Question

A potential energy function is given by U = A x3 + B x2 − C x + D . For positive parameters A, B, C, D this potential has two equilibrium positions, one stable and one unstable. Find the stable equilibrium position for A = 1.45 J/m3 , B = 2.85 J/m2 , C = 1.7 J/m , and D = 0.6 J . Answer in units of m

Answer 1

Answer:

7.8655

Step-by-step explanation:

Given the functionl

$U=Ax^3+Bx^2-cx+D$

Where,

$A=1.45J/m^3\\B=2.85J/m^2\\c= 1.7J/m\\D=0.6J$

We have, replacing,

$U=1.45x^3+2.85x^2-1.7x+0.6$

We need to derivate and verify for stable equilibrum that,

$\frac{d^2 U}{dx^2}>0$

Then,

$U'(x) = 4.32x^2+5.7x-1.7$

Roots in,

$x_1=-1.57008\\x_2=0.250636$

We obtain U"(x), then we have

$\frac{d^2U}{dx}=8.64x+5.7$

For $x=-1.57, U"(x)=-7.86549$ Unestable

For $x=0.250636 U"(x)=7.8655>0$ Stable