Laplace transforms: Critically Damped Motion: Difference between revisions
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Mark.bernet (talk | contribs) |
Mark.bernet (talk | contribs) |
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<math>\text {x(0)=0}\,</math> |
<math>\text {x(0)=0}\,</math> |
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<math>\dot{x}(0)= |
<math>\dot{x}(0)=-3</math> |
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<math>\text {Therefore the equation representing this system is}\,</math> |
<math>\text {Therefore the equation representing this system is}\,</math> |
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<math>\text {Now that we have the equation written in standard form we need to send}\,</math> |
<math>\text {Now that we have the equation written in standard form we need to send}\,</math> |
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<math>\text {it through the Laplace Transform}\,</math> |
<math>\text {it through the Laplace Transform}\,</math> |
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<math>\mathcal{L}_s\frac{d^2x}{dt^2}+8\frac{dx}{dt}+16x</math><br /><br /> |
Revision as of 18:37, 22 October 2009
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