# Uncertainty cannot be calculated?

I'm doing an experiment on resonance. The phase difference between the driving force and the one oscillating is given by $$\varphi=\arcsin\left(\frac{y_1}{y_2}\right)$$ where $$y_1$$ and $$y_2$$ are measurements of voltages. At resonance $$y_1=y_2$$ so $$φ=\frac{π}{2}$$.

However, I'm trying to calculate the uncertainty of $$\varphi$$ , I need to take the partial derivative of $$\varphi$$ with respect to $$y_1$$ and then $$y_2$$. In both cases , I end up with something in the form of $$\frac{\partial\varphi}{\partial y_1}\sim\frac{1}{\sqrt{1-\left(\frac{y_1}{y_2}\right)^2}}$$ multiplied by some other constants which don't really matter. So at resonance, my denominator will be zero. Any thoughts?

• To give you a rough idea, I have an LCR circuit and I'm measuring the voltage across the capacitor for low,medium and high damping (i.e. low,medium and high resistor). Using an oscilloscope, I can choose two displays : 1) one , displaying two sine waves 2) one displaying an ellipse. I could really go into more detail but I hope you get the idea. So , when I'm using the ellipse display , $y_1$ is the distance between the points where my ellipse crosses my y axes and $y_2$ is the vertical distance between the maximum and minimum point of my ellipse. Hope that makes sense :) – Jim Β Oct 11 '18 at 20:37
• I'm a 2nd year physics student . The fact that $φ=sin(\frac{y_1}{y_2})$ is given on the script. – Jim Β Oct 11 '18 at 20:39
• Looks like you might have a typo then - Note that $\frac{\pi}{2} \neq \sin\left(\frac{1}{1}\right)$. That should definitely be $\sin^{-1}$. – J. Murray Oct 11 '18 at 20:42
• Oh yeah sorry , my bad , what I meant is $φ=arcsin(\frac{y_1}{y_2})$. I'll edit my post now . So, for resonance where $y_1=y_2$ , I get $φ=\frac{π}{2}$ . Thanks for pointing out my typo , it didnt make any sense. – Jim Β Oct 11 '18 at 20:45
• You are trying to find experimental uncertainty for a theoretical situation? I would assume you have some way to see resonance in the experiment and that you will have values of $y_1$ and $y_2$ which are not exactly equal to each other. Otherwise you seem to have a perfect experiment – Triatticus Oct 11 '18 at 20:59

The idea that you can use partial derivatives multiplied by the uncertainty in the independent variable(s) to estimates uncertainty in the dependent variable is just an approximation.

It is related to a Taylor series expansion with an implicit assumption that the terms involving the second derivatives and higher can be neglected.

Obviously, for a sinusoidal function this isn't going to be true near $$\pi/2$$ since the first derivative becomes small.

A better way to proceed in your case is to do a Monte-Carlo simulation, allow $$y_1$$ and $$y_2$$ to be generated at random values determined by their own probability distributions and uncertainties, and then calculate the consequent values of $$\phi$$ and build up a probability distribution of $$\phi$$ from which an uncertainty can be calculated.

Edit: in response to your edited function, you now have $$\frac{y_1}{y_2} = \sin \phi$$ Thus $$\frac{\partial y_1}{\partial \phi} = y_2 \cos \phi$$ and the problem I mentioned above clearly applies at $$\phi = \pi/2$$.