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# Laplace in 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

$$f_s(t)=S\cdot x(t)\label{eq:spring}\tag{1}$$

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.

$$f_r(t)=R\cdot \frac{\mathrm{d}x(t)}{\mathrm{d}t}\label{eq:damper}\tag{2}$$

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

$$f_m(t)=M\cdot \frac{\mathrm{d}^2x(t)}{\mathrm{d}t^2}\label{eq:mass}\tag{3}$$

where $$M$$ is the mass [kg]

### First Order Example

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

$$f_{e}(t)=F_0\gamma(t)\label{eq:applied}\tag{4}$$

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{aligned} f_r(t) + f_s(t)&=f_e(t) \\ R\frac{\mathrm{d}x(t)}{\mathrm{d}t} + S x(t) &= F_0\gamma(t) \end{aligned}

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{aligned} X(s) (s R +S)&=F_0\frac{1}{s} \\ \Rightarrow\ X(s)&=\frac{F_0}{s(s R +S)} \end{aligned}

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]

\begin{aligned} X(s)&=\frac{F_0}{s(sR+S)}\equiv\frac{c_0}{s}+\frac{c_1}{sR+S} \end{aligned}\label{eq:heaviside}\tag{4}

The constants $$c_{0,1}$$ are found using Heaviside’s Cover-up Method [swarthmoreMIT-cu]:  multiply $$\eqref{eq:heaviside}$$ with respectively $$s$$  and $$(sR-S)$$.

\require{cancel} \left\{ \begin{aligned} \frac{\cancel{s}F_0}{\cancel{s}(sR+S)} &\equiv\frac{\cancel{s}c_0}{\cancel{s}}+\frac{sc_1}{sR+S} \\ \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}} \end{aligned} \right.

Given that these equations are true for any value of $$s$$, choose two convenient values to find $$c_0$$ and $$c_1$$

$$\require{cancel} \left\{ \begin{array}{ll} c_0=\left.\frac{F_0}{sR+S}-\frac{s\,c_1}{sR+S}\right|_{s=0}&=\frac{F_0}{S}-\frac{0}{0+S}&=\frac{1}{S}F_0\\ c_1=\left.\frac{F_0}{s}-\frac{(sR+S)c_0}{s}\right|_{s=-\frac{S}{R}}&=-\frac{R}{S}F_0-\frac{0}{-\frac{S}{R}}&=-\frac{R}{S}F_0 \end{array} \right. \label{eq:constants1}\tag{5}$$

The unit step response $$x(t)$$ follows from the inverse Laplace transform of $$\eqref{eq:heaviside}$$

\begin{aligned} x(t)&= \mathcal{L}^{-1}\left\{\frac{c_0}{s}\right\}+\mathcal{L}^{-1}\left\{\frac{c_1}{sR+S}\right\} & t\geq0\\ &= \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\\ &=c_0+\frac{1}{R}c_1e^{-\frac{S}{R}t} & t\geq0 \end{aligned}

Substituting the constants $$\eqref{eq:constants1}$$ gives the unit step response

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

### Coert Vonk

Independent Firmware Engineer at Los Altos, CA
Welcome to the things that I couldn’t find.This blog shares some of the notes that I took while deep diving into various fields.Many such endeavors were triggered by curious inquiries from students. Even though the notes often cover a broader area, the key goal is to help the them adopt, flourish and inspire them to invent new technology.