# Mechanical systems

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

$$f_s(t)=S\cdot x(t) \label{eq:spring}$$ where $$S$$ is the spring stiffness [N/m].

### Resistance

A dashpot provides friction force linear proportional with the velocity and opposite the direction of the force.

$$f_r(t)=R\cdot \frac{\mathrm{d}x(t)}{\mathrm{d}t} \label{eq:damper}$$ where $$R$$ is the resistance, or damping coefficient [Ns/m]

### 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

$$f_m(t)=M\cdot \frac{\mathrm{d}^2x(t)}{\mathrm{d}t^2} \label{eq:mass}$$ where $$M$$ is the mass [kg]

## First Order Example

Constant force on horizontal parallel damper and spring, starting at $$t=0$$

$$f_{e}(t) = F_0\gamma(t) \label{eq:applied}$$

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 $$x(0)=0$$.

$$R\,s\,X(s)+ S X(s) = F_0\frac{1}{s}$$

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 Laplace transform 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}