Numerically integrating Acceleration to get displacement?

2020 11 15
1 个回答
回答已采纳
更新时间：2020 11 15
22 次查看（30 天）

0 个投票

I want to rephrase my last question, as i was not very clear there.
What i am trying to do is integrate the following second order, nonlinear ode, which is an expression for angular acceleration,
twice, to get the displacement, and compare it with the actual expression for displacement:
BOTH ω and A are specified initially. They are constants.
------------------------------------equation for acceleration
(corrected from )
if initial conditions are required, then x(0)=pi/4 and dx/dt (t=0) =0
this has to be integrated w.r.t time, t, twice, between the limits t=0 and t=50 (it can be any time interval)
im not sure if the interval for x has to be defined as well.
the equation for displacement is
-----------------------------------------equation for displacement (corrected from
Any help would be appreciated!

请先登录，再回答此问题。

 采纳的回答

Ameer Hamza
Ameer Hamza 2020-11-15

0 个投票

Are you sure that the equations are correct? The solution you posted does not satisfy the initial conditions.
Generally, such a problem can be solved using the Symbolic toolbox
syms x(t) w A
dxdt = diff(x);
dx2dt2 = diff(x,2);
ode = dx2dt2 == -A*w*sin(dxdt);
cond = [x(0)==pi/4 dxdt(0)==0];
sol = dsolve(ode, cond)
or ode45() can be used for a numerical solution.

4 个评论
显示 2更早的评论 隐藏 2更早的评论

KLETECH MOTORSPORTS
KLETECH MOTORSPORTS 2020-11-15
编辑：KLETECH MOTORSPORTS 2020-11-15
Could you please clarify what solution you are talking about?
Also, i was thinkig more along the terms of using "integral" or "cumtrapz"to solve this.
Basically, i need to integrate the acceleration equation twice to get displacement, and then plot that against the existing espression for displaceent, i.e, x= A*sin(w*t)
MY Mistake was taking w=dx/dt.
Very sorry.
w and A are constants, they are specified initially.
I have already solves the acceleration expression using ode45. Itsimplifies the acceleration equation into 2 linear ode's.
What i want to do here is integrate it.
Ameer Hamza
Ameer Hamza 2020-11-15
编辑：Ameer Hamza 2020-11-15
Yes, your question makes more sense after the correction. But still, there is the issue of the initial condition. You mentioned that the equation for displacement is
But you also mentioned, x(0)=pi/4. As you can see, that if you put t=0 in above equation, you cannot get pi/4.
In any case, you can use ode45(), which is used to solve ODEs. The reason you cannot use integral() or cumtrapz() is that they don't support initial conditions.
ode45() can be used like this
IC = [0; 0];
tspan = [0 10];
[t, y] = ode45(@odeFun, tspan, IC);
plot(t, y)
legend({'$x$', '$\dot{x}$'}, 'Interpreter', 'latex', 'FontSize', 16, 'Location', 'best')
function dxdt = odeFun(t,x)
A = 1;
w = 3;
dxdt = zeros(2,1);
dxdt(1) = x(2);
dxdt(2) = -A*w^2*sin(w*t);
end
KLETECH MOTORSPORTS
KLETECH MOTORSPORTS 2020-11-15
I actually did use cumtrapz , and i think it worked, unless the logic is wrong:
i mistakenly used x=pi/4. what i should have said was A=pi/4. The amplitude is pi/4.
%this part integrates the acceleration(y1 = d^2(x)/dt^2) twice
% to get both velocity(y2) and displacement(y3)
t=0:0.01:2
w=15.24; %calculated using findwsq.m
A=pi/2;
y1=-A*(w.^2)*sin(w*t); %here y1 is diff(x,t,2) , i.e, accln
y2=cumtrapz(t,y1); %here y2 is diff(x,t) , i.e, velocity
y3=cumtrapz(t,y2); %here y3 is the
y3a= A*sin(w*t); %direct displacement expression
plot(t,y3, t,y3a)
Ameer Hamza
Ameer Hamza 2020-11-15
Yes, in that case, initial conditions will be zero and you can use cumtrapz().

请先登录，再进行评论。

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Numeric Solvers 的更多信息

产品

版本

R2020b

标签

Community Treasure Hunt

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

Start Hunting!

Translated by