Question

A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is U(x) = 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
• B. Independent of a
• C. Proportional to
• D. Proportional to

Answer 1

The correct option is A

Proportional to

U = $k|x{|}^3\Rightarrow F==\frac{dU}{dx}=-3k|x{|}^2$ ...(i)

Also, for SHm x = a sin $\omega$ t and $\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$

$\Rightarrow accelerationa=\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x\Rightarrow F=ma$

$=m\frac{d^{2}x}{d{t}^2}=-m{\omega }^{2}x$ ...(ii)

From equation (i) and (ii) we get $\omega =\sqrt{\frac{3kx}{m}}$

$\Rightarrow T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{3kx}}=2\pi \sqrt{\frac{m}{3k\left(asin\omega t\right)}}\Rightarrow T\propto \frac{1}{\sqrt{a}}$.