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 po

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 = 3.65 J/m3 , B = 3.25 J/m2 , C = 2.1 J/m , and D = 3.6 J . Answer in units of m.

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Kim Cúc 4 years 2021-08-25T11:07:04+00:00 2 Answers 67 views 0

Answers ( )

  1. King
    0
    2021-08-25T11:08:25+00:00
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    Answer:

    x \approx 0.232\,m.

    Explanation:

    First and second derivatives of the potential energy function are, respectively:

    \frac{dU}{dx} = 3\cdot A\cdot x^{2} + 2\cdot B\cdot x - C

    \frac{d^{2}U}{dx^{2}} = 6\cdot A \cdot x + 2\cdot B

    Particular forms of the function and its derivatives are:

    U = 3.65\cdot x^{3}+3.25\cdot x^{2}-2.1\cdot x+ 3.6

    \frac{dU}{dx} = 10.95\cdot x^{2} + 6.5\cdot x - 2.1

    \frac{d^{2}U}{dx^{2}} = 21.9 \cdot x + 6.5

    Let find the roots associated with the first derivative by using the general formula of the second-orden polynomial:

    10.95\cdot x^{2} + 6.5\cdot x - 2.1 = 0

    x_{1}\approx 0.232\,m, x_{2}\approx -0.825

    The values of the second derivatives evaluated in each solution are:

    \frac{d^{2}U}{dx^{2}} = 21.9 \cdot (0.232\,m) + 6.5

    \frac{d^{2}U}{dx^{2}} = 11.581

    \frac{d^{2}U}{dx^{2}} = 21.9 \cdot (-0.825) + 6.5

    \frac{d^{2}U}{dx^{2}} = -11.568

    x_{1} has an stable equilibrium solution. Hence, x \approx 0.232\,m.

  2. RobertKer
    0
    2021-08-25T11:08:34+00:00
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    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.

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