Solving a differential equations and finding the maximum amplitude
This is a practice exercise from my differential equation class:
Let $y'' + py' + qy = cos (wt)$ where $p,q,w$ are all positive and let $Acos(wt+\alpha)$ denote the particular solution. Find the value of $w$ so that A attains the maximum value.
It did not take too long to find that $ \displaystyle A = \frac {1} {\sqrt {(q-w^2)^2+(pw)^2}} $.
So I thought I can just find the minimum value of $g(w) = (q-w^2)^2+(pw)^2$ because $g(w)$ is always positive.
$g'(w) $
$= 2(q-w^2)*(-2w)+2p^2w$
$=2w(p^2-2(q-w^2))$
$=2w(2w^2+(p^2-2q))$
It is easy to see that $\displaystyle lim_{w \to \infty} g'(w)=\infty$, so there are two possibilities. 1) $g'(w)>0$ for all $w>0$. 2) There exists $w_0 >0$ such that $g'(w_0)=0$.
In the second case it is possible to decide the value of $w$ but if $p^2-2q>0$ then $g'(w)$ will not have minimum value. However the problem does not seem to provide any constraint about $p,q,r$.
What is wrong with my argument? Could you help me figure out how to find the solution?
p.s.
It seems that the A attains the maximum value when $w=0$, but the problem states that $w$ must be positive. If this were exam, how should I answer?
$\endgroup$1 Answer
$\begingroup$$p^2-2q \gt 0$ corresponds to an overdamped oscillator. It will have maximum forced amplitude at $w=0$ as your analysis shows. Intuitively, the resistance will cause it to lag behind the driving function, and the slower the driving function, the closer it comes to matching the amplitude of the driving function.
$\endgroup$ 5
