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

2 vues (au cours des 30 derniers jours)
Rhianne Tallarico
Rhianne Tallarico le 20 Fév 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 commentaires
dpb
dpb le 20 Fév 2018
Not sure precisely what you want to add where???
Rhianne Tallarico
Rhianne Tallarico le 20 Fév 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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Réponses (1)

Are Mjaavatten
Are Mjaavatten le 20 Fév 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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