### A simple pendulum consisting of a mass suspended in the vertical plane by a massless string.

A simple pendulum is consist of a 0.25 kg mass suspended in the vertical plane by a massless string of 0.5 m long. Assume that the pendulum moves in a plane. We need to numerically simulate the motion of the pendulum if it is released from rest with the string at 3 degrees from the vertical. We also need to plot the angle that the string makes with the vertical as a function of time. On the same axes, we need to plot the analytically solution for the angle obtained by linearizing the equations of motion assuming small oscillations.

### Linearizing the equation of motion by assuming small motion:

We have used three methods to plot the responses. Using Visual Basic's code in excel, using MatLAB code, and MatLAB using ode45 (ordinary differential equation)

### Numerical Solution vs. Theoretical Solution by Excel using Visual Basic in time steps of 0.00001 sec and step report of 0.1 sec

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 tstep 0.00001 t θ θ dot θ calc θ dot calc g 9.81 0 0.052359878 0 0.052359878 0 l 0.5 0.1 0.047309471 -0.099363422 0.047306808 -0.099403607 0.2 0.033130848 -0.179569434 0.033122908 -0.179621022 step 0.1 0.3 0.012560726 -0.225132799 0.012545855 -0.225169238 0.4 -0.010435845 -0.227254355 -0.010452712 -0.227256856 0.5 -0.031416854 -0.185522546 -0.031433769 -0.185480938 0.6 -0.046336661 -0.10800505 -0.046347699 -0.107904781 0.7 -0.052317983 -0.00966367 -0.052315919 -0.009501595 0.8 -0.048206739 0.090551683 -0.048186484 0.090735523 0.9 -0.03479602 0.173302551 -0.03475643 0.173459503 1 -0.014671658 0.222630468 -0.014617932 0.222703537 1.1 0.008286473 0.229000974 0.008342018 0.228962885 1.2 0.029643814 0.191179991 0.029691849 0.191029412 1.3 0.045283128 0.116471554 0.045310767 0.116224772 1.4 0.052185174 0.019311907 0.052184115 0.018987236 1.5 0.049020361 -0.081581489 0.04898525 -0.081915084 1.6 0.036397484 -0.166747352 0.036331592 -0.167006725 1.7 0.016752931 -0.219746566 0.016665464 -0.219863891 1.8 -0.006126974 -0.230346508 -0.006217316 -0.23028446 1.9 -0.027822822 -0.196494403 -0.027900073 -0.196257126 2 -0.044152002 -0.124728309 -0.044197753 -0.124349608

#### Numerical simlation vs. theoretical solution using MATLAB

clear all

% l*(theta doubledot)+g*theta=0 .... (thtea doubledot)=-g/l*theta
%[z1dot;z2dot] =[z2; -g/l*theta]

clear all
m=.25;
g=9.81;
l=0.5;
defn=inline('[z2;-9.81*0.25*.5/(.5^2*.25)*sin(z1)]','t','z1','z2');
a=0; b=2; N=2/0.001;
h=(b-a)/N; t=a+h*(0:N); LT=length(t);
x0=3*pi/180; y0=0;
w(:,1)=[x0;y0];
w(:,2)=[x0;y0];
for j=2:LT
w(:,j)=w(:,j-1)+h*defn(t(j-1),w(1,j-1),w(2,j-1));
end
z1=w(1,:);
z2=w(2,:);
thetat=3*pi/180*cos(sqrt(9.81/.5)*t);
xt=thetat
figure(1)
plot(t,z1,'b',t,xt,'r--')
xlabel('t -sec')
legend('\theta','\theta true')
title('Numerical Simulation vs. theoretical calculation')
grid on
figure(2)
plot(t,z2,'g',t,yt,'r--')
xlabel('t -sec')
legend('\theta_d_o_t','\theta_d_o_t true')
title('Numerical simulation of velocity vs. theoretical solution')
grid on



#### Numerical solution using MATLAB ode45



function zprime = ME716_6_1(t,z);
%SECONDODE: Computes the derivatives of y 1 and y 2,
%as a colum vector

g=9.81; l=0.5;

zprime = [z(2); -g/l*sin(z(1))];

The top and bottom part of this code from this section should be saved into different m-files

T=[0:0.1:2];
Z0=[85*pi/180 0]';

[t,z]=ode45(@ME716_6_1,T,Z0)

g=9.81
l=0.5
m=0.25

theta=85*pi/180*cos(sqrt(g/l)*t)
tdot=-85*pi/180*sqrt(g/l)*sin(sqrt(g/l)*t)
figure(1)
plot(t,z(:,1),'r',t,theta,'b--')
title('Numerical simulation vs. theoretical calc.')
xlabel('time t sec')
ylabel('angular position \theta rad')
legend('\theta Numerical', '\theta calc.')
grid on

figure(2)
plot(t,z(:,2),'r',t,tdot,'b--')
title('Numerical simulation vs. theoretical calc.')
xlabel('time t sec')
ylabel('angular velocity \theta_d_0_t rad/sec')
legend('\theta_d_o_t Numerical', '\theta_d_o_t calc.')
grid on


Note: When you want to change IVP (initial value problem) in the angle, just change the initial angle where you want to release the pendulum from the rest.

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