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, 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 $a^{\frac{3}{2}}$

Answer 1

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

$V=k|x{|}^{3}F=-\frac{dv}{dx}=-3k|x{|}^2$ ........(1)
$Theequationofsimpleharmonicmotionisgivenas$
$x=asin\omega t\Rightarrow m\frac{d^{2}x}{d{t}^2}=m(-a{\omega }^{2}sin\omega t)=-m{\omega }^{2}x$ ..........(2)
$Using\left(i\right)and\left(ii\right),weobtain3k|x{|}^2=m{\omega }^{2}x\Rightarrow \omega =\sqrt{\frac{3kx}{m}}\Rightarrow T=2\pi \sqrt{\frac{m}{3kx}}\Rightarrow T=2\pi \sqrt{\frac{m}{3kasin\omega t}}\Rightarrow T\propto \frac{1}{\sqrt{a}}$