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

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 $$\phi = \pi/2$$.