Question

Consider the forced vibration in Figure 1. We mass, m Figure 1: Forced Vibration 1. Use a free-body diagram and apply Newton's 2nd Law to show that the upward displacement of the mass, r(t), can be modelled with the ODE da da mdt2 + cat + kz = F(t) where k is the spring coefficient and c is the damping coefficient. = 2 kg, c = For the remainder of the questions, use the following values: m 8 Ns/m, k = 208 N/m. 2. Solve for x(0) given the mass is released from rest at = 0.2 m and no external force is applied (F(t) = 0). 3. Solve for r(t) given the mass is released from rest at 2 = 0.2 m and a ramping force of F(t) = 10 N is applied, where t is time measures in seconds. 4. The spring will fail once the force applied to it exceeds 100 N. In the sce- nario from Q3 (same force, initial conditions and parameters), determine how much time it takes for the spring to break. Your solution may be hand-worked or done in MATLAB.

Answer 1

  
Problem :) Free body diagram of the above big. 1 F0) "ку According to Newton's and law, mass x acceleration = force applied, - Force applied = f(t) - call a ka so; m . Kaspring Constank. = F(t) -en er ca damping constrand or, men inte + = FCE) Putting the given values above equation can be written as, 2 eet + 8 e + 208 x = F(t) = a + 1048 oul FC 2) The solution of the above equation when F(t) to is given by, C 8 x = ezt [G 2 V4_1016 to 2 14-10846 + 7 al- rest <=0, *20.2, po 0.2 = 4+62 in. delete-zēt [g, ev-ment Ge Twent'] +et+ Frete crie V-108t2 t 0-et-2 (9 +62). + 110812 (4-63) ev, 264+2) - i Jioene (4-6) y - Vioozt -Gel su, 0 +@ we get; 26, 2012 - Oyi Jiopia » 450.17 /10812 ✓ 10812 so, wit) - ét [101-pozi) el quent + (0-14 jei Jeenet] 0.2 10 M2

is applied the 3) when - above a ramping force of F(t) = 10th equation will look like , - dx +49 + 1099 25 t 104. ...(3) 1. difta 10 E + Ae Vlog E For this case as wo> b we have the solution, X(t) = 10€ + rezt cos [Joopt-y + -e] al tro, a 2012, 20 208 0,2 = ORS + A Cosa > A Crso 2012 oder eine r e znét cos (110842 + -) - sézt Joogia tin (roant -) 02 -28 Cos (8) + A Tingiz singo ar z Ålos q = A 210812 sina + 5 or,,2x012 - A V10 ste sing or Asing . . 04-0,05 Voogt2 TOBZ 104 i GOB 0.35- V1082 a tana = 0.35 7 - ✓ 10812 X 0.017 · 9 = tant (0.017) = 1 ✓ los at a sing = √ (0, 2) + 2 = 70.04 +1.13 x10-5 or, A = 0.2 so, r(t) = 5t + aizé 2t cos (110812 t - 1) (Ang 10Y