Question

Solve it with matlab

25.16 The motion of a damped spring-mass system (Fig. P25.16) is described by the following ordinary differential equation: d’x dx ++ kx = 0 m dr dt where x = displacement from equilibrium position (m), t = time (s), m 20-kg mass, and c = the damping coefficient (N · s/m). The damping coefficient c takes on three values of 5 (under- damped), 40 (critically damped), and 200 (overdamped). The spring constant k = 20 N/m. The initial velocity is zero, and the initial displacement x = 1 m. Solve this equation using a numerical method over the time period 0 st s 15 s. Plot the displacement versus time for each of the three values of the damping coefficient on the same curve.

Answer 1

program logic :

• Inorder to solve this 2nd order ODE , we have to form a state space first.
• The state space form can be obtained using the function odeToVectorField()
• Then, pass this state space to an ODE solve with the given initial conditions.
• using the obtained solution, plot the graph

program;

syms x(t)

sol1 = motiondamp(5);

sol2 = motiondamp(40);

sol3 = motiondamp(200);

fplot(@(x)deval(sol1,x,1),[0, 15]);

hold on

fplot(@(x)deval(sol2,x,1),[0, 15]);

fplot(@(x)deval(sol3,x,1),[0, 15]);

xlabel("t")

ylabel("x")

legend("c=5","c=40","c=200")

grid on

hold off

function sol = motiondamp(c)

m = 20;

k = 20;

x0 = 1;

v0 = 0;

syms x(t)

ss = odeToVectorField(m*diff(x,2)+c*diff(x)+k*x==0);

M = matlabFunction(ss,'vars',{'t','Y'});

sol = ode45(M,[0 15],[v0 x0]);

end

plot:

0.8 C=5 C=40 c=200 0.6 0.4 0.2 X -0.2 -0.4 5 10 15 t