Force-Driven Mass Analysis

Note that in the electrical equivalent circuit

• Driving force = voltage source emitting $f(t)$ volts
• Mass = inductor of $L=m$ Henrys.

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

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

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

• $F(s) =$ Laplace transform of the driving force $f(t)$ ,
• $x(0) =$ initial mass position, and
• ${\dot x}(0)\mathrel{\stackrel{\scriptscriptstyle\mathrm{\Delta}}{\scriptstyle=}}v(0) =$ initial mass velocity,

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