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# Roots and Optimization assignment | |
Demo of uploading assignment | |
## Solution to Gold Chain nonlinear spring constants | |
![Gold chain TEM image](au_chain.jpg) | |
![Gold chain model (artist rendition)](Auchain_model.png) | |
# Homework #3 plots of Gold chain F vs dx and Lennard-Jones Potential | |
```matlab | |
epsilon = 0.039; % kcal/mol | |
epsilon = epsilon*6.9477e-21; % J/atom | |
epsilon = epsilon*1e18; % aJ/J | |
% final units for epsilon are aJ | |
sigma = 2.934; % Angstrom | |
sigma = sigma*0.10; % nm/Angstrom | |
x=linspace(2.8,6,200)*0.10; % bond length in um | |
Ex = lennard_jones(x,sigma,epsilon); | |
%[Emin,imin]=min(Ex); | |
[xmin,Emin] = fminbnd(@(x) lennard_jones(x,sigma,epsilon),0.28,0.6) | |
h1=figure(1) | |
plot(x,Ex,xmin,Emin,'o') | |
ylabel('Lennard Jones Potential (aJ/atom)') | |
xlabel('bond length (nm)') | |
saveas(h1,'potential_energy.png') | |
Etotal = @(dx,F) lennard_jones(xmin+dx,sigma,epsilon)-F.*dx; | |
% Now with xmin determined find F vs dx for gold chain model | |
% with 50 steps from 0 to 0.0022 nN | |
N=50; | |
dx = zeros(1,N); % [in nm] | |
F_applied=linspace(0,0.0022,N); % [in nN] | |
for i=1:N | |
optmin=goldmin(@(dx) Etotal(dx,F_applied(i)),-0.001,0.035); | |
dx(i)=optmin; | |
end | |
h2=figure(2) | |
plot(dx,F_applied) | |
xlabel('dx (nm)') | |
ylabel('Force (nN)') | |
saveas(h2,'force_vs_dx.png') | |
``` | |
This script in included as `gold_chain_script.m` | |
Output is two plots, 'potential_energy.png' and 'force_vs_dx.png' | |
![Lennard-Jones potential energy for no applied force](potential_energy.png) | |
![Force vs displacement for gold chain](force_vs_dx.png) |