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# lecture_29 - MA 36600 LECTURE NOTES MONDAY APRIL 6...

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MA 36600 LECTURE NOTES: MONDAY, APRIL 6 Applications Spring-Mass System: One Mass, Two Springs. Now say that we have a mass m attached to two springs with constants k 1 and k 2 , respectively. We assume that these springs are attached to the opposite ends of a track of fixed length , with the mass m in between. We continue to assume that the mass is under the influence of an external force F ( t ). Denote x = x ( t ) as the displacement from equilibrium on the track at time t . We seek a differential equation for x . Let L 1 and L 2 denote the lengths to which the first and second springs are stretched at equilibrium, respectively. Using Hooke’s Law, we see that the force exerted on the mass from the first spring is - k 1 L 1 (pulling to the left), whereas the force exerted on the mass from the second spring is + k 2 L 2 (pulling to the right). Since we are at equilibrium, the sum of the forces must be zero: - k 1 L 1 + k 2 L 2 = 0 L 1 + L 2 = = L 1 = k 2 k 1 + k 2 ‘, L 2 = k 1 k 1 + k 2 ‘. In particular, if the springs are identical, then k 1 = k 2 , and so L 1 = L 2 = ‘/ 2. Denote u = u ( t ) as the position of the mass at time t . Then we have x = u - L 1 = L 2 - ( - u ) = u = x + L 1 = x + k 2 k 1 + k 2 ‘.

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