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Error calculations involving gradients 
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#1
Mar3108, 10:09 PM

P: 1

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(∆Tē/∆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 Tē would i square 0.001 and then double it. any explainations would be helpful Thanks ∆Tē=2.112 ∆L=0.51 


#2
Mar3108, 10:24 PM

HW Helper
P: 6,202

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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