A particle of unit mass movies on a straight line under a force having potential energy $$V(x)=\frac{bx^3}{x^4 + a^4}$$ where a,b are positive constants. Find the period of small oscillations about the position of stable equilibrium.
So, I differentiated $V(x)$ once, found the equilibrium points which were $x_1=0, x_2=a\sqrt[4]{3}, x_3=-a\sqrt[4]{3}$. By finding $V''(x)$, I verified that the STABLE equilibrium point was $c=x_3=-a\sqrt[4]{3}$ since $V''(c)=\frac{12b}{a^3}>0$ With the value for $V''(c)$ obtained I can finally wok out the period of small oscillations about the position of stable equilibrium using the formula $$2\pi \sqrt[2]\frac{m}{V''(c)}$$, where m is the mass of the particle and c is the abscissa of the stable equilibrium point. So I ended up with the period being $\frac{\pi}{3} \sqrt[2]{3a^3}$
Can someone check if this is correct? Didn't' have the patientce to write out $V'(x)$ fully. Does anyone know a good programme for that as wolfram doesn't let me insert constants like a,b in only actually numbers.
$V'(x) = -\frac{b\,{x}^{2}\,\left( {x}^{4}-3\,{a}^{4}\right) }{{\left( {x}^{4}+{a}^{4}\right) }^{2}}$. The stable equilibrium is at $\hat{x}=-\sqrt[4]{3}a$, and $V''(x) = \frac{2\,b\,x\,\left( {x}^{8}-12\,{a}^{4}\,{x}^{4}+3\,{a}^{8}\right) }{{\left( {x}^{4}+{a}^{4}\right) }^{3}}$, with $V''(\hat{x}) = \frac{3 \sqrt[4]{3} b}{ 4 a^3}$.
The linearized system around $\hat{x}$ becomes $\ddot{\delta} = - V''(\hat{x}) \delta$, hence the angular frequency is $\omega_0 = \sqrt{V''(\hat{x})} = \frac{3^\frac{5}{8}}{2}\sqrt{\frac{b}{a^3}}$, and so the period is $T = \frac{2 \pi}{\omega_0} = \frac{4 \pi}{3^\frac{5}{8}} \sqrt{\frac{a^3}{b}}$.
(Assuming I have made no mistakes, which is a big presumption.)