Hello,

Quote:

Originally Posted by

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

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

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.

any help gratefully received!

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

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

Differentiate :

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

Can you take it from here ?