Uncertainty with a simple pendulum

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aborder
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A simple pendulum is used to measure the acceleration of gravity using T=2pi(sqrt(L/g)) . The period T was measured to be 1.24 ± 0.02 s and the length L to be 0.381 ± 0.002 m. What is the resulting value for g with its absolute and relative uncertainty?

So the first thing I did was to isolate g. But to actually calculate the uncertainty, I am completely lost here. I am using the book "Experimentation" by D.C. Baird and nothing is making sense here. Most likely it talks about it in the book, but I am having a hard time understanding this. If someone could explain how to calculate uncertainty for this, it would probably help. Thanks.
 
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aborder said:
A simple pendulum is used to measure the acceleration of gravity using T=2pi(sqrt(L/g)) . The period T was measured to be 1.24 ± 0.02 s and the length L to be 0.381 ± 0.002 m. What is the resulting value for g with its absolute and relative uncertainty?

So the first thing I did was to isolate g. But to actually calculate the uncertainty, I am completely lost here. I am using the book "Experimentation" by D.C. Baird and nothing is making sense here. Most likely it talks about it in the book, but I am having a hard time understanding this. If someone could explain how to calculate uncertainty for this, it would probably help. Thanks.


Once transformed, you will have found g proportional to L/T2

to find the absolute largest value, you want to multiply by as much as possible, and divide by as little as possible.
So you will sub the upper limit of L, and the lower limit of T to get the largest possible g

To get the lowest possible g, use smallest L and largest T.

You will already have calculated the expected g, you can thus work out the absolute uncertainty.

For relative error [is that percentage error?] you would multiply the percentage error in L by the square of the percentage error in T.
 
That makes sense to divide by the highest and multiply by the lowest to get the low end and vice versa. Using the method you described, I got 9.79 +/- 0.184. The answer in the book gives 9.77 +/-0.04 with a relative uncertainty of 0.4%.

The relative uncertainty is given by this:

Relative uncertainty = Absolute Uncertainty / Measured Value
 
aborder said:
That makes sense to divide by the highest and multiply by the lowest to get the low end and vice versa. Using the method you described, I got 9.79 +/- 0.184. The answer in the book gives 9.77 +/-0.04 with a relative uncertainty of 0.4%.

The relative uncertainty is given by this:

Relative uncertainty = Absolute Uncertainty / Measured Value

Firstly, my error. I should have said you add the percentage errors not multiply them - haven't used percentage errors for a while.

Period T 1.24 +-0.02 means an error of 2 in 124 = 1.6%
Length L = 0.381 +- 0.004 means an error of 4 in 381 = 1.05%

So total error = 1.6 + 1.6 + 1.05 = 4.25%

so 9.78 +- 4% or 9.78 +- .04


I [almost]agree with your numbers, and would express it as 9.78 +- 0.04 or 9.78 +- 4%
I wonder if you mis-read the book and they actually had +- 4% not +- 0.4%

Note: I can only get your 9.79 if I assume pi = 22/7. I can only get their 9.77 if I assume pi = 3.14. Given that my calculator gives pi to about 10 decimal places, I used them all to get 9.78.
 
PeterO said:
Firstly, my error. I should have said you add the percentage errors not multiply them - haven't used percentage errors for a while.

Period T 1.24 +-0.02 means an error of 2 in 124 = 1.6%
Length L = 0.381 +- 0.004 means an error of 4 in 381 = 1.05%

So total error = 1.6 + 1.6 + 1.05 = 4.25%

so 9.78 +- 4% or 9.78 +- .04


I [almost]agree with your numbers, and would express it as 9.78 +- 0.04 or 9.78 +- 4%
I wonder if you mis-read the book and they actually had +- 4% not +- 0.4%

Note: I can only get your 9.79 if I assume pi = 22/7. I can only get their 9.77 if I assume pi = 3.14. Given that my calculator gives pi to about 10 decimal places, I used them all to get 9.78.


No, I wish I did mis-read it. The value is +/- 0.4%. I asked my instructor last night and he didn't give me a reply. He wasn't at school today either. When I get an answer, I'll post it. Thanks for the help.