Measurement uncertainty

1. Oct 15, 2005

Vegeta

The following question is about an experiment with the conical pendulum.
I have measured the length $l = 43\,cm$ the radius $r = 10\,cm$, I have 3 measurement of the period with the same radius, where the measurements are

$$T= [1.288, 1.285, 1.301] s$$

The uncertainties of the measurements are $\Delta l = \pm 1\,cm$, $\Delta r = \pm 1\,cm$ and $\Delta T = \pm 0.02\,s$.
I want to calculate the uncertainty in the measurement og $g$, when

$$g_i=4\pi^2\frac{\sqrt{l^2-r^2}}{T_i^2}$$

Can I calculate the uncertainty $\Delta g$ by

$$\Delta g = 4\pi^2\frac{\sqrt{(l+\Delta l)^2-(r-\Delta r)^2}}{(T_i-\Delta T)^2} - g_i$$

Where the expression $4\pi^2\left(\sqrt{(l+\Delta l)^2-(r-\Delta r)^2}\right)/(T_i-\Delta T)^2$ is the worst case scenario of the measuring $g$. Is that correct?
If that is how I can calculate the uncertainty in $g$, is the relative uncertainty then

$$\frac{\Delta g}{g_i} \qquad \mathrm{or} \qquad \frac{\Delta g}{\overline{g}}$$

Where $\overline{g}$ is the mean value. Which one is the correct one? The first expression has a relative uncertainty for each measurement.

Last edited: Oct 15, 2005
2. Oct 15, 2005

HallsofIvy

Yes, calculating the value for the largest possible and then smallest possible values of l, r, and T gives the possible error for the function.

The relative uncertainty is $$\frac{\Delta g}{g}$$.

3. Oct 15, 2005

Vegeta

Well in the equation

$$\Delta g_i = 4\pi^2\frac{\sqrt{(l+\Delta l)^2-(r-\Delta r)^2}}{(T_i-\Delta T)^2} - g_i$$

There is acctually an uncertainty of g, for every measurement $\Delta g_i$. Should I say

$$\frac{\Delta g_i}{g_i} \qquad \mathrm{or} \qquad \frac{mean(\Delta g_i)}{mean(g_i)}$$?

Where in the last equation I only have one value for the relative error. Im not quite certain of what to choose?

Last edited: Oct 15, 2005
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