将以下代码换成matlab语言def trj_v(): t = [0,dt]; a1 = [a0]; a2 = [a0]; vx1 = [-v0]; vx2 = [v0]; x1=(x0;x0+vx1[0]*dt+ k/m1*a1[0]*1/2*dt**2); x2=[-x0,-x0+vx2[0]*dt+ k/m2*a2[0]*1/2*dt** 2]; for (t(-1)<N): a1.append(-k / m1 * (x1[-1] - x2[-1] - (x1[0] - x2[0]))); a2.append(-k / m2 * (x2[-1] - x1[-1] - (x2[0] - x1[0]))); x1.append(2*x1[-1] -x1[-2] + ( a1[-1]*dt**2)); x2.append(2*x2[-1] -x2[-2] + (a2[-1]*dt**2)); vx1.append((x1[-1]-x1[-3])/(2*dt)); vx2.append((x2[-1]-x2[-3])/(2*dt)); t.append(t[-1]+dt); return x1[:-1],x2[:-1],t[:-1],vx1,vx2 
时间: 2023-03-31 12:00:25 浏览: 17
以下是将代码转换为 MATLAB 语言的结果:
function [x1,x2,t,vx1,vx2] = trj_v()
t = [,dt];
a1 = [a];
a2 = [a];
vx1 = [-v];
vx2 = [v];
x1=[x,x+vx1(1)*dt+(k/m1)*a1(1)*.5*dt^2];
x2=[-x,-x+vx2(1)*dt+(k/m2)*a2(1)*.5*dt^2];
while (t(end)<N)
a1(end+1)=-k/m1*(x1(end)-x2(end)-(x1(1)-x2(1)));
a2(end+1)=-k/m2*(x2(end)-x1(end)-(x2(1)-x1(1)));
x1(end+1)=2*x1(end)-x1(end-1)+(a1(end)*dt^2);
x2(end+1)=2*x2(end)-x2(end-1)+(a2(end)*dt^2);
vx1(end+1)=(x1(end)-x1(end-2))/(2*dt);
vx2(end+1)=(x2(end)-x2(end-2))/(2*dt);
t(end+1)=t(end)+dt;
end
x1=x1(1:end-1);
x2=x2(1:end-1);
t=t(1:end-1);
end
相关问题
将以下代码转换成matlab语言def trj_v(): t = [0,dt] a1 = [a0] a2 = [a0] vx1 = [-v0] vx2 = [v0] x1=[x0,x0+vx1[0]*dt+ k/m1*a1[0]*1/2*dt**2] x2=[-x0,-x0+vx2[0]*dt+ k/m2*a2[0]*1/2*dt** 2] while(t[-1]<N): a1.append(-k / m1 * (x1[-1] - x2[-1] - (x1[0] - x2[0]))) a2.append(-k / m2 * (x2[-1] - x1[-1] - (x2[0] - x1[0]))) x1.append(2*x1[-1] -x1[-2] + ( a1[-1]*dt**2)) x2.append(2*x2[-1] -x2[-2] + (a2[-1]*dt**2)) vx1.append((x1[-1]-x1[-3])/(2*dt)) vx2.append((x2[-1]-x2[-3])/(2*dt)) t.append(t[-1]+dt) return x1[:-1],x2[:-1],t[:-1],vx1,vx2
这段代码可以转换成Matlab语言,具体实现方法可以参考以下代码:
function [x1,x2,t,vx1,vx2] = trj_v()
t = [,dt];
a1 = [a];
a2 = [a];
vx1 = [-v];
vx2 = [v];
x1=[x,x+vx1(1)*dt+.5*k/m1*a1(1)*dt^2];
x2=[-x,-x+vx2(1)*dt+.5*k/m2*a2(1)*dt^2];
while(t(end)<N)
a1(end+1)=-k/m1*(x1(end)-x2(end)-(x1(1)-x2(1)));
a2(end+1)=-k/m2*(x2(end)-x1(end)-(x2(1)-x1(1)));
x1(end+1)=2*x1(end)-x1(end-1)+a1(end)*dt^2;
x2(end+1)=2*x2(end)-x2(end-1)+a2(end)*dt^2;
vx1(end+1)=(x1(end)-x1(end-2))/(2*dt);
vx2(end+1)=(x2(end)-x2(end-2))/(2*dt);
t(end+1)=t(end)+dt;
end
x1=x1(1:end-1);
x2=x2(1:end-1);
t=t(1:end-1);
end
在matlab中优化这条代码def trj_v(): t = [0,dt]; a1 = [a0]; a2 = [a0]; vx1 = [-v0]; vx2 = [v0]; x1=(x0;x0+vx1[0]*dt+ k/m1*a1[0]*1/2*dt**2); x2=[-x0,-x0+vx2[0]*dt+ k/m2*a2[0]*1/2*dt** 2]; for (t(-1)<N): a1.append(-k / m1 * (x1[-1] - x2[-1] - (x1[0] - x2[0]))); a2.append(-k / m2 * (x2[-1] - x1[-1] - (x2[0] - x1[0]))); x1.append(2*x1[-1] -x1[-2] + ( a1[-1]*dt**2)); x2.append(2*x2[-1] -x2[-2] + (a2[-1]*dt**2)); vx1.append((x1[-1]-x1[-3])/(2*dt)); vx2.append((x2[-1]-x2[-3])/(2*dt)); t.append(t[-1]+dt); return x1[:-1],x2[:-1],t[:-1],vx1,vx2
这是一个关于 Matlab 优化的问题,我可以回答。这段代码是用于模拟两个质点之间的运动,其中包括了加速度、速度和位移的计算。在代码中,使用了列表来存储不同时间点的加速度、速度和位移,并使用了循环来模拟整个过程。如果需要优化这段代码,可以考虑使用向量化的方法来替代循环,以提高计算效率。
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