MATH 1B Lecture Notes - Lecture 26: Dashpot, Klepton

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8 Apr 2015
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Math 1b: calculus - lecture 25: step-by-step solving 2nd order heterogeneous. Q: solve (d2y/dt2) = -5y, given that y=1 and (dy/dt) = 0 when t=0. A: we have (d2y/dt2) + 5y = 0. The auxiliary equation is: r2 + 5 = 0. r = 5 i. Think + i and - i. = 0 and y = e y = acos( 5 t) + bsin( 5 t). This has solutions r = 5 i, - 5 i. So, (dy/dt) = - 5 asin( 5 t) + 5 bcos( 5 t) Now y=1 and (dy/dt) = 0 when t = 0. A spring length of x at time t. This apparatus consists of a weight on the end of a spring in a dashpot. This weight has a force exerted on it by the spring, given by -k(x(t) - l). There is also a force exerted by the dashpot that opposes the direction of the motion.

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