Question

If you can provide the MatLab code too, thank you!

3. The equation of motion of a general mass-spring system subject to an external force F(t) is mr" + cr' + kr = F(t) where m, c and k are the mass, damping coefficient and spring constant respectively. Let's consider a simple undamped system where m = 1, k = 1, and c= 0 and with forcing function F(t) = 0.5 cos(t) (all in SI units) "+=0.5 cos(1) Find the exact analytic expression for the unique solution satisfying the initial con- ditions x(0) = 20) = 0 and use MATLAB to plot the function (t) for 0 <t<50. In the same figure, plot the lines r(t) = 0.25t and 12(t) = -0.25t as dashed lines. Save the figure as hw3q3.fig and submit it.

Answer 1

solution

Sold of d x lt) + X0= 0'5 cost is a(t) = G cos (t) + Ce Sin Lt) + sin(t): - te = complementary soln e = particular fit Iinitial condos: a (o)=0 and x'(o)=0 → G + C2 X0 = 0 ; x'(+) = -4 Sin (t) tezloslt => a=0 + * sin(t) + cosit) x'(o)= 0 => : Czx1 = 0 => C2 = 0 Putting G =C₂ = 0 in 7 we get - Hence, the final solution alt) = – 4 Sin (t).

d? ode is x(t) + x(t) =0.5.cos(t); ics := x(0) = 0, D(x) (0) = 0; 2 x(1) + x(t) = 0.5 cos(1) x(0) = 0, D(-x)(0) = 0 dsolve(ode) x(1) = sin(1) _c2 + cos(1) _C1 + 1 sin(t) = dsolvel {ode, ics}) x(t) = sin(e) { #This is the required solution. sol(t) = sin(t) 6; -sin() < plot(sol(),1=0..50) _ _ -10

x1(t) == 0.25-t, x2(t) ==-0.25-t, 1+0.251 1-(-1)-0.25 t plot([sol(1),x1(1), x2(1)], 1=0..50, linestyle=[solid, dash, dash]) OATHO 50 soluti

10 20 50