Laplace Transform Analysis

From the equivalent circuit, we derived

$$F_m(s) + 2F_{R}(s) = 0,$$

where

• $F_m(s)$ = Laplace transform of upward mass force
• $F_{R}(s)$ = Laplace transform of upward force on each string segment
• Recall:
  $$F_m(s) \mathrel{\stackrel{\mathrm{\Delta}}{=}}{\cal L}_s\{f_m\} \mathrel{\stackrel{\mathrm{\Delta}}{=}}\int_{0}^\infty f_m(t) e^{-st}dt$$

Equations of motion:

• String:
  $$\zbox{f_{R}(t) = R\,v(t) \quad\leftrightarrow\quad F_{R}(s) = R\,V(s)}$$
  where $V(s)\mathrel{\stackrel{\mathrm{\Delta}}{=}}{\cal L}_s\{v\}$ is the Laplace transform of $v(t)$

• Mass:
  $$\zbox{f_m(t) = m\,a(t)\;=\; m\,\dot v(t) \;\leftrightarrow\; F_m(s) = m\left[s\,V(s) - v_0\right]}$$
  (recall the Laplace-transform differentiation theorem)