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

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

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

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

First Order Example

Constant force on horizontal parallel damper and spring, starting at \(t=0\)

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Spring and resistance in parallel
$$ f_{e}(t)=F_0\gamma(t) \label{eq:applied} $$

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 \(c_1\)

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

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

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

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