Question

A particle of mass $m$ is executing oscillations about the origin on the x-axis. Its potential energy is $U\left(x\right)=k{\left(\left[x\right]\right)}^3$ , 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 to a
• C. Proportional to $\sqrt{a}$
• D. Proportional to $a^{\frac{3}{2}}$

Answer 1

Answer: (A)

Given,
$U=k{\left|x\right|}^3$
From conservative force and potential energy relation, we know,
$\Rightarrow F=-\frac{dU}{dx}=-3k{\left(\left|x\right|\right)}^2...\left(1\right)$
Also, for SHM $x=asin\omega t$ and $\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$
$\Rightarrow$ Acceleration $a=\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$
$\Rightarrow F=ma$
$\Rightarrow F=-m{\left(\omega \right)}^{2}x...\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$ we get $\omega =\sqrt{\frac{3kx}{m}}$
$\Rightarrow T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{3kx}}=2\pi \sqrt{\frac{m}{3k\left(asin\omega t\right)}}$ $\Rightarrow T\propto \frac{1}{\sqrt{a}}$ .