[Physics] How to find when an LRC circuit is critically damped mathematically, given a set of voltage/time data


In an undergraduate-level experiment to approximate the resistance at which an LRC circuit system is critically damped, I found the resistance range within which the system is likely to be critically damped. The setup used allowed me to vary the resistance used, keeping all other variables (C, V) constant. I found this range by noting the resistance used when the graph (on a display of voyage against time) stopped showing underdamped behavior (when I continually increased the resistance from $0\ \Omega$), and when it stopped showing overdamped behavior (when I decreased the resistance from a very high value).

Now that was settled and I found the range of resistances within which the critical resistance is likely to be within.

However, how could I do this mathematically? Critical damping occurs when the oscillating frequency $\omega= 0$. How do I find mathematically, using any combination of a pair of datasets (voltage, time, resistance, driving frequency, inductance, current) gotten from experiments on changing resistance, the resistance at which critical damping occurs? I thought that, if the frequency $\omega$ should be zero, then some derivative of the voltage/time data should either tend to infinity or be null. I chose to try using a data set of voltage against time.

I have $V(t) = V_0 \cos\omega t$ and $\dot V(t)= -V_0 \omega \sin \omega t = -V/RC\dots$ where $V$ = voltage, $\omega$ = frequency, $R$ = resistance, and $C$ = capacitance.

$$\dot V(t) = -V_0 \omega^2 \sin \omega t = ?$$

How do I find, using the second derivative, the resistance at which critical damping occurs? I certainly am missing something very obvious.

Best Answer

Critical damping occurs when the oscillating frequency ω=0.

Not so.

Assuming a series LCR circuit you have a second order differential equation of the form $L\ddot q +R\dot q +\dfrac q C=0$, where $q$ is the charge, which is of the form $\ddot x + \beta \dot x + \omega_o^2 = 0$.
Note that you can set up equivalent equations for a current $I$.
There are three possibilities:
$\beta^2 > 4 \omega_o^2$ where the solutions are non-oscillatory - system is over damped.
$\beta^2 < 4 \omega_o^2$ where the solutions are oscillatory - system is overdamped.

and then $\beta = 2 \omega_o$ which is the condition for critical damping.
This is the transition frequency between oscillatory and non oscillatory behaviour and it is the condition for which the system when displaced will return to its equilibrium position and stay there in the shortest time.

The solution to the differential is slightly different in that instead of the sum of two exponential functions it is of the form $x = (A + Bt)e^{-\omega_o t}$

In the electrical case $\beta = \dfrac R L$ and $\omega^2_o = \dfrac {1}{LC}$

So the condition for critical damping is $R^2 = \dfrac {4L}{C}$.

Experimentally by trial and error it is probably best to start with a small value of resistance so the system is under damped and then increase the resistance until you observe no overshot.
This is not the easiest of things to do.