Question

A particle of mass 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 proportional

• A. proportional to $\frac{1}{\sqrt{a}}$
• B. proportional to $\sqrt{a}$
• C. Independent $a^{\frac{3}{2}}$
• D. None of these

Answer 1

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

$V\left(x\right)=k|x{|}^3$
Since, $F=-\frac{dV\left(x\right)}{dx}=-3k|x{|}^2...\left(i\right)$
$x=a\sin \left(\omega t\right)$
This equation always fits to the differential equation
$\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$ or $m\frac{d^{2}x}{dt}=-m{\omega }^{2}x$
$\Rightarrow F=-m{\omega }^{2}x...\left(ii\right)$
Equations $\left(i\right)$ and $\left(ii\right)$ give
$-3k|x{|}^2=-m{\omega }^{2}x$
$\Rightarrow \omega =\sqrt{\frac{3kx}{m}}=\sqrt{\frac{3ka}{m}}\left[\sin \right(\omega t){]}^{1/2}$
$\Rightarrow \omega \propto \sqrt{a}$
$\Rightarrow T\propto \frac{1}{\sqrt{a}}$