%% Lab 5 - Kiele Mahan - MAT 275 Lab
% The Mass-Spring System
%% EX 1
%% A)
% y=y(t) is represented b ythe blue curve because it uses the initial
% condition y=-0.8 and can be seen as the blue line's y-intercept on graph.
%% B)
% Taking the difference in the x-coordinates of two peaks on the graph, the
% period is approximately 3.685. Knowing the period is 2*pi/omega0 and
% omega0=5/3, we know the period is equal to 6*pi/5.
%% C)
% Without dampening, the mass will not come to rest. This can be seen on the
% graph as the amplitude reamining unchanged with each oscillation.
%% D)
% The amplitude of y is 0.8.
%% E)
% List the t values either in decimal format or as a fractions
% involving pi.
% Maximal velocity in magnitude is seen at
% t=1.098+(n*pi)/omega0=1.098+(n*3pi)/5, for n=0,1,2,3...
% When the mass is at the equilibrium point, the velocity is at its peak.
%% F)
% Run LAB05ex1 with the given values of m and k. Include the two
% distinct graphs, each with y(t) and v(t) plotted.
% Comment on the results.
type LAB05ex1_1.m
LAB05ex1_1
type LAB05ex1_2.m
LAB05ex1_2
%%
% The period is equal to 2pi/omega0= 2pi/sqrt(k/m)=2pi*sqrt(m/k). Looking
% at the final equation listed, we can see that an increase in m lengthens
% the period while an increase in spring stiffness k shortens the period.
%% EX 2
%% A)
% add commands to LAB05ex1 to compute and plot E(t). Then use ylim([~,~])
% to change the yaxis limits.
% Include the code, at least one plot of E(t) and a comment.
type LAB05ex1_3.m
LAB05ex1_3
%%
% The curve of E is seen to vary when the bounds of the y-ais are very
% small. After y-limits were set from 7.5 to 9.5, the line representing E
% appears constant
%% B)
% write out main steps here
% first differentiate E(t) with respect to t using the chain rule. Then
