Laplace transforms:Mass-Spring Oscillator: Difference between revisions
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'''Solution:''' |
'''Solution:''' |
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We first begin by setting up a few equations from Newton's laws. |
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By Newton's first law: |
By Newton's first law: |
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<math>\mathbf{F}=m |
<math>\mathbf{F}=m{a} ~~~~~ \Rightarrow ~~~~~ \mathbf{f}_m(t)=m\ddot{x}</math> |
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By Hooke's law: |
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<math>\mathbf{F}=k{x} ~~~~~ \Rightarrow ~~~~~ \mathbf{f}_k(t)=mx</math> |
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By Newton's third law of motion that states ''every action produces an equal and opposite reaction'', we have f_''k'' = -f_''m''. That is, the force f_''k'' applied by the mass to the spring is equal and opposite to the accelerating force f_''m'' exerted in the -''x'' direction by the spring on the mass. |
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<math>\mathbf{f}_m(t)+\mathbf{f}_k(t)=0 ~~~~~~ \Rightarrow ~~~~~ m\ddot{x}(t)+k{x}(t)=0</math> |
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We now have a second order differential equation that governs the motion of the mass. Taking the Laplace transform of both sides gives: |
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<math>0=\mathcal{L}_s\left\{m\ddot{x}(t)+k{x}(t)\right\} |
Revision as of 15:36, 19 October 2009
Problem Statement:
An ideal mass m sliding on a frictionless surface, attached via an ideal spring k to a rigid wall. The spring is at rest when the mass is centered at x=0. Find the equation of motion that the spring mass follows.
Solution:
By Newton's first law:
By Hooke's law:
By Newton's third law of motion that states every action produces an equal and opposite reaction, we have f_k = -f_m. That is, the force f_k applied by the mass to the spring is equal and opposite to the accelerating force f_m exerted in the -x direction by the spring on the mass.
We now have a second order differential equation that governs the motion of the mass. Taking the Laplace transform of both sides gives:
<math>0=\mathcal{L}_s\left\{m\ddot{x}(t)+k{x}(t)\right\}