By studying the dynamics of a pendulum like that shown in Figure 8.6-1, we can better understand the dynamics of machines such as a robot arm. The pendulum shown consists of a concentrated mass m attached to a rod whose mass is small compared to m. The rod’s
length is L. The equation of motion for this pendulum is
(j + ! sin 0 = 0 (8.6-3)
L
Suppose that L = 1m and g = 9.81 m/s2. Use MATLAB to solve this equation for OCt) for two cases: 0(0) = 0.5 rad and 0(0) = 0.8rr rad. In both cases 0(0) = O.Discuss how to check the accuracy of the results .
• Solution
If we use the small angle approximation sin 0 ~ 0, the equation becomes .. g.
0+ -0 = 0 L
which is linear and has a solution given by (8.4-16):
(8.6-4)
OCt) = 0(0) cos J[t (8.6-5)
Thus the amplitude of oscillation is 0(0) and the period is P = 2rr/ Jg/ L = 2 s. We can use this information to select a final time and to check our numerical results. First rewrite the pendulum equation (8.6-3) as two first-order equations. To do
this,let

Thus
XI = iJ =X2
X2 = iJ = -!sin r,
L
The following function file is based on the last two equations. Remember that the output xdot must be a column vector.
function xdot = pend l(t,x)
global 9 L xdot = [x(2); – (g/L) *sin(x(1))];
It is called as follows. The vectors ta and xa contain the results for the case where 8(0) = 0.5. The vectors tb and xb contain the results for 0(0) = 0.8Jl’.
global 9 L
9 = 9.81;L = 1;
[ta, xa] = ode45(‘pendul’, [0,5]. [0.5,0]);
[tb,.xb] = ode45(‘pendul’, [0,5]. [0.8*pi, 0]);
plot (ta,xa (:,1), tb,xb(:, 1)) ,xlabel (‘Time (s)’), …
ylabel(‘Angle (rad) ‘),gtext(‘Case 1’),gtext(‘Case 2’)
The results are shown in Figure 8.6-2. The amplitude remains constant, as predicted by the small angle analysis. and the period for the case where 0(0) = 0.5 is a little longer

The pendulum angle as a ~ of time for two starting positions.

than 2 s, the value predicted by the small angle analysis. Therefore, we can place some confidence in the numerical procedure. For the case where 8(0) = O.Srr, the period of the numerical solution is about 3.3 s. This longer period illustrates an important property
of nonlinear differential equations. The free response of a linear equation has the same period for any initial conditions; however, the form of the free response of a nonlinear equation oftens depends on the particular values of the initial conditions