Question

establish the energy test for stability.

Answer 1

$\textbf{energy test for stability}$

energy and stability are interrelated to each other. stability can be defined with help of second derivatives of energy . yeah of course we have to use $\textbf{calculus}$ to define it,
as you know, when potential energy of system is function of position e.g., $P.E=U(x)$
then, negative of second derivatives equals force .
mathematically, $F=-\frac{d^2U(x)}{dx^2}$
you also know, mechanical equilibrium depends on net force acts on system of bodies.

case 1 :- when second derivatives < 0 e.g., $\frac{d^2U(x)}{dx^2}$ < 0
then, potential energy is at a local maximum, { e.g., system has maximum amount of energy} which means that the system is in an unstable equilibrium state. see 1st figure to understand unstable equilibrium.

case 2 :- when second derivatives > 0 e.g., $\frac{d^2U}{dx^2}$ > 0
then, potential energy is at a local minimum {e.g.,system has minimum amount of energy} which means that the system is in a stable equilibrium state. see 2nd figure to understand stable equilibrium

case 3 :- when second derivatives = 0 or doesn't exist e.g., $\frac{d^2U(x)}{dx^2}=0$ or doesn't exist.
then, potential energy is at critical point. which means that system is in a neutral equilibrium.
see 3rd figure to understand neutral equilibrium.

Answer 2

Answer:

Explain energy test for stability