\(\)Using Laplace transforms to solve mechanical ordinary differential equations.
Elements
Before we look at examples of mechanical systems, let’s recall the equations of mechanical element.
Spring
According to Hooke’s law, the reactive force is linear proportional to the displacement and opposite the direction of the force
Resistance
A dashpot provides friction force linear proportional with the velocity and opposite the direction of the force.
Mass
According to Newton’s second law of motion the reactive force is linear proportional to the acceleration and opposite the direction of the force
First Order Example
Constant force on horizontal parallel damper and spring, starting at \(t=0\)
Sum of the forces must be zero. This combine the equations for the external force \(\eqref{eq:applied}\) with the equations for spring \(\eqref{eq:spring}\) and damper \(\eqref{eq:damper}\). $$ \begin{align} f_r(t) + f_s(t) &= f_e(t) \nonumber \\ R\frac{\mathrm{d}x(t)}{\mathrm{d}t} + S x(t) &= F_0\gamma(t) \end{align} $$
Transform this to the ordinary differential equation, knowing that the system starts from rest, therefor \(\frac{\mathrm{d}x(0)}{\mathrm{d}t}=0\) and \(x(0)=0\).
Solve for \(X(s)\) $$ \begin{align} X(s) (s R +S)&=F_0\frac{1}{s} \nonumber \\ \Rightarrow\ X(s)&=\frac{F_0}{s(s R +S)} \end{align} $$
To return back to the time domain \(x(t)\), we need to find a reverse Laplace transform. There is none. According to Heaviside, this can be expressed as partial fractions. [swarthmore] $$ X(s)=\frac{F_0}{s(sR+S)}\equiv\frac{c_0}{s}+\frac{c_1}{sR+S} \label{eq:heaviside} $$
The constants \(c_{0,1}\) are found using Heaviside’s Cover-up Method [swarthmore, MIT-cu]: multiply \(\eqref{eq:heaviside}\) with respectively \(s\) and \((sR-S)\). $$ \left\{ \begin{align} \frac{\cancel{s}F_0}{\cancel{s}(sR+S)} &\equiv\frac{\cancel{s}c_0}{\cancel{s}}+\frac{sc_1}{sR+S} \nonumber\\ \frac{\cancel{(sR+S)}F_0}{\cancel{(sR+S)}s}&\equiv\frac{(sR+S)c_0}{s}+\frac{\cancel{(sR+S)}c_1}{\cancel{sR+S}}\nonumber \end{align} \right. $$
Given that these equations are true for any value of \(s\), choose two convenient values to find \(c_0\) and \(c_1\) $$ \left\{ \begin{eqnarray} c_0=\left.\frac{F_0}{sR+S}-\frac{s\,c_1}{sR+S}\right|_{s=0}&=\frac{F_0}{S}-\frac{0}{0+S}&=\frac{1}{S}F_0\label{eq:constants1}\\ c_1=\left.\frac{F_0}{s}-\frac{(sR+S)c_0}{s}\right|_{s=-\frac{S}{R}}&=-\frac{R}{S}F_0-\frac{0}{-\frac{S}{R}}&=-\frac{R}{S}F_0\nonumber \end{eqnarray} \right. $$
The unit step response \(x(t)\) follows from the inverse
of \(\eqref{eq:heaviside}\) $$ \begin{align} x(t)&= \mathcal{L}^{-1}\left\{\frac{c_0}{s}\right\}+\mathcal{L}^{-1}\left\{\frac{c_1}{sR+S}\right\} & t\geq0 \nonumber \\ &= \mathcal{L}^{-1}\left\{\frac{c_0}{s}\right\}+\frac{1}{R}\mathcal{L}^{-1}\left\{\frac{c_1}{s+\frac{S}{R}}\right\} & t\geq0 \nonumber \\ &= c_0+\frac{1}{R}c_1e^{-\frac{S}{R}t} & t\geq0 \end{align} $$Substituting the constants \(\eqref{eq:constants1}\) gives the unit step response $$ \begin{align} x(t)&=\frac{1}{S}F_0+\frac{1}{\cancel{R}}(-\frac{\cancel{R}}{S}F_0)e^{-\frac{S}{R}t} & t\geq0 \nonumber \\ &=\frac{1}{S}F_0\left(1-e^{-\frac{S}{R}t}\right) & t\geq0\\ \end{align} $$