Motion from Initial Conditions

From the equivalent circuit, we derived

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

where

$$\begin{eqnarray*} F_{R}(s) &=& R\,V(s)\\ F_m(s) &=& ms\,V(s) - m\,v_0 \end{eqnarray*}$$

Substituting gives

$$m\,s\,V(s) - m\,v_0 + 2R\,V(s) = 0$$

Solving for $V(s)$ gives

$$\zbox{V(s) = \frac{m\,v_0}{ms + 2R}}$$

Since

$$e^{-at}u(t) \;\longleftrightarrow\; \frac{1}{s+a}$$

(where $u(t) =$ unit step function), we find the velocity of the mass-string contact point to be

$$\zbox{v(t) = v_0\, e^{-{\frac{2R}{m}t}}, \quad t\ge 0}$$