Answer to Question #101611 in Mechanics | Relativity for John

Question

Find the positions at which a particle of mass m move on the x-axis with energy v=E/q^4(x^4+4ax^3-8a^2x^2) is in stable equilibrium

Answer 1

$V(x)=\frac{E}{a^4}(x^4+4ax^3-8a^2x^2)$

*q isn’t given, so I think there should be a (in this case we know, that a ≠ 0)

For the equilibrium positions, $\frac{d V}{d x }$ will be zero.

$\frac{d V}{d x}=\frac{d}{d x}(\frac{E}{q^4}(x^4+4ax^3-8a^2x^2))=\frac{E}{q^4}(4x^3+12ax^2-16a^2x)=0$

$\frac{E}{a^4} 4x(x^2+3ax-4a^2)=0$

$x(x+4a)(x-a)=0$

$x_1=0, x_2=-4a, x_3=a$ (if a≠0, then all these points are different)

For stable equilibrium $\frac{d ^2 V}{d x^2}$ must be greater than zero so that potential energy is a minimum.

$\frac {d^2 V}{d x ^2}=\frac{d}{dx}(\frac{E}{a^4}(4x^3+12ax^2-16a^2x))=\frac{E}{a^4}(12x^2+24ax-16a^2)$

At $x=0$ gives $\frac{d ^2 V}{d x^2}=-\frac{16E}{a^2}$

At $x=-4a:$ $\frac{d ^2 V}{d x^2}=\frac{80E}{a^2}$

At $x=a: \frac{d ^2 V}{d x^2}=\frac{20E}{a^2}$

Answer:

if $E>0: \; x_1=-4a,x_2=a$ are positions of stable equilibrium

if $E<0: \;x=0$ is the point of stable equilibrium

if $E=0 \Rightarrow V(x)=0$ neutral equilibrium.