Q. A particle of mass $m$ is executing oscillations about the origin on the $x$-axis. Its potential energy is $U(x) = k\lbrack x\rbrack^{3},$ 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 $a^{3/2}$
Explanation
(a) $U = k|x|^{3} \Rightarrow F = - \frac{dU}{dx} = - 3k|x|^{2}$ …(i)
Also, for SHM $x = a\sin{\omega t}$ and$\ \frac{d^{2}x}{dt^{2}} + \omega^{2}x = 0$
$\Rightarrow$ acceleration $a = \frac{d^{2}x}{dt^{2}} = - \omega^{2}x \Rightarrow F = ma$
$= m\frac{d^{2}x}{dt^{2}} = - m\omega^{2}x\ \ \ \ \ \ $ …(ii)
From equation (i) & (ii) we get $\omega = \sqrt{\frac{3kx}{m}}$
$\Rightarrow T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{3kx}} = 2\pi\sqrt{\frac{m}{3k(a\sin{\omega t})}} \Rightarrow T \propto \frac{1}{\sqrt{a}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.