\(\)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 $$ \begin{align} x(t)&=\frac{1}{S}F_0+\frac{1}{\cancel{R}}(-\frac{\cancel{R}}{S}F_0)e^{-\frac{S}{R}t} & t\geq0 \nonumber \\ &=\frac{1}{S}F_0\left(1-e^{-\frac{S}{R}t}\right) & t\geq0\\ \end{align} $$