Laplace transforms: Critically Damped Motion

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

Problem Statement

An 8 pound weight is attached to a spring with a spring constant k of 4 lb/ft. The spring is stretched 2 ft and rests at its equilibrium position. It is then released from rest with an initial upward velocity of 3 ft/s. The system contains a damping force of 2 times the initial velocity.

Solution

Things we know

$m={\frac {8}{32}}={\frac {1}{4}}slugs$

${\text{k=4}}\,$

${\text{Damping constant C=2}}\,$

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

${\dot {x}}(0)=-3$

Solving the problem

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

${\frac {1}{4}}{\frac {d^{2}x}{dt^{2}}}=-4x-2{\frac {dx}{dt}}$

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

${\frac {d^{2}x}{dt^{2}}}+8{\frac {dx}{dt}}+16x=0$

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

${\mathcal {L}}{\frac {d^{2}x}{dt^{2}}}+8{\frac {dx}{dt}}+16x$

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

$\mathbf {s} ^{2}{X}(s)+8\mathbf {s} {X}(s)+16\mathbf {X} (s)=-3$

$\mathbf {X} (s)(s^{2}+8s+16)=-3$

$\mathbf {X} (s)=-{\frac {3}{(s+4)^{2}}}$

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