A particle of mass $m$ is executing oscillations about the origin on the X-axis. Its potential energy is $U(x)=k|x|^3$, where $k$ is a positive constant. If the amplitude of oscillation is $a$, then its time period $T$ is proportional to $a^p$. Find $p$.
I’ll use the standard SHM derivation that we are taught. $$F=-\frac{\partial U}{\partial x}=-3k\frac{x^3}{|x|}=3k\operatorname {sgn}(-x)x^2$$ Then, acceleration $$\omega=v\frac{dv}{dx}= 3\frac km\operatorname {sgn}(-x)x^2 $$$$\implies \int_0^v vdv= \operatorname {sgn}(-x)\int_a^x 3\frac km x^2dx$$ so that $$v=\sqrt{\frac{2 \operatorname {sgn}(-x) k}{m} (x^3-a^3)}$$ Thus, $$T=\int_0^T dt \propto \int_0^x \frac{dx}{\sqrt{ \operatorname {sgn}(-x) (x^3-a^3)}}.$$ Now the question is, how do I find its dependency on $a$?I could, of course, take $a^3$ out of the integral, but I can’t guarantee that the remaining integral won’t give me any $a$ terms.
The original solution does the problem based on dimensional analysis, but it’s not so appealing that way. Also, I plugged this integral into Integral Calculator and it showed no elementary anti-derivative.
I doubt the exercise wishes for you to solve the integral, mainly because it is just asking for what power of the Amplitude the time period depends on.
You can simply note that something of dimensions $L^{-\frac32}$ when integrated wrt $L$ produces something of dimensions $L^{-\frac32+1}=L^{-\frac12}$. Since the only length based term the Time period of this motion can depend on is the amplitude, $T\propto a^{-\frac12}\implies p={-\frac12} $.
Side note, any oscillatory motion which has a potential energy defined as $U=k|x|^n$ has $T \propto a^\frac{2-n}2$, something you can derive using similar considerations.