# Solving coupled second order differential equations

15 ビュー (過去 30 日間)
Abraham Hartley 2022 年 9 月 6 日
コメント済み: Alan Stevens 2022 年 9 月 6 日
Hi,
I constantly get an error that I do not know how to resolve when I run this code. Any help would be greatly appreciated. The Force function applies a force of 50N @ t = 0 and equals 0 at all other time steps.
syms c k m1 m2 x1(t) x2(t) t F Y;
% 1st and 2nd derivative
dx1 = diff(x1);
d2x1 = diff(x1,2);
dx2 = diff(x2);
d2x2 = diff(x2,2);
% Defining equations
Eq1 = d2x1 == -(c/m1)*(dx1-dx2) - (k/m1)*(x1(t)-x2(t)) + F/m1;
Eq2 = d2x2 == -(c/m2)*(dx1-dx2) + (k/m2)*(x1(t)-x2(t));
[VF,subs] = odeToVectorField(Eq1, Eq2);
ftotal = matlabFunction(VF,'Vars',{t,Y,F,c,k,m1,m2});
m1 = 10;
m2 = 20;
c = 30;
k = 5;
F = @(t) Force(t);
tspan = [0 1];
ic = [8 0 0 0];
[t,Y] = ode45(@(t,Y) ftotal(t,Y,F,c,k,m1,m2), tspan, ic);
figure
plot(t, Y)
grid
legend(string(subs))
##### 1 件のコメント-1 件の古いコメントを表示-1 件の古いコメントを非表示
Torsten 2022 年 9 月 6 日
The value of the "Force" function at t=0 must somehow be part of the initial conditions ic.
As part of the differential equations, it won't influence the result.

サインインしてコメントする。

### 採用された回答

Alan Stevens 2022 年 9 月 6 日
Something like this?
m1 = 10;
m2 = 20;
c = 30;
k = 5;
tspan = [0 1];
ic = [8 0 0 0];
[t,Y] = ode45(@(t,Y) ftotal(t,Y,c,k,m1,m2), tspan, ic);
figure
plot(t, Y)
xlabel('t'), ylabel('y and dydt')
grid
legend('y1','dy1/dt','y2','dy2dt')
function dYdt = ftotal(t,Y,c,k,m1,m2)
y1 = Y(1); dy1dt = Y(2);
y2 = Y(3); dy2dt = Y(4);
dYdt = [dy1dt;
-(c/m1)*(dy1dt - dy2dt) - (k/m1)*(y1 - y2) + 50*(t==0)/m1;
dy2dt;
-(c/m2)*(dy1dt - dy2dt) + (k/m2)*(y1 - y2)];
end
##### 1 件のコメント-1 件の古いコメントを表示-1 件の古いコメントを非表示
Alan Stevens 2022 年 9 月 6 日
Instead of 50*(t==0)/m1; you could try 50*(t<0.01)/m1; but you still won't see an effect. However, if you try 5000*(t<0.01)/m1; you start to see an effect. If you try decreasing the 5000 to smaller values, towards 50, you see the effect getting smaller and smaller.

サインインしてコメントする。

### カテゴリ

Help Center および File ExchangeProgramming についてさらに検索

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by