As a lab assignment I have to replicate experiment with spring oscillations, and eventually calculate spring constant. The measurements I took are: spring displacement from equilibrium $x$, mass of weights $m$, and period $T$. And presumed uncertainties: $\Delta_\alpha(x)=1[mm]$, $\Delta_e(x)=2[mm]$, $\Delta_\alpha(t)=0.1[s]$, $\Delta_e(t)=1[s]$. Where $\alpha$ stands for uncertainty of used equipment, and $e$ is readings uncertainty made by me.
I calculated the standard uncertainty using this formula: $$\sqrt\frac{\Delta_\alpha(x)^2+\Delta_e(x)^2}{3}$$
Which produced $u(x)=1.29[mm]$. Then using the following substitutions:
I assert that:
Because in this part of experiment I plotted $x=f(m)$, using excel's linest, I get the slope $s$
Now, my question is, since I had uncertainty for $x$ introduced, and it is used to compute $tan(a)$, then the latter is affected and should have uncertainty of itself? How in this case should I go about propagating uncertainty to $tan(a)$?


The only way to propagate uncertainty through a trig function is to use the expression for general error propagation of an arbitrary function. It's quite easy, actually...

For any function


where $X$ is a set of $n$ parameters

$X = \{x_1, x_2,... x_n\}$

the relative uncertainty in $g(X)$ depends on the error in each parameter $x_i$ via the following form:

$\dfrac{\delta g(X)}{g(X)} = \sqrt{\left( \dfrac{\partial g}{\partial x_1}\cdot\dfrac{\delta x_1}{x_1}\right)^2 + \left( \dfrac{\partial g}{\partial x_2}\cdot\dfrac{\delta x_2}{x_2}\right)^2 + ... \left( \dfrac{\partial g}{\partial x_n}\cdot\dfrac{\delta x_n}{x_n}\right)^2}$

$ = \sqrt{\sum_{i=1}^n \left( \dfrac{\partial g}{\partial x_i}\cdot\dfrac{\delta x_i}{x_i}\right)^2}$

So, in your case, it is simply

$\dfrac{\delta \tan(a)}{\tan(a)} = \sqrt{\left( \dfrac{\partial \tan(a)}{\partial a}\cdot\dfrac{\delta a}{a} \right)^2}$

where of course

$\dfrac{d}{dx}\tan(x) = \sec^2(x)$,

and you should calculate $\delta a$ using the standard rules of error propegation.

EDIT: Sorry, I just realized my notation is different than yours... note that $\delta x = \Delta x$, going from my notation to yours. Let me know if any of this is beyond your knowledge level.

EDIT 2: Everything I said above is correct, but may not actually pertain to your issue at all. Why are you asserting $tan(a) = g/k$? I can't think of any reason that a tangent function should appear anywhere in the context of this experiment.

  • $\begingroup$ Do I understand correctly, for δa, a is arctan of my tan function? $\endgroup$ – Misha.P May 5 '17 at 23:50
  • 1
    $\begingroup$ @Michael.P Wait, I'm sorry, I answered your question too hastily; where is this tangent function even coming from..? Why do you assert $tan(a) = g/k$? Why can't you just solve the spring constant as $mg/x = k$, and propagate the error through that? $\endgroup$ – user97626 May 5 '17 at 23:55
  • $\begingroup$ Also your formula for uncertainty in $x$ doesn't make sense. Why are you dividing by 3? $\endgroup$ – user97626 May 5 '17 at 23:56
  • $\begingroup$ $x=\frac{x}{k}m+0$ is a line equation, y=ax+b. From which it flows that $\frac{x}{k}=tan(\alpha)$. As for why I did not go with mg/x=k, I have no idea $\endgroup$ – Misha.P May 5 '17 at 23:59
  • $\begingroup$ the formula was outlined in a physics scipt $\endgroup$ – Misha.P May 6 '17 at 0:02

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