Question

A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is $V\left(x\right)=k|x{|}^3$ where k is a positive constant. If the amplitude of oscillation is a, them 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 $\frac{1}{\sqrt{a}}$

$U=k|x{|}^3$
$⟹F=-\frac{dU}{dx}=-3k|x{|}^{2}equation\left(1\right)$
For SHM $x=a\sin \omega t$
and $\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$
$⟹$ acceleration, $a=\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$
$⟹F=ma$
$\therefore m\frac{d^{2}x}{d{t}^2}=-m{\omega }^{2}xequation\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$\omega =\sqrt{\frac{3kx}{m}}$
$\therefore T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{3kx}}=2\pi \sqrt{\frac{m}{3ka\sin \omega t}}$
$⟹T\propto \frac{1}{\sqrt{a}}$