Question

The spring-mass system is at rest when the force P(t) is applied, where P(t) = 100 N 20 N when t < 2 s when t>2 s The differential equation for the ensuing motion is P(t) k mm Determine the maximum displacement of the mass. Use m= 2.5 kg and k= 75 N/m.

NOTE:Follow the Runge-Kutta Method.Any programming language(Matlab) can be used.Plot Position-Time and Velocity-Time graphs.

Answer 1

k=75;
m=2.5;
f=@(t,y)[y(2) (10*t*(t<2)+20*(t>=2))/m-k*y(1)/m];
t=0:.01:5;
h=.01;
y0=[1 0];%initial condition
y=zeros(length(t),length(y0));
y(1,:)=y0;%initial value put
for i = 1:length(t)-1%this calulates y using RK4
k1=f(t(i),y(i,:));
k2=f(t(i)+h/2,y(i,:)+k1*h/2);
k3=f(t(i)+h/2,y(i,:)+k2*h/2);
k4=f(t(i)+h,y(i,:)+k3*h);
phi = (k1+2*k2+2*k3+k4)/6;
y(i+1,:) = y(i,:) + phi*h;
end
plot(t,y(:,1))
xlabel('time')
ylabel('displacement')
figure
plot(t,y(:,2))
xlabel('time')
ylabel('Velocity')
fprintf('Max displacement between 0 to 5second: %f\n',max(y(:,1)))

>> rungeK Max displacement between 0 to 5second: 1.291302 fx >> A Figure 2 A Figure 1 Desktop Window Help Tools Desktop Window Help File Edit View Insert Tools File Edit View Insert 6. 1.5 0.5 -0.5F -6 -1 1.5 0.5 2 2.5 3 3.5 4 4.5 0.5 1.5 2.5 3.5 4 4.5 5 time time displacement Velocity