How to find the frequency of small oscillation of a particle under gravity that moves along curve $y = a x^4$ where $y$ is vertical height and $(a>0)$ is constant?
I tried comparing $V(x) = \frac 1 2 V''(0) x^2 + \mathscr O(x^3) = \frac 1 2 kx^2$ (assuming $V(0)$ is ground state and $V'(0)$(slope) remains is horizontal at extrememum, but unfortunately $V''(0) = 0$. I am pretty much clueless. Thanks for your help.
ADDED:: Letting $m=1$ the Lagrangian is $L = \frac 1 2 (\dot x^2 + \dot y^2) - gax^4 = \frac 1 2 (\dot x^2 + (4 ax^3 \dot x)^2)-gax^4 = \frac 1 2 \dot x^2(1 + 16a^2x^6)-gax^4$
The above Lagrangian gives the equation of motion as $$\ddot x(1+16a^2x^6)+\dot x^2 96 a^2x^5 + 4 gax^3 = 0$$
Since we are considering the system a small oscillation whose potential is of order 4, $\mathscr{O}(x^{k>4})$ can be ignored which reduces into $\ddot x = -4agx^3$.
To solve this, $$\frac 1 2 \frac{d}{dt}(\dot x^2) = -\frac{d}{dt}(agx^4)$$ which gives $\dot x = \sqrt{k - 2agx^4}$, Assuming the system begins from $t=0$ at $x=x_0$ with $\dot x = 0$, $k = 2agx_0$, which turns the integral into $$\int_0^{T/4} dt= \int_0^{x_0} \frac 1 {\sqrt{2agx_0^4-2agx^4}} dx=\frac{1}{\sqrt{2ag}x_0}\int_0^1\frac{1}{\sqrt{1-y^4}}dy$$
Which gives $$T = 2 \sqrt 2 \sqrt{\frac{\pi}{ag}}\cdot \frac{\Gamma(5/4)}{\Gamma(3/4)}\cdot \frac 1 {x_0}$$
which is a dubious result. Please someone verify it. Any other methods are welcome.
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