\magnification=\magstep1
\centerline{Mathematics 304 -- Ordinary Differential Equations}
\centerline{Lab/Problem Set 7 -- Solutions and Comments}
\centerline{November 8, 2004}
\bigskip
\noindent
B)  When $R = 0$, the equation becomes 
$$LC v'' + v = a \sin(\omega t)$$
The natural frequency of the solutions of the corresponding
homogeneous equation
$$LC v'' + v = 0$$
comes from the roots of the characteristic polynomial:
$LC \lambda^2 + 1 =0$, so $\lambda = \pm i {1\over \sqrt{LC}}$.
If $\omega \ne {1 \over \sqrt{LC}}$, then as particular solution
we take $v_p(t) = A \cos(\omega t) + B \sin(\omega t)$.
Then 
$$LC v_p''(t) + v_p(t) = (1-LC \omega^2) A \cos(\omega t) + (1 - LC\omega^2)\sin(\omega t)$$
Setting this equal to $a \sin(\omega t)$, we get
$$B = {a \over 1 - LC \omega^2}, \qquad A = 0.$$
So 
$$v_p(t) = {a \over 1 - LC\omega^2} \sin(\omega t).$$


If $\omega = {1\over \sqrt{LC}}$, then we must take a different 
form for the particular solution:
$$v_p(t) = A t\cos(\omega t) + B t\sin(\omega t)$$
Then 
$$
\eqalign{
v_p'(t) &= (A + B\omega t)\cos(\omega t) + (B - A\omega t)\sin(\omega t)\cr
v_p''(t) &= (2B\omega - A\omega^2 t) \cos(\omega t) + (-2A\omega - B\omega^2 t)\sin(\omega t)
\cr}$$
So using $\omega^2 LC + 1 = 0$, we get
$$LC v_p'' + v_p = 2B LC \omega \cos(\omega t) - 2A LC \omega \sin(\omega t) = a \sin(\omega t)$$
which implies:
$$B = 0, \qquad A = {-a \over 2 LC \omega} =  {-a \over 2 \sqrt{LC}}$$
The particular solution in this case
looks like
$$v_p(t) = {a t \over 2\sqrt{LC}} \cos(\omega t).$$

\noindent
Comment: It is the extra factor of $t$ that produces the resonance when 
$\omega = {1\over \sqrt{LC}}$.  The amplitude of the oscillations of $v_p(t)$
increases without bound as $t \to \infty$.  In a physical system, of course,
this would not be possible because the object undergoing this type
of oscillation would never be able to withstand it forever; eventually
something must ``break'' and the situation described by the differential equation
would not hold any longer.   Resonant situations tend to be ``very bad'' 
for this reason.  With sound vibrations, for instance, it is possible for
a very strong singer to create resonance in objects by hitting exactly the
right frequency.  It is possible to shatter glasses, windows, etc. this way(!)

\bigskip
\noindent
E)  Does resonance also occur in the presence of damping ($R \ne 0$), say
in the underdamped case?  
For the damped equation, 
$$LC v'' + RC v' + v = a \sin(\omega t),$$
the characteristic polynomial is 
$$LC \lambda^2 + RC \lambda + 1 = 0$$
and 
$$\lambda = {-R\over 2L} \pm {\sqrt{(RC)^2 - 4LC} \over 2LC}.$$
Write $\alpha = -R/(2L)$ and $\beta = {\sqrt{4LC - (RC)^2} \over 2LC}$.
We see that the solutions of the homogenous
equation are 
$$v(t) = c_1 e^{\alpha t} \cos(\beta t) + c_2 e^{\alpha t} \sin(\beta t)$$
and $\sin(\omega t)$ is {\it never a solution} (because it does not contain
the exponential factor).  This means that a particular solution of the 
form 
$$v_p(t) = A \cos(\omega t) + B \sin(\omega t)$$
{\it always exists} (without the extra factor of t).  Thus, we never
get the resonant solutions with unbounded amplitude as $t \to +\infty$.
On the other hand, the amplitude of the particular solution (the 
steady state - what remains after the transients decay to close to 
zero) can be significantly larger than $a$ (the amplitude
of the forcing term).  The formulas work out like this:    
$$LC v_p'' + RC v_p' + v_p = ((1 - LC \omega^2) A + RC \omega B)\cos(\omega t) + ((1 - LC \omega^2) B - RC\omega A)\sin(\omega t)$$
So
$$\eqalign{(1 - LC \omega^2) A + RC \omega B &= 0\cr
                   - RC \omega A + ((1 - LC \omega^2) B &= a\cr}$$
By Cramer's Rule, we get
$$A = {-a \omega RC \over (1 - LC \omega^2)^2 + (RC\omega)^2}; \qquad
B = {a(1 - LC\omega^2) \over (1 - LC \omega^2)^2 + (RC\omega)^2}.$$
The amplitude of $v_p$ is
$$\sqrt{A^2 + B^2} = {a \over \sqrt{(1 - LC \omega^2)^2 + (RC\omega)^2}}$$
which has a {\it maximum} at $\omega = {1\over \sqrt{LC}}$.
\bye  
