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Springs

\begin{figure}\centerline{
\psfig{file=oscillations.1.eps,height=2.5in}
}\end{figure}

We will work in one dimension (call it $x$) and will for the time being place the spring equilibrium at the origin. Its equation of motion is thus:


\begin{displaymath}
F = - k x = m a = m \frac{d^2x}{dt^2}
\end{displaymath} (31)

Rearranging:


$\displaystyle \frac{d^2x}{dt^2} + \frac{k}{m} x$ $\textstyle =$ $\displaystyle 0$ (32)
$\displaystyle \frac{d^2x}{dt^2} + \omega^2 x$ $\textstyle =$ $\displaystyle 0$ (33)

where $\omega^2 = k/m$ must have units of inverse time squared (why?).

This latter form is the standard harmonic oscillator equation (of motion). If we solve it once and for all now, whenever we can put an equation of motion into this form in the future we can just read off the solution by identifying similar quantities.

To solve it, we note that it basically tells us that $x(t)$ must be a function that has a second derivative proportional to the function itself. We know at least three functions whose second derivatives are proportional to themselves: cosine, sine and exponential. To learn something very important about the relationship between these functions, we'll assume the exponential form:

\begin{displaymath}
x(t) = x_0 e^{\alpha t}
\end{displaymath} (34)

(where $\alpha$ is an unknown parameter). Substituting this into the differential equation and differentiating, we get:
$\displaystyle \frac{d^2 x_0 e^{\alpha t}}{dt^2} + \omega^2 x_0 e^{\alpha t}$ $\textstyle =$ $\displaystyle 0$ (35)
$\displaystyle \left(\alpha^2 + \omega^2\right) x_0 e^{\alpha t}$ $\textstyle =$ $\displaystyle 0$ (36)
$\displaystyle \left(\alpha^2 + \omega^2\right)$ $\textstyle =$ $\displaystyle 0$ (37)

where the last relation is called the characteristic equation for the differential equation. If we can find an $\alpha$ such that this equation is satisfied, then our assumed answer will indeed solve the D.E.

Clearly:

\begin{displaymath}
\alpha = \pm i \omega
\end{displaymath} (38)

and we get two solutions! We will always get $n$ independent solutions for an $nth$ order differential equation, so this is good:
$\displaystyle x_+(t)$ $\textstyle =$ $\displaystyle x_{0+} e^{+i\omega t}$ (39)
$\displaystyle x_-(t)$ $\textstyle =$ $\displaystyle x_{0-} e^{-i\omega t}$ (40)

and an arbitrary solution can be made up of a sum of these terms:
\begin{displaymath}
x(t) = x_{0+} e^{+i\omega t} + x_{0-} e^{-i\omega t}
\end{displaymath} (41)


next up previous contents
Next: Math: Complex Numbers and Up: Oscillations Previous: Oscillations   Contents
Robert G. Brown 2004-04-12