Question

when U=x^3 /3 - 4x +8, then find position of equilibrium and also type of equilibrium
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Answer 1

Given that,

$\sf \: U(x) = \dfrac{ {1} }{3} {x}^{3} - 4x + 8$

Since,

$\sf \: F = - \dfrac{dU}{dx}$

At equilibrium, net force is zero.

$\sf \dfrac{dU}{dx} = 0 \\ \\ \implies \sf \: x {}^{2} - 4 = 0 \\ \\ \implies \sf \boxed{ \boxed{ \sf \: x = \pm \: 2}}$

To determine the type of equilibrium, we need to find double derivative of U.

$\sf \: \dfrac{d {}^{2} U}{dx {}^{2} } = 2x$

Condition for stable equilibrium,

$\boxed{\boxed{\sf \: \dfrac{d {}^{2} U}{dx {}^{2} } > 0}}$

Condition for unstable equilibrium,

$\boxed{\boxed{\sf \: \dfrac{d {}^{2} U}{dx {}^{2} } < 0}}$

At x = 2, particle is at stable equilibrium.

$\sf \: \dfrac{d {}^{2} U}{dx {}^{2} } \bigg|_{x = 2}= 4$

At x = -2, particle is at unstable equilibrium.

$\sf \: \dfrac{d {}^{2} U}{dx {}^{2} } \bigg|_{x = - 2}= - 4$

ALITER,

Substitute x = 2 and x = - 2 in U(x).

• If U(x) is at its minima for some value of x, the particle is at stable equilibrium.
• If U(x) is at its maxima for some value of x, the particle is at unstable equilibrium.

Likewise, at x = 2 and x = - 2 particle is at stable and unstable equilibrium respectively.

Answer 2

Answer:

x=+2 and -2

Explanation: