_{x,y}s_z^2 }$$
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
$$s = \sqrt{ \left( nt \right)^2 s_t^2 }$$
where s is the total uncertainty, n is number of cycles, t is cycle time and s sub t is uncertainty per cycle.

 

[GingerBread27]

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:

 

[GingerBread27]

last minute hope lol can anyone help?