# Pendulum errors ?

Apr 11, 2017

$T = 2.9954$[s]

#### Explanation:

The pendulum period for small oscillations is given by

$T = 2 \pi \sqrt{\frac{l}{g}}$

applying the logarithm $\log$ to both sides

$\log \left(T\right) = \log \left(2 \pi\right) + \frac{1}{2} \left(\log l - \log g\right)$

deriving totally both sides

$\frac{\mathrm{dT}}{T} = \frac{1}{2} \left(\frac{\mathrm{dl}}{l} - \frac{\mathrm{dg}}{g}\right)$ or for finite differences

$\frac{\Delta T}{T} = \frac{1}{2} \left(\frac{\Delta l}{l} - \frac{\Delta g}{g}\right)$

so we have:

$\Delta l = 0$
$\Delta g = 9.82 - 9.79 = 0.03$

and consequently

$\Delta T = - \frac{T}{2} \left(\frac{\Delta g}{g}\right) = - 0.00459653$

so at the new location the period was shortened and now is

$T = 2.9954$[s]

The claimed precision regarding $T$ has not sense because the precision in $g$ is much lesser than the demanded precision on $T$.

Apr 11, 2017

The new period is $= 2.99541 s$

#### Explanation:

The period of a pendulum is

$T = 2 \pi \sqrt{\frac{l}{g}}$

In the first place, we have

${T}_{1} = 3.00000 = 2 \pi \sqrt{\frac{l}{9.79}}$

In the second place, we have

${T}_{2} = 2 \pi \sqrt{\frac{l}{9.82}}$

Therefore,

${T}_{2} / 3.00000 = \sqrt{\frac{9.79}{9.82}}$

${T}_{2} = 3.00000 \cdot \sqrt{\frac{9.79}{9.82}}$

$= 2.99541$