# Error propagation in pendulum ?

May 2, 2017

See below.

#### Explanation:

We have

$T = 2 \pi \sqrt{\frac{l}{g}}$ Applying $\log$ to both sides

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

Deriving totally

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

or

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

which is the formula for error propagation concerning $T$ or

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

May 2, 2017

The previous post demonstrates the effect of the measurement errors that can creep in if you are predicting the period using the given formula $T = 2 \pi \sqrt{\text{l/g}}$.

Another thing to bear in mind is the linearisation that occurs in deriving that formula.

In pendulum motion the DE is:

$\ddot{\theta} = - \frac{g}{l} \textcolor{red}{\sin \theta}$

That is not easily solveable; it's non-linear and requires the use of non elementary functions. But, because for small $\theta$, we can say that $\theta \approx \sin \theta$, we can re-write the DE as:

$\ddot{\theta} = - \frac{g}{l} \theta$

Much easier, that is now simple harmonic motion, and the approximation is what tells us that $T = 2 \pi \sqrt{\text{l/g}}$.

Working out how much this assumption costs in terms of accuracy is beyond my abilities, but there is much on the web on the subject.

But it's also not clear from your question if you're actually doing this using a stopwatch, because that's a totally different question.

Reaction time etc....reliability kicks in. Sure, you can do it over a long time-period, using averaging to reduce the effect of one-off reaction-time-type errors.

But you also have to think about entropy and friction. In that case, theory predicts a DE of: $\ddot{\theta} + \text{b/m" \ dot theta +"g/l} \setminus \theta = 0$, for the usual type of frictional resistance $b$, so that:

T= (2 pi)/{\sqrt((g)/(l)-(b^2)/(4color(red)(m)^2))

So, we think the period will reduce over time due to friction! Of course it will, that is what friction does. The reason for mentioning this bit is that at long last mass does seem to make a difference. In our predictive model, a sufficiently heavy pendulum can begin to annul the effect of a simply-imagined friction fudge-factor, $b$.