Can you help me please !

I have 1 code vibration.

function main
d1=1;
v=10;
t_end=3;
t=0:0.01:t_end;
y0=[0;0;0;0];
[t,y]=ode45(@daodong,t,y0);
hold on;grid on;
plot(t,y(:,1))
end
function yp = daodong(t,y)
c=10;
vt=10;
k1=84.24;
m1=5.265;
I1=0.1594;
a=0.174;
yp = zeros(4,1);
yp(1) = y(2);
yp(2) = ((-(c*(y(2)-0.05*pi*vt*(sin(4*pi*vt*t))-(a*y(4)/4))+k1*(y(1)-(0.05*(sin(2*pi*vt*t))^2)-(a*y(3)/4))))/m1);
yp(3) = y(4);
yp(4) = ((-(a*c*(y(2)-0.05*pi*vt*(sin(4*pi*vt*t))-(a*y(4)/4))+a*k1*(y(1)-(0.05*(sin(2*pi*vt*t))^2)-(a*y(3)/4))))/(2*I1));
end

please help me determine acceleration at t=0 and t=t_end.

[Walter Roberson]

Unfortunately you did not document which of the four components corresponds to position, speed, or acceleration.

It looks to me as if your acceleration is a = 0.174

[Johnny Vendetta]

Thank you Walter Roberson, can you help me a code calculate acceleration, please.

[KALYAN ACHARJYA]

Walter alteady commented "Unfortunately you did not document which of the four components corresponds to position, speed, or acceleration"

Which variables refer the acceleration in your code?

[David Goodmanson]

Hi JV,

it seems pretty likely that y(1),y(3) are positions and y(2),y(4) are velocities so I will assume that to be the case. If you change the plot command to plot(t,y), you can see all four variables. There is some slowly varying stuff, plus a wicked 20 Hz forced vibration due to the sine terms. You might want to think about the model, although such behavior is certainly possible.

From y(2) and y(4) it's clear that ode45 is not doing a very good job of making the time intervals small enough to accurately describe the velocities. It does give the basic idea though. The accelerations at the end points are a separate consideration, but for, say, the velocity y(4), its amplitude in the 20 Hz component is roughly .05, so the max acceleation is around .05*2*pi*20 ~~ 6.

[Johnny Vendetta]

Thank you KALYAN ACHARJYA and David Goodmanson very much.

I try used method different.

[Walter Roberson]

it seems pretty likely that y(1),y(3) are positions and y(2),y(4) are velocities

I though that at first. The first output depends upon y(2) and not directly on any other variables, which is consistent with a simple acceleration leading to the integrated first output being position and the second output being velocity. And at first glance, the third output has the same form, making it appear that the third output is also a position and the fourth output being velocity. However, the second output depends upon y(1), y(2), y(3), y(4), so the second output depends upon the integral of an integral and so if the first output is position and the second is velocity, the third would have to be accelaration, and the fourth would have to be rate of change of accelaration. But the fourth output also depends upon y(1), y(2), y(3), y(4) so you induce an infinite hierachy of accelations and so all of the variables become variables representing rate of change of an infinite chain of values, and none of them become acceleration in the classical sense.

I am getting the impression that this is a system with two coupled springs ??

Johnny Vendetta will have to define the variables for us to proceed any further.

[Walter Roberson]

Wait a second, though. The first column of output of the ode would be the integration of the first output of the function. It is unlikely that the integral of the position is desired. It is more likely that the first output of the function is a velocity which the ode would integrate to position. The second output of the function would then be an acceleration which the ode would integrate to velocity.

Although it might be possible to add another layer of variables to capture the acceleration directly in the y output as an integral of something else, it would probably be easier to just use gradient() of the time and the second column of the y output.