next up previous contents
Next: Relations Involving Up: Simple Harmonic Oscillation Previous: Simple Harmonic Oscillation   Contents

Solution

We generally are interested in real part of $x(t)$ when studying oscillating masses, so we'll stick to the following solution:
\begin{displaymath}
x(t) = X_0 \cos(\omega t + \phi)
\end{displaymath} (57)

where $X_0$ is called the amplitude of the oscillation and $\phi$ is called the phase of the oscillation. The amplitude tells you how big the oscillation is, the phase tells you when the oscillator was started relative to your clock (the one that reads $t$). Note that we could have used $\sin(\omega t + \phi)$ as well, or any of several other forms, since $\cos(\theta) = \sin(\theta + \pi/2)$. But you knew that.

$X_0$ and $\phi$ are two unknowns and have to be determined from the initial conditions, the givens of the problem. They are basically constants of integration just like $x_0$ and $v_0$ for the one-dimensional constant acceleration problem. From this we can easily see that:

\begin{displaymath}
v(t) = \frac{dx}{dt} = - \omega X_0 \sin(\omega t + \phi)
\end{displaymath} (58)

and
\begin{displaymath}
a(t) = \frac{d^2x}{dt^2} = - \omega^2 X_0 \cos(\omega t + \phi) =
-\frac{k}{m} x(t)
\end{displaymath} (59)

(where the last relation proves the original differential equation).


next up previous contents
Next: Relations Involving Up: Simple Harmonic Oscillation Previous: Simple Harmonic Oscillation   Contents
Robert G. Brown 2004-04-12