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 $a^{3/2}$

Answer 1

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

$\text{Dimensions of time period}\left(T\right]=\left(M^0{L}^0{T}^1\right]\text{Time period}T\text{can depend on the three variables}m,k\text{and}a.\left(m\right]=\left(M^1{L}^0{T}^0\right]\left(a\right]=\left(M^0{L}^1{T}^0\right]\text{We are given}U=k{x}^3.\left(U\right]=\left(M^1{L}^2{T}^{-2}\right]\text{and}\left(x\right]=\left(M^0{L}^1{T}^0\right]\text{Therefore,}\left(k\right]=\left(M^1{L}^{-1}{T}^{-2}\right]\text{Let}T\propto {\left(m\right]}^a{\left(k\right]}^b{\left(a\right]}^c\text{i.e}T\propto {\left(M\right]}^{a+b}{\left(L\right]}^{-b+c}{\left(T\right]}^{-2b}\propto {\left(M\right]}^0{\left(L\right]}^0{\left(T\right]}^1\Rightarrow a+b=0-b+c=0-2b=1\text{Solving the equations, we get}c=-\frac{1}{2}\text{Therefore},T\propto \frac{1}{\sqrt{a}}$