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Error calculations involving gradients

  • Thread starter acidburner
  • Start date
I'm trying to work out the percentage error in working out the value of gravity,g, from a pendulums motion.
i know that percentage error is (possible error/value used)*100 however im using multiple values multiple times and its getting a little confusing.
In the investigation g=k(∆T²/∆L), my problem is that i have the error for working out T which is 0.001 seconds. The error in L is 0.001m. As i'm using a difference of two values for each of the ∆'s would i double each error and for the error of T² would i square 0.001 and then double it.
any explainations would be helpful
Thanks
∆T²=2.112 ∆L=0.51
 

Answers and Replies

rock.freak667
Homework Helper
6,230
31
You know that

[tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]

so that

[tex]T^2=4\pi^2 \frac{L}{g}[/tex]

and that means that

[tex]g=\frac{4\pi^2}{T^2}L[/tex]

To find the error you do this.

[tex]\frac{\delta g}{g}=2\frac{\delta T}{T} + \frac{\delta L}{L}[/tex]

[itex]\delta T[/itex] would be the error in T and similarly for [itex]\delta L[/itex] is the error in L.
 

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