Approximating a quartic equation for small oscillations

Question

I have a potential:

V(x) = $(x^2-1)^4$

The question is:

Consider a particle released with small velocity near x = 1. Writing x = 1+X, show that X satisfies the approximation

m$\ddot X$+ $64X^3$ ≈ 0.

As far as I'm aware, the standard approximation for x near c would be

V(x) = $\frac{1}{2}$V''(x)(x-c)$^2$

but this doesn't seem to work in this case as V''(1)=0.

I can't seem to understand what I'm supposed to do in this question, I see that F=m$\ddot{X}$=-64X$^3$ means that V(X)=16X$^4$ but I don't understand how this relates to the V given in the question, or how I am supposed to use x=1+X in this case.

Answer 1

What you are overlooking is the $(X+2)^4$ term in that $$ V(X) = (x^2-1)^4 = (x+1)^4 (x-1)^4 =(X+2)^4 X^4 $$ Near $x=4$ (or $X=0$), the $(X+2)^4$ can be approximated as $2^4=16$ so

$$F(X) \approx 16(4X^3) = 64X^3$$

Answer 2

Writing $x = 1 + X$, we have $V(x) = X^8+8 X^7+24 X^6+32 X^5+16 X^4$. Thus Newton's law $F=ma$ gives us

$$ m \ddot{X} + \dfrac{dV}{dX} = m \ddot{X} + 8 X^7 + 56 X^6 + 144 X^5 + 160 X^4 + 64 X^3 = 0$$

When $X$ is near $0$, the higher powers of $X$ are small compared to $X^3$, so we have

$$ m \ddot{X} + 64 X^3 \approx 0 $$