Next  |  Prev  |  Up  |  Top  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search

Force-Driven Mass Analysis

Note that in the electrical equivalent circuit

From Newton's second law of motion ``$ f=ma$ '', we have

$\displaystyle f(t) = m\,a(t) \mathrel{\stackrel{\mathrm{\Delta}}{=}}m\,{\dot v}(t) \mathrel{\stackrel{\mathrm{\Delta}}{=}}m\,{\ddot x}(t).
$

Taking the unilateral Laplace transform and applying the differentiation theorem twice yields

\begin{eqnarray*}
F(s) &=& m\,{\cal L}_s\{{\ddot x}\}\\
&=& m\left[\,s {\cal L}_s\{{\dot x}\} - {\dot x}(0)\right]\\
&=& m\left\{\,s \left[s\,X(s) - x(0)\right] - {\dot x}(0)\right\}\\
&=& m\left[s^2\,X(s) - s\,x(0) - {\dot x}(0)\right].
\end{eqnarray*}

Thus, given

we can solve algebraically for $ X(s)$ , the Laplace transform of the mass position for all $ t\ge 0$


Next  |  Prev  |  Up  |  Top  |  JOS Index  |  JOS Pubs  |  JOS Home  |  Search

Download Laplace.pdf
Download Laplace_2up.pdf
Download Laplace_4up.pdf

``The Laplace Transform'', by Julius O. Smith III, (From Lecture Overheads, Music 420).
Copyright © 2015-03-30 by Julius O. Smith III
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
CCRMA  [Automatic-links disclaimer]