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, we’ll 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 Hooke’s law force exerted by the spring spring force =- ky (Hooke) from Friday’s lecture, and plug it all into Newton’s 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 doesn’t matter; the equation makes sense in any case. Sometimes it’s convenient to write the equation as my ′′ + cy ′ + ky = 0 . (Damped spring normalized) In order to solve this, I’ll 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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