Answer:
[tex]x \approx 0.232\,m[/tex].
Explanation:
First and second derivatives of the potential energy function are, respectively:
[tex]\frac{dU}{dx} = 3\cdot A\cdot x^{2} + 2\cdot B\cdot x - C[/tex]
[tex]\frac{d^{2}U}{dx^{2}} = 6\cdot A \cdot x + 2\cdot B[/tex]
Particular forms of the function and its derivatives are:
[tex]U = 3.65\cdot x^{3}+3.25\cdot x^{2}-2.1\cdot x+ 3.6[/tex]
[tex]\frac{dU}{dx} = 10.95\cdot x^{2} + 6.5\cdot x - 2.1[/tex]
[tex]\frac{d^{2}U}{dx^{2}} = 21.9 \cdot x + 6.5[/tex]
Let find the roots associated with the first derivative by using the general formula of the second-orden polynomial:
[tex]10.95\cdot x^{2} + 6.5\cdot x - 2.1 = 0[/tex]
[tex]x_{1}\approx 0.232\,m, x_{2}\approx -0.825[/tex]
The values of the second derivatives evaluated in each solution are:
[tex]\frac{d^{2}U}{dx^{2}} = 21.9 \cdot (0.232\,m) + 6.5[/tex]
[tex]\frac{d^{2}U}{dx^{2}} = 11.581[/tex]
[tex]\frac{d^{2}U}{dx^{2}} = 21.9 \cdot (-0.825) + 6.5[/tex]
[tex]\frac{d^{2}U}{dx^{2}} = -11.568[/tex]
[tex]x_{1}[/tex] has an stable equilibrium solution. Hence, [tex]x \approx 0.232\,m[/tex].
Given Information:
Function = U = Ax³ + Bx² − Cx + D
A = 3.65 J/m³
B = 3.25 J/m²
C = 2.1 J/m
D = 3.6 J
Required Information:
Stable equilibrium position = X = ?
Answer:
Stable equilibrium position = X₂ = 0.232
Explanation:
Stability can be determined by analyzing the slope of the function. First we will find out the first derivative of the given function and the equilibrium position will be at dU/dx = 0.
U = Ax³ + Bx² − Cx + D
U = 3.65x³ + 3.25x² − 2.1x + 3.6
dU/dx = 3*(3.65)x² + 2*(3.25)x - 2.1 + 0
dU/dx = 10.95x² + 6.5x - 2.1
10.95x² + 6.5x - 2.1 = 0
Solving the above quadratic equation yields,
X₁ = -0.826
X₂ = 0.232
The stable point is the one for which the second derivative dU²/dx² > 0 and the unstable point is the one for which dU²/dx² < 0.
dU/dx = 10.95x² + 6.5x - 2.1
dU²/dx² = 2*(10.95)x + 6.5 - 0
dU²/dx² = 21.9x + 6.5
For X₁ = -0.826
dU²/dx² = 21.9(-0.826) + 6.5
dU²/dx² = -18.09 + 6.5
dU²/dx² = -11.59
Since dU²/dx² < 0, X₁ = -0.826 is the unstable point.
For X₂ = 0.232
dU²/dx² = 21.9(0.232) + 6.5
dU²/dx² = 5.08 + 6.5
dU²/dx² = 11.58
Since dU²/dx² > 0, X₂ = 0.232 is the stable point.
