# Critical damping, overshooting

For an over damped system, the position is given by:

$$x(t) = \frac{1}{2\kappa}e^{-\gamma\cdot t}[(\kappa\,x_0+ \gamma\,x_0 + v_0 )e^{\kappa t}+(\kappa\,x_0 - \gamma\,x_0 - v_0 )e^{-\kappa t}]$$

with $\kappa = \sqrt{\gamma^2 - \omega_0^2}$ and over damping meaning $\gamma > \omega_0$

Now, overshooting happens when:

$$x(t = t_O) = 0 \iff \frac{(\kappa\,x_0+ \gamma\,x_0 + v_0 )}{(\kappa\,x_0 - \gamma\,x_0 - v_0)} =- e^{-2\kappa t_0} \iff -\frac{1}{2\kappa}\ln\left(-\frac{(\kappa\,x_0+ \gamma\,x_0 + v_0 )}{(\kappa\,x_0 - \gamma\,x_0 - v_0)}\right)= t_O$$

Therefore: $$\frac{(\kappa\,x_0+ \gamma\,x_0 + v_0 )}{(\kappa\,x_0 - \gamma\,x_0 - v_0)} < 0 \text{ and }\left|\frac{(\kappa\,x_0+ \gamma\,x_0 + v_0 )}{(\kappa\,x_0 - \gamma\,x_0 - v_0)}\right| < 1$$

Now, according to one of my lectures, overshooting is supposed to happen when $|v_0| > |(\gamma + \kappa)x_0|$

I don't really see how that follows from above though, especially the second condition. For large $t$ we can neglect the second term. Then the condition becomes $(\kappa + \gamma)x_0 + v_0 = 0$; however, I am not sure if we can easily say it like that.

Since the thing inside the absolute value is negative by the first inequality, the second inequality can be written as: $$-\frac{(\kappa\,x_0+ \gamma\,x_0 + v_0 )}{(\kappa\,x_0 - \gamma\,x_0 - v_0)} < 1$$ Doing the algebra and dividing by $x_0$ we obtain: $$\frac{2\kappa}{\kappa\, - \gamma\, - v_0/x_0} > 0$$ The numerator is always positive, so the denominator must be positive. Since $\kappa < \gamma$ we have obtained two results: $$v_0/x_0<0\,\,\,\,\,\,\,\,\mathrm{and}\,\,\,\,\,\,\,\,\kappa\, - \gamma\, - v_0/x_0>0$$ We've exhausted the second inequality. Now turn to the first inequality and again divide by $x_0$: $$\frac{(\kappa+\gamma + v_0/x_0 )}{(\kappa-\gamma - v_0/x_0)} < 0$$ We've just shown that the denominator of this is positive, so the numerator must be negative: $$\kappa+\gamma + v_0/x_0<0$$ Since $\kappa+\gamma>0$ and $v_0/x_0<0$, $$|\kappa+\gamma| - |v_0|/|x_0|<0$$ which is equivalent to the result you're quoting: $$|v_0|>|(\kappa+\gamma)x_0|$$