1. ## maximum error percentage

period T of a simple pendulum with small oscillations is calculated:

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

L = length of pendulum, with 0.5% max error value
g = acceleration due to gravity, with 0.1% max error value

Use differentiation to approximate the max % error in the calculated value of T.

2. Hello,
Originally Posted by Dr Zoidburg
period T of a simple pendulum with small oscillations is calculated:

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

L = length of pendulum, with 0.5% max error value
g = acceleration due to gravity, with 0.1% max error value

Use differentiation to approximate the max % error in the calculated value of T.

We have $T=2\pi \sqrt{\tfrac {L}{g}}$ so

$\ln T=\ln\left( 2\pi\right) + \ln \sqrt{L}-\ln \sqrt{g}=\ln\left( 2\pi\right) + \frac{\ln L}{2}-\frac{\ln g}{2}$.

Differentiate :

$\frac{\mathrm{d}T}{T} = 0 + \frac12\cdot\frac{\mathrm{d}L}{L} - \frac12\cdot\frac{\mathrm{d}g}{g}=\frac12\cdot\fra c{\mathrm{d}L}{L} - \frac12\cdot\frac{\mathrm{d}g}{g}$

Can you take it from here ?

3. I think so.
Is
$\frac{dL}{L} = \frac{1}{2L}$
and
$\frac{dg}{g} = -\frac{1}{2g}$
?

4. Originally Posted by Dr Zoidburg
I think so.
Is
$\frac{dL}{L} = \frac{1}{2L}$
and
$\frac{dg}{g} = -\frac{1}{2g}$
?
No, you're given relative errors, not absolute ones. From

$\frac{\mathrm{d}T}{T}=\frac12\cdot\frac{\mathrm{d} L}{L} - \frac12\cdot\frac{\mathrm{d}g}{g}$

we derive

$\frac{\Delta T}{T}=\frac12\cdot\frac{\Delta L}{L} {\color{blue}+} \frac12\cdot\frac{\Delta g}{g}$

where $\Delta T$, $\Delta L$, $\Delta g$ are the absolute error in $T$, $L$ and $g$, respectively. By definition, the ratios $\tfrac{\Delta T}{T}$, $\tfrac{\Delta L}{L}$ and $\tfrac{\Delta g}{g}$ are relative errors ; as we are given $\tfrac{\Delta L}{L}=0.5\%$ and $\tfrac{\Delta g}{g}=0.1\%$, we can compute the value of $\tfrac{\Delta T}{T}$.