lecture_21 - MA 36600 LECTURE NOTES: FRIDAY, MARCH 6...

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MA 36600 LECTURE NOTES: FRIDAY, MARCH 6 Mechanical and Electrical Vibrations Hooke’s Law. Say that we have a mass m which is attached to a spring. Consider four forces on the mass m : Gravity: Newton’s Law of Gravity states that F g = mg . Restoring Force: Hooke’s Law states that F s = - k ( L + u ) for some positive constant k . Damping: Say that the mass is in a viscous fluid, one which gives a type of damping. For example, the mass may be affected by air resistance, or the mass may be in a liquid. Then the force of damping is proportional to the velocity i.e., F d = - γ u 0 for some positive constant γ . External: Say that the mass is acted upon by an outside mechanical force. We simply write this as F e = F ( t ) for some function. Newton’s Second Law of Motion states that F = ma . That is, mu 00 + γ u 0 + k u = F ( t ) . Hence the position u = u ( t ) satisfies a linear second order differential equation. Note that in general, if we have an equation in the form ay 00 + by 0 + cy = f ( t ) where a , b , and c are constants, we identify m = a as the “mass,” γ = b as the “damping constant,” k = c as the “spring constant,” and F ( t ) = f ( t ) as the “external force.” Analysis of Solutions. We focus on solutions to the differential equation mu 00 + γ u 0 + k u = F ( t ) . We know that the general solution is in the form u ( t ) = Au 1 ( t ) + B u 2 ( t ) + U ( t ) for some constants A and B , where we have denoted the functions u 1 ( t ) = ( e λt cosh μt if γ 2 - 4 mk > 0, e λt cos μt if γ 2 - 4 mk < 0. u
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This note was uploaded on 05/24/2011 for the course MA 36600 taught by Professor Ma during the Spring '09 term at Purdue University-West Lafayette.

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lecture_21 - MA 36600 LECTURE NOTES: FRIDAY, MARCH 6...

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