91Ó°ÊÓ

Over damped$R^2>\frac{4L}{C}$

Critical damped$R^2=\frac{4L}{C}$

Under damped $R^2<\frac{4L}{C}$

Auxiliary equation:

It is an algebraic equation of degreeupon which depends the solution of a given nth-order differential equation or difference equation.

Take eq. (5.33) and solve it (notice that the solution of eq. (5.34) is like what to do now). Eq. (5.33) is

$\begin{array}{l}L\frac{d^{2}q}{d{t}^2}+R\frac{dq}{dt}+\frac{1}{C}q=0\\ \frac{d^{2}q}{d{t}^2}+\frac{R}{L}\frac{dq}{dt}+\frac{1}{LC}q=0\end{array}$

Nowlet ${\mathrm{Ó\neg }}^2=1/LC$, and $2b=R/L$, and use the differential operator $D=d/dt$ the differential equation will become

$(D^2+2bD+{\mathrm{Ó\neg }}^2)q=0$

Now need to find the root of the auxiliary equation$D^2+2bD+{\mathrm{Ó\neg }}^2=0$,

$\begin{array}{c}D=\frac{−2bÂ\pm \sqrt{4b^2−4{\mathrm{Ó\neg }}^2}}{2}\\ \text{ }D=−bÂ\pm \sqrt{b^2−{\mathrm{Ó\neg }}^2}\end{array}$

There are three possible cases,

The first one when$b^2>{\mathrm{Ó\neg }}^2$and called it overdamped oscillations (notice that the auxiliary equation has two different roots). In this case$\sqrt{b^2−{\mathrm{Ó\neg }}^2}$ is real, and so the general solution would be in the form of eq.(5.11).

$q=A{e}^{−(b−\sqrt{b^2−{\mathrm{Ó\neg }}^2})t}+B{e}^{−(b+\sqrt{b^2−{\mathrm{Ó\neg }}^2})t}$

-The second one is when$b^2={\mathrm{Ó\neg }}^2$and called it critically damped oscillations, but the auxiliary equation has only one root, so the solution of form of eq.(5.17)

$q=e^{−bt}[A\sin \left(\sqrt{b^2−{\mathrm{Ó\neg }}^2}t\right)+B\cos \left(\sqrt{b^2−{\mathrm{Ó\neg }}^2}t\right)]$

In terms of,$R,L$and $C$, for the overdamped motion the condition is $R^{2}C>4L$, and the solution for is

$q=A{e}^{−(\frac{R}{2L}−\sqrt{\frac{R^2}{4L^2}−\frac{1}{LC}})t}+B{e}^{(\frac{R}{2L}+\sqrt{\frac{R^2}{4L^2}−\frac{1}{LC}})t}$

For the critically damped oscillations the condition is $R^{2}C=4L$, and the solution is

$q=(A+Bt)e^{−\frac{R}{2L}t}$

Here for the underdamped damped oscillations the condition is $R^{2}C<4L$, and the solution is

localid="1664292088513" $q=e^{−\frac{R}{2L}t}[A\sin \left(t\sqrt{\frac{R^2}{4L^2}−\frac{1}{L^2{C}^2}}\right)+B\cos \left(t\sqrt{\frac{R^2}{4L^2}−\frac{1}{L^2{C}^2}}\right)]$