Question

A particle of mass $m$ is executing oscillations about the origin on the $x$-axis. It's potential energy is $U\left(x\right)=k{\left|x\right|}^3$, where $k$ is a positive constant. If the amplitude of oscillation is $a$, then its time period $T$ is:

• A. proportional to $\frac{1}{\sqrt{a}}$
• B. independent of $a$
• C. proportional to $\sqrt{a}$
• D. proportional to $a^{3/2}$

Answer 1

The correct option is A proportional to $\frac{1}{\sqrt{a}}$

$U=k|x{|}^3\Rightarrow F=-\frac{dU}{dx}=-3k|x{|}^2...\left(i\right)$
Also, for SHM $x=a\sin \omega t$ and $\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$
$\Rightarrow accelerationa=\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x\Rightarrow F=ma$
$=m\frac{d^{2}x}{d{t}^2}=-m{\omega }^{2}x...\left(ii\right)$
From equation $\left(i\right)$ & $\left(ii\right)$ we get $\omega =\sqrt{\frac{3kx}{m}}$
$\Rightarrow T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{3kx}}=2\pi \sqrt{\frac{m}{3k\left(a\sin \omega t\right)}}\Rightarrow T\propto \frac{1}{\sqrt{a}}$