Update 2013 5.4-18 (a) 2x'' + 12x' + 50x=0, x(0)=0, x'(0)=-8; solve the equation. ====== Classify as overdamped, critically damped, underdamped. Write x(t) in phase=amplitude form. (b) 2u'' + 0u' + 50u=0, u(0)=0, u'(0)=-8; solve the equation. Write u(t) in phase=amplitude form. (a) 2x'' + 12x' + 50x=0, x(0)=0, x'(0)=-8; The characteristic equation for the damped system is 2(r^2+6r+25)=0 with complex conjugate roots r=-3+4i,r=-3-4i. Solution Atoms=e^{-3t}cos 4t, e^{-3t}sin 4t. Underdamped. (b) 2u'' + 0u' + 50u=0, u(0)=0, u'(0)=-8; The characteristic equation for the undamped system is 2(r^2+0+25)=0 with complex conjugate roots r=5i,r=-5i. Solution Atoms=cos 5t, sin 5t. Solve each homogeneous system, as a linear combination of the solution atoms. Evaluate the constants in the linear combination, in each of the two cases, using the initial conditions x(0)=0, x'(0)=-8. There are two linear algebra problems to solve. Write each solution in phase-amplitude form, a trig problem. Plot the solutions so obtained, on a hand calculator or maple, and draw by pencil a graphic replica next to your hand-written solution. 5.4-20 (a) 2x'' + 16x' + 40x=0, x(0)=5, x'(0)=4; solve the equation. ====== Classify as overdamped, critically damped, underdamped. Write x(t) in phase=amplitude form. (b) 2u'' + 0u' + 40u=0, u(0)=5, u'(0)=4; solve the equation. Write u(t) in phase=amplitude form. 2x'' + 16x' + 40x=0 The characteristic equation for the damped system is 2(r^2+8r+20)=0 with complex conjugate roots r=-4+2i,r=-4-2i. Solution Atoms=e^{-4t}cos 2t, e^{-4t}sin 2t. Underdamped. 2x'' + 0x' + 40x=0 The characteristic equation for the undamped system is 2(r^2+0+20)=0 with complex conjugate roots r=sqrt(20)i,r=-sqrt(20)i. Solution Atoms=cos( sqrt(20)t), sin( sqrt(20)t). Solve each homogeneous system, as a linear combination of the solution atoms. Evaluate the constants in the linear combination, in each of the two cases, using the initial conditions x(0)=5, x'(0)=4. There are two linear algebra problems to solve. Answers: Coefficients 5, 2 for 2x'' + 16x' + 40x=0 Amplitude sqrt(5^2 + 12^2) = 13 Coefficients 5, 2/sqrt(5) for 2x'' + 0x' + 40x=0 Amplitude sqrt(5^2 + 4/5) = sqrt(129/5) Plot the solutions so obtained, on a hand calculator or maple, and draw by pencil a graphic replica next to your hand-written solution. Don't plot from the phase-amplitude conversions, because with computing assist there is no reason to do that. Write each solution in phase-amplitude form, a trig problem. See section 5.4 for specific instructions on how to find the amplitude C and phase shift alpha. The book's answers are correct. (a) tan(alpha) = 5/12 (b) tan(alpha) = 5 sqrt(5)/2 5.4-34 100 x'' + cx' + kx = 0. Given a solution x(t) such that its ====== first two maxima are t=0.34 seconds, x=6.73 inches and t=1.17 seconds, x=1.46 inches. Find c and k. Use the results freely from problems 32 and 33 to solve problem 34. Let x1=6.73/12 and x2=1.46/12 in problem 33. Use times t1=0.34 and t2=1.17 in problem 32. Problems 5.4-32 and 5.4-33 are applied to this problem to determine the values for symbols p and omega_1; these symbols will now be defined, using the characteristic equation and the quadratic formula. The characteristic equation: m r^2 + c r + k = 0 Because of the oscillation described in 5.4-34, this has to be the underdamped case with solution (21) in section 5.4. In particular, the results of problems 5.4-32 and 5.4-33 apply. The quadratic formula: r = [-c/(2m)] + [(1/(2m)) sqrt(c^2 - 4km)] r = [-c/(2m)] - [(1/(2m)) sqrt(c^2 - 4km)] Because of the underdamped case, the argument of the square root is negative. The book defines in section 5.4 equations (21) to (23) symbols p and omega_1 via the equation r = [-p] + [omega_1 sqrt(-1)] r = [-p] - [omega_1 sqrt(-1)] Then m = 3.125 [given in 5.4-34], p = (1/2)(c/m), omega_1 = sqrt(4km-c^2)/(2m). Once you know both symbols p and omega_1, then you also know c and k. Equations (21) to (23) in section 5.4 describe the relations between m,c,k and the symbol omega_1 used in problems 32 and 33. The phase-amplitude form of the solution, as described in section 5.4, equations (12) to (14), is used to solve problems 32 and 33. Because problem 5.4-33 is solved in the student solution manual, you are able to reference details for that solution. No details are expected for the solution to 5.4-32, especially, do not include them with your solution to 5.4-34. =======end========