Question

m d2x dx + B + kx = 0 dt dt m = 0.25 pulg k = 4 lb/ft B = 2 Condiciones iniciales: *(0) = 0 x'(0) = -3 ft/s (r + 4)2 = 0 r = -4 x(t) = Ge-4 + cate 4 x(0) = 0 G = 0 x'(t) = -4tc2e-4 + cze-4-4c2te-46 x'(0) = -3 C2 = -3 x(t) = -3te-4

A mass weighing 8 pounds lengthens a spring by 2 feet assuming that a damped force equal to 2 times the instantaneous velocity and acting on the system determines the equation of motion if the initial mass is released from the equilibrium position with a velocity ascending 3 ft / s. Solve the previous exercise with La Place transforms

Answer 1

solution: Given dt? tp I B=2 m = 0.25 » 0.25 > du 2 oda +km=0 dt pulg ,K=4 lblft dx ta Im + 40 =0 de² dt d²x +8 +16x=0 dE² de d²x dt² = 5 x(s) - 5X(0) - *'(o) d[ dat = SX(s) - X(0) [Take laplace Iranyfosas on both sides] $*657 -5X60) - x'60) +8 [5x(s)-x()]+16 X(s)=0 3* x26) 546) - x86 +8[/SMS). + $XC) -56)-(-3) +8 [ sx (s)-o] +16 X) = → [6 +88+18]*()=-3 -3 X(5) - Ś +85 +16 (S+4² Take inverse laglace Transfory -3 I [X] cs Scanne with (5+49 CamScanner 3 2 both side. on 3 .]-ita $+85 +16

] cat th n!. (Sta)"+1 i [sto -3[*** ] => X() -46 216) = -3% e #. CS Scanned with CamScanner