Equilibrium and concept of potential well

In all conservative field, potential energy U = U (x, y, z)

$\because\;\;\;\;\;\overrightarrow{F}=-\nabla U=-\left ( \widehat{i}\frac{\partial U}{\partial x}+ \widehat{j}\frac{\partial U}{\partial y}+ \widehat{k}\frac{\partial U}{\partial z} \right )$

If particle moves only along x-axis, then force F_x = – (∂U/∂x)
Similarly F_y = – (∂U/∂y) and F_z = – (∂U/∂z)

Force = slope of tangent at any point of the curve
Since the tangent at P, Q, R and S are parallel to x-axis

$\therefore \;\;\;\;\; \left.\frac{\partial U}{\partial x} \right \}_{P}= \left.\frac{\partial U}{\partial x} \right \}_{Q}= \left.\frac{\partial U}{\partial x} \right \}_{R}= \left.\frac{\partial U}{\partial x} \right \}_{S}=0$

$or \;\;\;\;\; F_{P}=F_{Q}=F_{R}=F_{S}=0$

These positions are known as equilibrium positions.
If a particle is slightly displaced from stable equilibrium position P, then it starts to oscillate between points A and B until it crosses point B.
P is the position having minimum potential energy and is called stable equilibrium.
The region of minimum potential energy bounded between points A and B is called potential well .
The difference between the maximum and minimum potential energy of a potential well is called the binding energy of potential well.
If the energy of particle is less than the B.E. of well, then it is always bound in the potential well and such state is called bound state .
Let a particle be slightly displaced from its mean position P = (x = x₀) then from Taylor series expansion, its potential energy at any point x is

$\therefore \;\;\;\;\; U(x)=U(x_{0})+(x-x_{0})\left. \frac{\partial U}{\partial x}\right \}_{x_{0}}+\frac{(x-x_{0})^{2}}{2!}\left. \frac{\partial ^{2}U}{\partial x^{2}}\right \}_{x_{0}}+...$

For stable equilibrium position P

$\;\;\;\;\; U(x_{0})=min.,\;\;\;\; \left. \frac{\partial U}{\partial x}\right \}_{x_{0}}=0$

$\;\;\;\;\; \left. \frac{\partial^{2} U}{\partial x^{2}}\right \}_{x_{0}}>0=k(say);\;\;\;\;\left. \frac{\partial^{3} U}{\partial x^{3}}\right \}_{x_{0}}>0=k_{1}(say)$

$\because \;\;\;\;\; U(x)=U(x_{0})+(x-x_{0})\left. \frac{\partial U}{\partial x}\right \}_{x_{0}}+\frac{(x-x_{0})^{2}}{2!}\left. \frac{\partial ^{2}U}{\partial x^{2}}\right \}_{x_{0}}$

$+\frac{(x-x_{0})^{3}}{3!}\left. \frac{\partial ^{3}U}{\partial x^{3}}\right \}_{x_{0}}+...$

$\therefore \;\;\;\;\; U(x)=U(x_{0})+(x-x_{0})\times 0 + \frac{(x-x_{0})^{2}}{2!}\times k+\frac{(x-x_{0})^{3}}{3!}\times k_{1}+...$

$or \;\;\;\;\; U(x)=U(x_{0})+ \frac{1}{2}k(x-x_{0})^{2}+\frac{1}{6}k_{1}(x-x_{0})^{3}+...$

If P lies at origin i.e., x₀ = 0 and U(x₀) = 0

$\therefore \;\;\;\;\; U(x)= \frac{1}{2}kx^{2}+\frac{1}{6}k_{1}x^{3}+...$

$or \;\;\;\;\; F_{x}=-\frac{\partial U}{\partial x}=-kx-\frac{1}{2}k_{1}x^{2}+...$

For small displacement x³ → 0, x⁴ → 0, …

$U(x)=\frac{1}{2}kx^{2}$

$\therefore\;\;\;\;\; F_{x} =-kx$

In this position a curve between displacement and potential energy will be a parabola and F ∝ x
Therefore the motion of particle in a parabolic potential well is always oscillatory and is simple harmonic .