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\begin{document}
\input{epsf.sty}

\title{Springs - Forced Oscillations}
\date{}
\author{}
\maketitle


\section{No damping}

The differential equation we start with is
\begin{equation} \label{ode - no c}
mx'' + kx = \cos \omega t
\end{equation}
We define $\omega_0$ as
\begin{equation}
\omega_0 = \sqrt{\frac{k}{m}}
\end{equation}
This leads to complimentary solution:
\begin{eqnarray}
x_c & = & c_1 \cos\omega_0t + c_2 \sin \omega_0t \\
	& = & C \cos(\omega_0t-\alpha)
\end{eqnarray}

To find a particular solution, we use the method of undetermined coefficients.



\subsection{Assuming $\omega_0 \neq \omega$, beats}

In this case, we assume that our particular solution looks like
\begin{equation}
x_p = A \cos\omega t + B \sin \omega t
\end{equation}
We solve for $A$ and $B$ and see that $B=0$ (not a surprise, we discussed this in class) and 
$A = F_0/(m(\omega_0^2 - \omega^2))$:
\begin{equation}
x_p = \frac{F_0}{m(\omega_0^2 - \omega^2)} \cos \omega t
\end{equation}
and the general solution is:
\begin{eqnarray} 
x & = & x_c + x_p \\
\label{gen sol 1}
	& = & c_1 \cos\omega_0t + c_2 \sin \omega_0t + \frac{F_0}{m(\omega_0^2 - \omega^2)} \cos \omega t 
\end{eqnarray}

\subsubsection{Initial conditions}

Assume that we now have the following initial conditions (the mass starts at rest):
\begin{equation}
x(0)=0 \qquad x'(0)=0
\end{equation}
We can then solve for $c_1$ and $c_2$ in equation~\ref{gen sol 1}.
This gives $c_1 = -F_0/(m(\omega_0^2-\omega^2))$ and $c_2=0$:
\begin{equation} \label{beat one}
x = \frac{F_0}{m(\omega_0^2-\omega^2)} (\cos \omega t - \cos \omega_0 t)
\end{equation}
We now apply a trig trick to $\cos \omega t - \cos \omega_0 t$.
If we let $A=(\omega_0+\omega)t/2$ and $B=(\omega_0-\omega)t/2$ then 
$A+B = \omega_0t$ and $A-B = \omega_t$.
We now apply the trig identity $2\sin A \sin B = \cos(A-B) -cos(A+B)$.
This transforms equation~\ref{beat one} to:
\begin{equation} \label{beat two}
x = \frac{2F_0}{m(\omega_0^2-\omega^2)} \sin \left(\frac{1}{2}(\omega_0 - \omega)t \right) 
	\sin \left(\frac{1}{2}(\omega_0 + \omega)t \right)
\end{equation}

Now, we consider an example where $\omega_0$ and $\omega$ are close in magnitude so that
$\omega_0 - \omega$ is small and $\omega_0+\omega$ is big.
If $m=.1, F_0=50, \omega_0=55, \omega=45$ then our solution (equation~\ref{beat two}) looks like:
\begin{equation}
x = \sin5t \sin50t
\end{equation}
We can graph this to get the following graph:
%
\epsfysize=8 cm
\begin{figure}[htb]
\center{
\leavevmode
\epsfbox{beats.eps}
\caption{The graph $x(t)=\sin5t \sin50t$}
}
\end{figure}

Notice how you see an ``envelopping amplitude.''


\subsection{Resonance}

We now assume that $\omega_0=\omega$.
In this case, $A\cos\omega_0t+B\sin\omega_0t$ will be a solution to the homogeneous equation.
Therefore, we cannot try this for a particular solution.
So, we try the following for a particular solution:
\begin{equation}
x_p = t(A\cos\omega_0t+B\sin\omega_0t)
\end{equation}
We plug this into the differential equation (\ref{ode - no c}) and solve for $A$ and $B$ to get
\begin{equation}
A = 0 \qquad B=\frac{F_0}{2m\omega_0}
\end{equation}
and our general solution looks like:
\begin{eqnarray}
x & = & x_c + x_p \\
	& = &  c_1 \cos\omega_0t + c_2 \sin \omega_0t + \frac{F_0}{2m\omega_0} t \sin\omega_0t \\
	& = & C \cos(\omega_0t-\alpha)+ \frac{F_0}{2m\omega_0} t \sin\omega_0t
\end{eqnarray}

If we set up the problem as:
\begin{eqnarray}
m=.1 & & k=25 \\
F_0 = 10 & & \omega_0 = 50 \\
x(0)=0 && x'(0)=0
\end{eqnarray}
the the resulting equation is:
\begin{equation}
x = t \sin 50t
\end{equation}
If we graph this we see the resonance:
%
\epsfysize=8 cm
\begin{figure}[htb]
\center{
\leavevmode
\epsfbox{resonance.eps}
\caption{The graph $x(t)=t\sin50t$}
}
\end{figure}


\newpage


\section{Damping}

We now assume that our differential equation is of the form
\begin{equation}
mx'' + cx' + kx = \cos \omega t
\end{equation}
The complimentary solution is of one of the following forms:
\begin{eqnarray}
x_c & = & c_1e^{-r_1t} + c_2e^{-r_2t} \\
x_c & = & e^{-r_1t}(c_1+c_2t) \\
x_c & = & e^{-rt_1}{c_1 \cos bt + c_2 \sin b_t}
\end{eqnarray}
where $r_1, r_2 >0$ are positive real numbers (and cannot be zero).
Regardless of which case we are in, we have
\begin{equation}
\lim_{t \ra \infty} x_c = 0
\end{equation}
This solution is called \emph{transient} because it dies out and doesn't affect the 
behavior as $t \ra \infty$.
We will only study the behavior of the particular solution for this reason.

To find the particular solution, we try $x_p = A\cos\omega t+ B\sin\omega t$.
Solving for $A$ and $B$ gives:
\begin{equation}
A = \frac{(k-m\omega^2)F_0}{(k-m\omega^2)^2 + (c\omega)^2}
	\qquad 
B = \frac{c\omega F_0}{(k-m\omega^2)^2 + (c\omega)^2}
\end{equation}
If we apply our usual trig trick to convert $A\cos\omega t+ B\sin\omega t$ to $C \cos(\omega t-\alpha)$ we have
$C^2=A^2+B^2$ and
\begin{equation}
x_p = \frac{F_0}{\sqrt{(k-m\omega^2)^2 + (c\omega)^2}} \cos(\omega t - \alpha)
\end{equation}

Notice that regardless of anything else, the particular solution will be simple periodic.
There will be no resonance.
But, if you notice, the amplitude of the motion is given by $C$.
On can try to maximize this quantity (maximize it as a function of the inputs, especially $\omega$).

Here is an example.
\begin{eqnarray}
m=1 && c=2 \\
k=26 && F_0=82
\end{eqnarray}
This gives the differential equation:
\begin{equation}
x'' + 2x' +26x = 82 \cos\omega t
\end{equation}
The particular solution looks likes:
\begin{equation}
x_p = \frac{82}{\sqrt{\omega^4 - 48 \omega^2+676}} \cos(\omega t - \alpha)
\end{equation}
If we plot the function $C(\omega)$ we get the graph:
\epsfysize=8 cm
\begin{figure}[htb]
\center{
\leavevmode
\epsfbox{p_resonance.eps}
\caption{The graph $x(t)=t\sin50t$}
}
\end{figure}

This shows that even though we won't see resonance, we can still get a very large amplitude.

\end{document}


