12 damping, complex roots

12 damping, complex roots - 18.03 Lecture #12 Oct. 5, 2009:...

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Unformatted text preview: 18.03 Lecture #12 Oct. 5, 2009: notes Today will be about the nice little spring system from Friday, complicated with the addition of a dashpot. This is a mechanical damping device (typically a piston moving inside a cylinder filled with viscous oil). The word damping means slowing down; so a damping device should act always to decelerate. According to Newton, that means it should apply a force pointed away from the velocity. In the case of a dashpot, well assume (as on page 101 of EP) that the damping force is proportional to the velocity: dashpot damping force =- cy (Dashpot) When we combine this damping with the Hookes law force exerted by the spring spring force =- ky (Hooke) from Fridays lecture, and plug it all into Newtons law F = ma , we get my =- cy - ky. (Damped spring) y ( t ) = y (initial position) , y ( t ) = y (initial velocity) . (Spring initial conditions) The physical setting suggests that the constants m , c , and k should all be positive, but mathe- matically that doesnt matter; the equation makes sense in any case. Sometimes its convenient to write the equation as my + cy + ky = 0 . (Damped spring normalized) In order to solve this, Ill rewrite it once more. Define D = d dt = operation of taking a derivative . The idea is that D takes a function (of one independent variable t ) and spits out its derivative: Df = df dt , D cos =- sin , D ( t 11 ) = 11 t 10 ....
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12 damping, complex roots - 18.03 Lecture #12 Oct. 5, 2009:...

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