# Matlab Diff EQ help

1. Jul 6, 2005

### gob71

I am new to matlab and need help with a problem. we have tried
several different methods and each time cant make sense of what is going on wrong

" A mass spring motion is governed by the ordinary differential
equation m(dx^2/dt^2) + b(dx/dt) + kx=F(t) , where m is the mass, b
is the damping constant, k is the spring constant, and F(t) is the
external force."
Assume the following m=1 kg, k= 16 N/m, and F(t)=0, x(0)=0, x'(0)=0.
Solve the ODE on the interval [0,20] for the following cases.

a). b=0 -no damping
b). b=2- under damping
c). b=8 - critical damping
d). b=10 - over damping

2.
Set the damping constant equal to 0, b=0, and consider an agiing spring whose spring constant depends on time as follows

k(t)=16e^-LT

Predict the motion of the mass, i.e. show time plots and phase plots and discuss your results for the following cases.

a. L=0
b. L=0.1
c. L=0.3

3.
Set b = 1/5 k=1/5 and assume a forcing F(t) = coswt,

a. Use ODE45-solver to obtain the solution curves for several values of w between .5 and 2. Plot the solutions and estimate the amplitude A of the steady response in each case.

b. Using the data from part (a) plot the graph of A vs. w. For what frequency W is the amplitude greatest?

2. Jul 6, 2005

### saltydog

Hey Gob, you got to push that spring dude. Think about it: You have a mass on a spring sitting there. It has no external force, you don't displace it, and you don't give it an initial velocity. It's not going anywhere, i.e. the only solution is y(x)=0. Are you sure you have the initial conditions right? How about just a little push, say y'(0)=0.1. That'll get it going.

3. Jul 6, 2005

### gob71

your right, the intial conditions are x(0)=1, x'(0)=0

the first answer is solved.....we are having problems with #2

or m file looks something like this

function out=spring2(t,u)
global l m b k
out(1)= u(2);
out(2)= 16 exp(-l*t)
out=out';

trying to run it it gives us u is undefined and we cant figure out what is wrong, and we cant even get to the global part to where we can sub in our different l values

4. Jul 6, 2005

### saltydog

Sorry, can' help with MatLab. I use Mathematica.

5. Jul 6, 2005

### saltydog

You know Gob, it makes sense: as you increase the value of L, the rate at which the spring constant decreases, increases right for:

$$k(t)=e^{-Lt}$$

and with no dampening, as the spring "constant" decreases, the oscillations will get larger, quicker for increasing L. That's what it looks like to me anyway. Can't you just do a numerical solve of the three ODEs with the three values of L in MatLab and plot the results?. In Mathematica it's a breeze.

6. Jul 6, 2005

### gob71

matlab blows a big one, heck i even like maple more....this thing is killin me and we are about to go nuts

7. Jul 6, 2005

Blasphemy!

I don't really understand the matlab code you provided. Is that your function for the ode solver? If you could post anything else you have or more specific problems it may help.

A quick starter for you
Code (Text):

% PF for gob71
% x(t) is position, y(t)=dx/dt=velocity
clear all;

dt = 0.0005; % Time step size
t = 0:dt:20; % Vector of time values

% Constants & Parameters
m = 1; k = 16; b = 0; w = 0; F = 0; L = 0.0;
y(1)=0; yold = y(1); % Initial Velocity
x(1)=1; xold = x(1); % Initial Position

% Real men write their own ODE solvers
for i = 2:length(t)
k = exp(-L*t(i));
% F = cos(w*i);
x(i) = (m*xold/dt^2 + b*xold/dt + m*yold/dt + F) / (m/dt^2 + b/dt + k);
y(i) = (x(i)-x(i-1))/dt;
xold = x(i); yold = y(i);
end

%figure;plot(t,x,t,cos(1*t))
figure;plot(t,x)

A few quick checks for your solution.
x'' + b/m*x' + k/m*x = F/m
If b=F=0 and m=1, then when k=16, x(t)=cos(4t), and when k=exp(0*t), x(t)=cos(1*t)

8. Jul 7, 2005

### saltydog

Why, when you can do this: 1,2,3, done . . .
Code (Text):

L=0.1
sol1 = NDSolve[{x''[t] + 16 e^{-L t} x[t] == 0,
x[0] == 1, x'[0] == 0}, x, {t, 0, 20}]
f[t_] := Evaluate[x[t] /. Flatten[sol1]];
Plot[f[u], {u, 0, 20}]
The plots are for L=0.1, 0.2, 0.3 respectively.

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Last edited: Jul 7, 2005