Laplace transforms: Critically Damped Spring Mass system

Using the Laplace Transform to solve a spring mass system that is critically damped

Problem Statement

An 98 Newton weight is attached to a spring with a spring constant k of 40 N/m. The spring is stretched 4 m and rests at its equilibrium position. It is then released from rest with an initial upward velocity of 2 m/s. The system contains a damping force of 40 times the initial velocity.

Solution

Given

$m=\frac{98}{9.81}kg$

$\text{Spring Constant k=40}N/m$

$\text{Damping Constant C=40}$

$\text{x(0)=0}$

$\dot{x}(0)=-4$

$\text{Standard equation: }$

$m\frac{d^{2}x}{d{t}^2}+C\frac{dx}{dt}+khx=0$

Solving the problem

$\text{Therefore the equation representing this system is.}$

$\frac{98}{9.8}\frac{d^{2}x}{d{t}^2}=-40x-40\frac{dx}{dt}$

$\text{Now we put the equation in standard form}$

$\frac{d^{2}x}{d{t}^2}+\frac{40}{10}\frac{dx}{dt}+\frac{40}{10}x=0$

$\text{Now that we have the equation written in standard form we need to send}$ $\text{it through the Laplace Transform.}$

${ℒ}[\frac{d^{2}x}{d{t}^2}+\frac{40}{10}\frac{dx}{dt}+\frac{20}{5}x]$

$\text{And we get the equation (after some substitution and simplification)}.$

${\mathbf{s}}^{2}X(s)+4\mathbf{s}X(s)+4\mathbf{X}(s)=-4$

$\mathbf{X}(s)(s^2+4s+4)=-4$

$\mathbf{X}(s)=-\frac{4}{(s+2{)}^2}$

$\text{Now that we have completed the Laplace Transform}$ $\text{and solved for X(s) we must so an inverse Laplace Transform. }$

${{ℒ}}^{-1}[-\frac{4}{(s+2{)}^2}]$

$\text{and we get}$

$\mathbf{x}(t)=-4t{e}^{-2t}$

$\text{So there you have it the equation of a Critically Damped spring mass system.}$

Apply the Initial and Final Value Theorems to find the initial and final values

Initial Value Theorem

$\underset{s\to \mathrm{\infty }}{\lim }sF(s)=f(0)$

Final Value Theorem

$\underset{s\to 0}{\lim }sF(s)=f(\mathrm{\infty })$

Applying this to our problem

$\text{The Initial Value Theorem}$

$\underset{s\to \mathrm{\infty }}{\lim }\mathbf{s}\mathbf{X}(s)=-\frac{4}{(s+2{)}^2}$

$\underset{s\to \mathrm{\infty }}{\lim }\mathbf{s}X(s)=-\frac{4}{(\mathrm{\infty }+2{)}^2}=0$

$\text{So as you can see the value for the initial position will be 0.}$ $\text{Because the infinity in the denominator always makes the function tend toward zero.}$

$\text{Which makes sense because the system is initially in equilibrium. }$

$\text{The Final Value Theorem}$

$\underset{s\to 0}{\lim }\mathbf{s}X(s)=-\frac{4}{(s+2{)}^2}$

$\underset{s\to 0}{\lim }\mathbf{s}X(s)=-\frac{4}{(0+2{)}^2}=-\frac{4}{4}$

$\text{This shows the final value to be}$ $-\frac{4}{4}m$

$\text{Which appears to mean the system will be right below equilibrium after a long time. }$

Bode Plot of the transfer function

Transfer Function

$\mathbf{X}(s)=-\frac{4}{(s+2{)}^2}$

Bode Plot

$\text{This plot is done using the control toolbox in MatLab. }$

Fig (1)

Break Points

$\text{Find the Break points using the transfer function.}$

Transfer fucntion

$\mathbf{X}(s)=-\frac{4}{(s+2{)}^2}$

$\text{The equation above contains break points but only in the denominator.}$

$\text{There is only the variable s in the denominator so only those types of break point exist}$

$\text{The break points are asymtotes at the point -2 which occurs twice in this particular equation}$

Convolution

$\text{The convolution equation is as follows: }$

$x(t)=x_{in}(t)*h(t)=\underset{0}{\overset{t}{\int }}x(t_0)h(t-t_0)d{t}_0$

$\text{It does basically the same thing as the Laplace Transform. }$ $\text{To start we must inverse transform our transfer function }$

$\mathbf{X}(s)=-\frac{4}{(s+2{)}^2}$

$\text{Which once more yields: }$

$\mathbf{x}(t)=-4t{e}^{-2t}$

$\text{Then we put this into the convolution integral: }$

$x(t)=x_{in}(t)*h(t)=\underset{0}{\overset{t}{\int }}-4(t-t_0)e^{-2t-t_0}d{t}_0$

$\text{Which once more yeilds: }$

$\mathbf{x}(t)=(-ct{e}^{-2t})$

$\text{Not exactly the same but remember initial conditions arnt used}$

State Space

$\text{Using state equatons is just another way to solve a system modeled by an ODE }$

$\text{First we need to add an applied force so u(t)=2N }$

$m=\frac{98.1}{9.81}=10kg$

$\text{k=40}$

$\text{C=40}$

$\text{x(0)=0}$

$\dot{x}(0)=-4$

$\ddot{x}(0)=0$

$\left[\begin{array}{c}\dot{x}\\ \ddot{x}\end{array}\right]=\left[\begin{array}{cc}0& 1\\ -k/m& -C/m\end{array}\right]\left[\begin{array}{c}x\\ \dot{x}\end{array}\right]+\left[\begin{array}{c}0\\ 1/m\end{array}\right]u(t)$

Created by Greg Peterson

Checked by Mark Bernet