How do I plot a point at the x intercept on this graph?

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Rhianne Tallarico
Rhianne Tallarico el 20 de Feb. de 2018
Respondida: Are Mjaavatten el 20 de Feb. de 2018
e=1;
sigma=0.154;
r=linspace(0.154,1,100);
U = @(e,sigma,r) 4*e*((sigma./r).^12)-((sigma./r).^6);
norm_r = r./sigma;
figure
plot(norm_r,U(e,sigma,r))
title('Lennard-Jones Potential')
xlabel('Distance (nm)')
ylabel('Potential Energy')
  2 comentarios
dpb
dpb el 20 de Feb. de 2018
Not sure precisely what you want to add where???
Rhianne Tallarico
Rhianne Tallarico el 20 de Feb. de 2018
This code plots a curved line, I am looking to find the x intercept value of the line, and plot a marker at that point. Thanks!

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Respuestas (1)

Are Mjaavatten
Are Mjaavatten el 20 de Feb. de 2018
Let U be a function of r only and find the solution to U(r) = 0:
e=1;
sigma=0.154;
r=linspace(0.154,1,100);
U = @(r) 4*e*((sigma./r).^12)-((sigma./r).^6);
norm_r = r./sigma;
figure
plot(norm_r,U(r))
title('Lennard-Jones Potential')
xlabel('Distance (nm)')
ylabel('Potential Energy')
norm_r_0 = fzero(U,1)/sigma;
hold on; plot(norm_r_0,0,'*r')

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