MATH 164, Spring 2001 Due Date: Name(s): Honors Project 4: Spring-Mass Damping Introduction In this project we investigate solutions to the spring-mass damping equation mx + ex According to Hooke's Law, the force exerted by an ideal spring in restoring the spring to its equilibrium position is F = —kx, where A; is a positive constant, and x is the distance the spring is stretched out of equilibrium. If an object of mass m is attached to the spring then over time t, then the force exerted on the spring on the object is F = ma = mx where the distance x = x (t) the spring is stretched out of equilibrium is a function of time. If these forces balance each other then the position of the object at time t can be described by the second order differential equation in t: kx = 0. -kx mx + kx = 0. (1) Such an object exhibits what is known as simple harmonic motion. If the motion of the object is damped by a force (such as air resistance) proportional in magnitude to the velocity of the object, then equation (1) becomes -kx or mx + ex + kx = 0 where c is again a positive constant. Finally, if some external force of magnitude / time, acts on (or drives) the motion of the object, then mx + ex + kx = /(í). (2) f (t) depending on (3) Equation (3) is known as the spring-mass damping equation. In this project we investigate the solutions of equations (1) and (2). (In this project we do not derive the solutions of these equations: we simply verify and investigate them. The fact that these are the only solutions to (1) is proved in a course on differential equations.) Equation (1) Any solution to (1) may be written in the form x = Acos(ujt + ) where A is the maximum displacement from equilibrium, to = y/k/m is the frequency, and (f> is the phase. The values of A and (f> depend on the initial values x(0) and x(0) of the displacement x and the velocity x: in general, A= \ x(0)2 + i(0) and 4> = Tan -i(0) ujx(0) Task 1: Using Maple, verify that x = Acos(ujt + ) is a solution to (1). Task 2: Using Maple, graph x = A cos(u>t + = tt/4, find x(0) and x(0). Equation (2) The character of solutions to equation (2) depends on the sizes of m, c, and k. In general, they are of the form: • if c2 — Amk ^ 0 then x = c\erit + c2er2t where —c ± \/c2 — 4mA; ri = --------------------------- 2m are the solutions to the algebraic equation mX2 + cX + k = 0 and c\ and c2 are constants depending on x(0) and x(0), and • if c2 — 4mA; = 0 then x = (ci + c2t)e~ct/2m where ci and C2 are constants depending on x(0) and x(0). Equation (2), Case 1: c2 — 4mA; > 0 (over damping) Task 4: Using Maple, graph x = c\erit + c2er'2t for t G [0, 4ir] assuming that m = 2.0, c = 5.0, k = 3.0, ci = 0.4, and C2 = 1.0. Task 5: Using Maple, and assuming that m = 2.0, c = 5.0, k = 3.0, c\ = 0.4, and C2 = 1.0, find x(0) and x(0). Task 6: Using Maple, and assuming that m = 2.0, c = 5.0, k = 3.0, x(0) = 1.0 and x(0) = —1.5, find ci, and C2. Equation (2), Case 2: c2 — 4mA; < 0 (under damping) If c2 — 4mA; < 0 then the roots r j of the equation mX2 + cX + k = 0 are complex, and we may write x = cieTlt + c2eT2t = e~ct/2m(c! cosujt + c2 sinwi) where u> = ^/Amk — c2/2m. Task 7: Using Maple, graph x = e~ct/2m(ci coswi + C2sinwi) for í G [0,47t] assuming that m = 2.0, c = 2.0, k = 3.0, c\ = 0.4, and C2 = 1.0. Using these same parameters, also plot x = [c\ + C2Í)e~ct/2m, x = \Jc\ + cle-0'/2™ and x = y^i + cle-0'/2™ for t G [0, 47t] on a single set of coordinate axes. Task 8: Using Maple, and assuming that m = 2.0, c = 2.0, k = 3.0, c\ = 0.4, and c2 = 1.0, find x(0) and x(0). Task 9: Using Maple, and assuming that m = 2.0, c = 2.0, k = 3.0, x(0) = 0.4 and x(0) = 1.0, find ci, and C2- Equation (2), Case 3: c2 — Amk = 0 (critical damping) Task 10: Using Maple, graph x = (ci + c2i)e~ct/2m for í G [0, 47t] assuming that m = 2.0, c = 2.0, A; = 3.0, c\ = 0.4, and c2 = 1.0. Task 11: Using Maple, and assuming that m = 2.0, c = 2.0, k = 3.0, c\ = 0.4, and C2 = 1.0, find x(0) and x(0). Task 12: Using Maple, and assuming that m = 2.0, c = 2.0, k = 3.0, x(0) = 0.4 and x(0) = 0.6, find ci, and c2.