1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

It's slightly annoying, but the result you state can be obtained from the two inequalities that you have correctly obtained.

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: $$ |v_0|>|(\kappa+\gamma)x_0|$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.