ENGI 3424 Mid Term Test Solutions Page 1 of 9

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1 ENGI 344 Mid Term Test 07 0 Solutions Page of 9. The displacement st of the free end of a mass spring sstem beond the equilibrium position is governed b the ordinar differential equation d s ds 3t 6 5s 3e dt dt s 0 s 0 0 together with the initial conditions (a) B an valid method, find the complete solution to this initial value problem. [0] (b) Is the sstem under-damped, criticall damped or over-damped? [] (a) Chapter method: A.E.: j 3 4 j 3t C.F.: s e Acos 4t Bsin 4t C P.S. (method of undetermined coefficients): 3t r 3e which is not a constant multiple of either part of the C.F. 3t 3t s e cos 4 t, s e sin 4t Therefore tr Substitute in 3 t 3 t 3 t s c e s 3c e s 9c e P P P s 6s 5s 3e P P P 3t 3 t 3 t 3 t ce ce 3e c s P 3t e G.S.: 3t st e Acos 4t Bsin 4t I.C.: 3t 3t st 3e A cos 4t B sin 4t e 4A sin 4t 4B cos 4t 0 3t st e 3A 4B cos 4t 3B 4A sin 4t 6 s 0 0 A 0 0 A s B B 0 C.S.: 3t cos 4 s t e t Alternative methods follow on the net page. (b) 3 4j, a comple conjugate pair the sstem is under-damped

2 ENGI 344 Mid Term Test 07 0 Solutions Page of 9 (a) Alternate solutions: P.S. (b the method of variation of parameters): 3t 3t 3t s e cos 4 t, s e sin 4 t, r 3e W 6t s s 3t 3t e cos 4t e sin 4t s 3t 3 s e t t e t t t 3cos 4 4sin 4 3sin 4 4cos 4 e 3cos 4t sin 4t 4cos 4t 3cos 4t sin 4t r s 4sin 4t 4 e 6 0 s 3t 3t 6t W s r e sin 4t 3e 3e sin 4t 6t W 3e sin 4t u 8sin 4t u cos 4t W 6t 4e s 0 3t 3t 6t W sr e cos 4t 3e 3e cos 4t s r 6t W 3e cos 4t W 6t 4e v 8cos 4t v sin 4t 3t 3t 3t s u s v s cos 4t e cos 4t sin 4t e sin 4t e P t Chapter 3 method (Laplace transforms): Y u L s t then the initial value problem becomes Let 3 u Y 0u 0 6uY 0 5Y u 3 3 u 6u 5Y u 3 3 a bu 3 4c Y u 3 u 3 6 u 3 u a u 3 4 b u 3 4c u 3 u 3: 3 a a u : 0 b 0 b u 0 : c c 0 u 3 3t 3t L s t L e e cos 4t u 3 u 3 4 3t cos 4 s t e t

3 ENGI 344 Mid Term Test 07 0 Solutions Page 3 of 9 (a) Use the RK4 (Runge-Kutta) method with a single step (h = 0.5) to estimate the [9] value of at =.5 to three decimal places, given that and that is the solution of the ordinar differential equation d Note: the RK4 algorithm for,, f o o d includes k f, n, n n, 3 n n k f h, h k 4 n n 3 n k f h h k k f h h k h n n k k k3 k4 6 (b) Confirm our answer to question (a) b solving the initial value problem [9], d eactl. (a) k f, k f k 3 0.5, f 0.5, k4 f 0.5, Note that premature roundoff of intermediate values can cause errors in the final answer.

4 ENGI 344 Mid Term Test 07 0 Solutions Page 4 of 9 (b) d d or 0 [but the ODE is linear, so there is no singular solution] d ln C e e e [and 0 is part of this general solution, when A 0 ] A e / / / / e e and / e.5 / e e The error in the RK4 estimate is less than 0.%! / C / C / /C / Common error: ln C e e C Alternate solution to part (b): The ODE is linear: 0 d d P, R 0 h P d d h e R d 0d 0 h h / / e e R d C e 0 C C e C e C e / / / / e e and / e.5 / e e

5 ENGI 344 Mid Term Test 07 0 Solutions Page 5 of 9 3. B an valid method, find the general solution of the ordinar differential equation [] d and find the complete solution given the additional information 0 0 This ODE is not linear. Use the method of separation of variables: d d d (provided 0 and ) Note that 0 and are both solutions to the ODE and therefore ma be part of the general solution. For the complete solution, 0 is consistent with the initial condition but is not. a b B the cover-up rule, a 0 0 and b d ln ln C d [Note that ln ] C C ln C e e e Setting A 0 leads to 0, so that 0 is part of this general solution, but no finite value of A leads to, so that is a singular solution. Therefore the general solution is Complete solution: A A 0 A or. Also is inconsistent with 0 onl 0 0

6 ENGI 344 Mid Term Test 07 0 Solutions Page 6 of 9 3. (continued) Alternative solution: d d d (provided 0 and ) a b B the cover-up rule, a 0 0 and b d ln ln C C C ln C e e e B e B e B e B e Setting B 0 leads to, so that is part of this general solution, but no finite value of B leads to 0, so that 0 is a singular solution. Therefore the general solution is Be or 0 This alternate solution is equivalent to the previous solution upon assigning Complete solution: B, which has no finite solution for B. However is consistent with A. B 0 Note: This ODE is a Bernoulli ODE (beond the scope of this course). A change of variables will transform this ODE into a linear form.

7 ENGI 344 Mid Term Test 07 0 Solutions Page 7 of 9 3. (continued) Additional Notes: Graphs of some members of the famil of solutions or Graphs of some members of the famil of solutions Be or 0

8 ENGI 344 Mid Term Test 07 0 Solutions Page 8 of 9 4. A set of equipotentials in the plane has the equation 9 A where A is an real number. (a) Find the equation of the set of lines of force associated with these equipotentials [9] (that is, find the orthogonal trajectories). BONUS QUESTION (b) Sketch on one diagram an two equipotentials and an two lines of force. [+4] (a) For the equipotentials, 9 A 8 0 d d 9 The orthogonal trajectories (lines of force) must therefore be the general solution of 9 d d d ln 9 ln C ln ln B ln B Therefore the lines of force are B 9 Common error: 9 3! (the onl eception is when and/or is zero)

9 ENGI 344 Mid Term Test 07 0 Solutions Page 9 of 9 4 (b) The equation a b [or 0, b if b a]. represents an ellipse, centre the origin, vertices at a,0 9 A A A /9 A A /3 A These are ellipses, with aes intercepts at A,0 and 0, 3 n The lines of force are simple curves of the tpe when n is odd. Additional notes: Ever line of force intersects ever equipotential at right angles. 9 Even the degenerate line of force 0 0 intersects ever ellipse at right angles. The onl eception is A 0, which is a point at the origin. Back to the inde of solutions