Question

To find the value of g using simple pendulum, $T=2.00s$ and $l=1.00m$ were measured. Estimate maximum permissible error in g. Also find the value of g.

Answer 1

$T=2\pi \sqrt{\frac{l}{g}}\Rightarrow g=\frac{4{\pi }^{2}l}{T^2}$
${\left(\frac{dg}{g}\right)}_{max}=\frac{△l}{l}+2\frac{△T}{T}=(\frac{0.01}{1.00}+2\frac{0.01}{2.00})\times 100\%=2\%$
$g=\frac{4{\pi }^{2}l}{T^2}=\frac{4\times 10\times 100}{(2.00{)}^2}=10.0m{s}^{-2}$
${\left(\frac{dg}{g}\right)}_{max}=\frac{2}{100}or\frac{d{g}_{max}}{10.0}=\frac{2}{100}$
$\left(dg\right)max=0.2$ = max error in g
So $g=(10.0\pm 0.2)m{s}^{-2}$