I would like to sole the spring mass equation of the form: $M\ddot{Y}+B\dot{Y}+KY = F_y(t)$, where $M :$ mass, $B :$ Damping, $K : $ stiffness, $F_y :$ external force and $Y :$ displacement in transverse direction; using Runge-Kutta 4th-order method. I modify RK4 C++ code as:
/*
Definition of the y'(t) = f1(t,y,y') = y' by the definition
*/
double f1(double t, double y, double v)
{
double d1y;
d1y = v;
return d1y;
}
/*
Definition of the y"(t) = f2(t,y,y')
*/
double f2(double t, double y, double v, double Fy)
{
double d2y;
double M = 0.03715; // 10.0;
double B = 0.0;
double K = 0.149;
d2y = (-Fy-B*v-K*x)/M;
return d2y;
}
/*-----------------------------------------------------------------------
Program to solve 1st order Ordinary Differential Equations
y'(t) = d1y(t,y)
y"(t) = d2y(t,y)
method: 4th-order Runge-Kutta method */
double rk4_2nd(double(*d1x)(double, double, double),
double(*d2x)(double, double, double, double),
double ti, double xi, double vi, double tf, double Fy,
double& xf, double& vf)
{
double h,t,k1x,k2x,k3x,k4x,k1v,k2v,k3v,k4v;
h = tf-ti;
t = ti;
k1x = h*d1x(t,xi,vi);
k1v = h*d2x(t,xi,vi,Fy);
k2x = h*d1x(t+h/2.0,xi+k1x/2.0,vi+k1v/2.0);
k2v = h*d2x(t+h/2.0,xi+k1x/2.0,vi+k1v/2.0,Fy);
k3x = h*d1x(t+h/2.0,xi+k2x/2.0,vi+k2v/2.0);
k3v = h*d2x(t+h/2.0,xi+k2x/2.0,vi+k2v/2.0,Fy);
k4x = h*d1x(t+h,xi+k3x,vi+k3v);
k4v = h*d2x(t+h,xi+k3x,vi+k3v,Fy);
xf = xi + (k1x + 2.0*(k2x+k3x) + k4x)/6.0;
vf = vi + (k1v + 2.0*(k2v+k3v) + k4v)/6.0;
return 0.0;
}
The problem I am facing is that:
RK4 is called inside a function e.g.Function(void) and this Function() is called inside main where a time loop runs as form of iterations, How can I define ti, tf, to RK4 function for my problem.
void Function() { Y0[n] = Y[n] - YC; // n = number of objects i.e. n = 1 at the moment // Y : current displacement; YC: Initial/reference position = constant V0[n] = V[n]; // Velocity at (t-1) = velocity at (t) FY[n] += -k * (Y0[n]); // external force; k : spring constant rk4_2nd(f1, f2, t0, Y0[n], V0[n], t1, FY[n], Y[n], V[n]); } int main() { N = 0; do { N = N + 1; Function(); } while (N < NMAX); return 0; }