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POEquation of motion by Runge-Kutta continuation method in Matlab
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  1. POEquation of motion by Runge-Kutta continuation method in Matlab
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    copied!<ul> <li>Equation of motion is given by:</li> </ul> <p><img src="https://i.stack.imgur.com/GhLU3.png" alt="equation"> ,where m, b are stationary values of mass and damping. The time varying term f(t) is excitation power and q(t) is generalized displacement.</p> <ul> <li>I solved that:</li> </ul> <p><img src="https://i.stack.imgur.com/NwV1n.png" alt="enter image description here"></p> <p><img src="https://i.stack.imgur.com/k5ZLy.png" alt="enter image description here"></p> <ul> <li>And I should to solve in MatLab via [t,x]=ode23('rightside',tspan,x0).</li> <li>f(t) and k(t) I solved in complex numbers via Fourier series like that in Matlab:</li> </ul> <p><i> % complex fourier series for f(t)</p> <pre><code>ft=zeros(size(t)); for j=1:2*N+1 n= j-(N+1); if n==0 f(j)=f0/2; else f(j)=f0*( (exp(-i*n*2*pi)*(i*2*pi*n+1)-1)/(4*pi^2*n^2)); end ft=ft+f(j)*exp(i*n*om*t); end </code></pre> <p></i></p> <p><i> % complex fourier series for k(t)</p> <pre><code>kt=k0*ones(size(t)); for s=1:2*N+1 n= s-(N+1); if n==0 c(s)=k0; else c(s)=i*(k0+ktyl)/n/pi*(1-cos(n*pi)); end kt=kt+c(s)*exp(i*n*om*t); end </code></pre> <p></i></p> <ul> <li>And we know:</li> </ul> <p><i></p> <pre><code>T=30; dt=0.01; t=0:0.01:5*T; k0=1e6; om=2*pi/T; ktyl=0.5e6; N=10; m=1; ks=1e4; D=0.01; OMG=sqrt(ks/m); b=2*D*OMG*m; f0=100; </code></pre> <p></i></p> <p>Thank you.</p> <ul> <li><strong>It should be similar princip like that:</strong></li> </ul> <p></p> <pre><code>function v=prst1(t,y) global m b k Om D F omeg v(1)=....; v(2)=y(1); v=v(:); </code></pre> <ul> <li><strong>and:</strong></li> </ul> <p></p> <pre><code>global m b k Om D F omeg m=1; b=10; k=1000; F=10; Om=sqrt(k/m); omeg=1*Om; D=b/(2*Om*m); x0=[0;0]; [t,x]=ode23('prst1',0:0.01:10,x0); plot(t,x) </code></pre> <ul> <li><strong>BUT I don't how to get there f(t) and k(t).</strong></li> </ul>
 

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