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In one of my labs I had to calculate the an expression for a constant $m$, where $m$ is defined as $m = \frac{T}{L^2v^2}$, with $T$ being tension (constant), $L$ being length and $v$ being frequency. Now in order to do this accurately, I found 6 values for $L$ and $v$ - the tension was constant for all of these values. Then I used excel to plot a graph of $T$ against $L^2v^2$ and the gradient of the line of best fit gave me my value for $m$. Now the thing that I'm having problems with is finding out if it's possible for me to find the uncertainty on $m$. I know that I can find the uncertainty for $m$ if I know the uncertainties for $T$, $L^2$ and $v^2$ using the formula below:

$$\left. \begin{array}{l} Z = A \times B \\ or \\ Z = \frac{A}{B}\\ \end{array} \right\} \Rightarrow \biggl (\frac{\Delta Z}{Z} \biggr )^2 = \biggl (\frac{\Delta A}{A} \biggr )^2 + \biggl (\frac{\Delta B}{B} \biggr )^2$$

I'm able to calculate the uncertainty on $T$ but I'm not sure if I can calculate the uncertainties for $v^2$ and $L^2$. I know that the uncertainties for each value of $v$ and $L$ were $\pm 0.1$ Hz and $\pm 0.05$ Hz. However, in order to calculate the uncertainties for $v^2$ and $L^2$ I need to use the formula:

$$Z = A^n \Rightarrow \frac{\Delta Z}{Z} = n \biggr ( \frac{\Delta A}{A} \biggl )$$

which requires values for $v$ and $L$, which I do not have since I used the gradient of the line of best fit to find $m$ and not specific values of $L$ and $v$. So I'm wondering if it would be sensible for me to perhaps use the average of my values for $L$ and $v$ in order to calculate the uncertainties for $L^2$ and $v^2$ or if it is sensible at all to even calculate an uncertainty for $m$ in this situation.

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  • $\begingroup$ You can estimate the uncertainty in the slope from the deviations of the measured points to the fitted line. $\endgroup$
    – user137289
    Commented Mar 5, 2018 at 17:14
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    $\begingroup$ In general, the uncertainty of $Z$ (let it be $\Delta z$) is defined as $\Delta z=\Sigma_{i}\big|\frac{\partial Z}{\partial x_i}\big|\Delta x_i$ if $Z=Z(\{x_i\})$ $\endgroup$
    – MatMorPau22
    Commented Mar 5, 2018 at 17:21
  • $\begingroup$ @Pieter How exactly would I do that? Is there a reference for that sort of technique? $\endgroup$
    – Hai
    Commented Mar 5, 2018 at 17:22
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    $\begingroup$ Method of least squares, minimizing the sum of the squares of deviations. For a straight line it is called linear regression. These are terms you can google. An alternative is to make a graphic estimate by eye, see what slope is roughly consistent with data points. $\endgroup$
    – user137289
    Commented Mar 5, 2018 at 17:26
  • $\begingroup$ Closely related: physics.stackexchange.com/q/772859/226902, physics.stackexchange.com/q/98584/226902 $\endgroup$
    – Quillo
    Commented Jul 21, 2023 at 13:13

1 Answer 1

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Absolute error of function is defined as : $$ \Delta_f = \sum_n \left| \frac{\partial f(x_1,x_2,\ldots, x_n)} {~\partial x_n} \right| \cdot \Delta {x_n} $$

And relative error of function is just : $$ \delta_f = \frac{\Delta_f}{\left|f(x_1,x_2,\ldots, x_n)\right|} $$

So for your function : $$ m = \frac{T}{L^2v^2} $$ with two variables $L,v$, find partial derivatives with respect to them and substitute into first formula for absolute error :

$$ \Delta_m = \frac {2~T}{L^3 v^2} \Delta L ~+~ \frac{2~T}{L^2 v^3}\Delta v $$

Substitute average values of $L,v,T$ and measurement errors $\Delta L,\Delta v$ and you will get $m(L,v)$ function absolute error. If you need a relative error, then according to second formula - just divide found absolute error by $m$ function value.

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