Question

A particle of mass m is executing oscillations about the origin on the x-axis. Its potential energy is
$U\left(x\right)=k|x{|}^3$ 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 $\frac{1}{\sqrt{a}}$

$U\left(x\right)=k|x{|}^3$

$\therefore \left[k\right]=\frac{\left[U\right]}{\left[x^3\right]}=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[L^3\right]}=\left[M{L}^{-1}{T}^{-2}\right]$

$T\propto {\left(mass\right)}^x{\left(amplitude\right)}^y{\left(k\right)}^z$

$\therefore \left[M^0{L}^{0}T\right]={\left[M\right]}^x{\left[L\right]}^y{\left[M{L}^{-1}{T}^{-2}\right]}^z=\left[M^{x+z}{L}^y{T}^{-2z}\right]$

Equating the powers, we get

$-2z=1orz=-\frac{1}{2}$

y - z = 0 or y = z = $-\frac{1}{2}$

$\therefore$ T $\propto {\left(amplitude\right)}^{\frac{1}{2}}$

or T $\propto \frac{1}{\sqrt{a}}$