Project 4, Numerical Analysis

Solutions Project 4 Sostegni 996088

Contents

• 1 Case 1 2
• 1.1 Analytical solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
• 1.2 Numerical solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
• 2 Case 2 4
• 2.1 Analytical solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
• 2.2 Numerical solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
• 3 Case 3 7
• 3.1 Analytical solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
• 3.2 Numerical solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
• 4 Qualitative comparison: from physics, to mathematics and numerics 9

1 Case 1

1.1 Analytical solution

4. 2 2∆ = 5 Kg /s −1−3.819660112501051 ·r = 10 1/s1 0−2.618033988749895 ·r = 10 1/s2 0·c = 1.170820393249937 10 m1 −1−1.708203932499369 ·c = 10 m2 −0.3820 −2.6180r t r t t t−x(t) = c e + c e = 1.1708 e 0.1708 e m1 21 2 −0.3820 −2.6180r t r t t t−0.4472v(t) = r c e + r c e = e + 0.4472 e m/s1 21 1 2 2KM 2 2E(t) (v(t)) + (x(t)) == 2 2 2 2−0.3820 −2.6180 −0.3820 −2.6180t t t t 2 2 −0.4472 −= 0.5 e + 0.4472 e + 0.5 1.1708 e 0.1708 e Kg m /s2−0.3820 −2.61802 t t 2 3D(t) −0.4472= C (v(t)) = 3 e + 0.4472 e Kg m /s7. 29.

1.2 Numerical solution

8. 310.

2 Case 2

2.1 Analytical solution

4. 2 2−3.9600∆ = Kg /s −1 −1−1.000000000000000 · · ·r = 10 + i 9.949874371066200 10 1/s1 −1 −1−1.000000000000000 · − · ·r = 10 i 9.949874371066200 10 1/s2c = 1 m1 −1·c = 1.005037815259212 10 m2 −0.1α t tx(t) = e (c cos(β t) + c sin(β t)) = e (cos(0.9950 t) + 0.1005 sin(0.9950 t)) m1 2 −0.1α t t−v(t) = e [(α c + β c ) cos(β t) + (α c β c ) sin(β t)] = e [−1.0050 sin(0.9950 t)] m/s1 2 2 1M K2 2E(t) = (v(t)) + (x(t)) =2 2 2 2−0.1 −0.1t t 2 2 = 0.5 e [−1.0050 sin(0.9950 t)] + 0.5 e (cos(0.9950 t) + 0.1005 sin(0.9950 t)) Kg m /s2−0.12 t 2 3D(t) = C (v(t)) = 0.2 e [−1.0050 sin(0.9950 t)] Kg m /s47.9. 5

2.2 Numerical solution

8.10. 6

3 Case 3