Question Number 28929 by NECx last updated on 01/Feb/18
$${If}\:{T}=\mathrm{2}\pi\left(\frac{{L}}{{g}}\right)^{\frac{\mathrm{1}}{\mathrm{2}\:}} \:{and} \\ $$$${L}=\mathrm{100}\pm\mathrm{0}.\mathrm{1}\:{cm}\left({limit}\:{standard}\:\right. \\ $$$$\left.{error}\right) \\ $$$${T}=\mathrm{2}.\mathrm{01}\pm\mathrm{0}.\mathrm{01}\:{s}\:\left({limit}\:{standard}\right. \\ $$$$\left.{error}\right) \\ $$$${Calculate}\:{the}\:{value}\:{of}\:{g}\:{and}\:{its} \\ $$$${standard}\:{error}. \\ $$
Answered by Tinkutara last updated on 02/Feb/18
$${g}=\frac{\mathrm{4}\pi^{\mathrm{2}} {L}}{{T}^{\mathrm{2}} } \\ $$$${L}=\mathrm{1}\pm\mathrm{10}^{−\mathrm{3}} \:{m} \\ $$$${T}=\mathrm{2}.\mathrm{01}\pm\mathrm{0}.\mathrm{01}\:{s} \\ $$$${g}\approx\mathrm{9}.\mathrm{77}\:{m}/{s}^{\mathrm{2}} \\ $$$$\frac{\Delta{g}}{{g}}=\frac{\Delta{L}}{{L}}+\frac{\mathrm{2}\Delta{T}}{{T}} \\ $$$$\Delta{g}=\mathrm{9}.\mathrm{77}\left(\frac{\mathrm{10}^{−\mathrm{3}} }{\mathrm{1}}+\frac{\mathrm{2}×\mathrm{0}.\mathrm{01}}{\mathrm{2}.\mathrm{01}}\right)\approx\mathrm{0}.\mathrm{107} \\ $$$$\Rightarrow{g}=\mathrm{9}.\mathrm{77}\pm\mathrm{0}.\mathrm{107}\:{m}/{s}^{\mathrm{2}} \\ $$
Commented by NECx last updated on 03/Feb/18
$${wow}...\:{Thanks} \\ $$
Commented by NECx last updated on 08/Feb/18
$${i}\:{think}\:{this}\:{was}\:{for}\:{maximum} \\ $$$${error}. \\ $$