Question

A particle having a mass $m$ is executing oscillation about the origin on the X-axis. The potential energy has been given as $V\left( x \right)=K{{\left| x \right|}^{3}}$. Where $K$ be a positive constant. When the amplitude of oscillation be $a$, then its time period $T$ will be proportional to,
A. proportional to $\dfrac{1}{\sqrt{a}}$
B. independent of $a$
C. proportional to $\sqrt{a}$
D. proportional to ${{a}^{\dfrac{3}{2}}}$

Answer 1

The dimensional analysis of the potential energy of the particle is to be found. The dimension of potential will be the product of the mass, length raised to negative of one and the time period raised to negative of two. This time period will be dependent upon mass, amplitude and $K$. This will help you in answering this question.

Complete step by step answer:
the potential energy of the particle has been mentioned in the question as,
$V\left( x \right)=K{{\left| x \right|}^{3}}$
Rearranging this equation in terms of the positive constant can be shown as,
$K=\dfrac{V}{{{x}^{3}}}$
The dimensional analysis of this equation can be written as,
$K=\dfrac{\left[ M{{L}^{2}}{{T}^{-2}} \right]}{\left[ {{L}^{3}} \right]}=\left[ M{{L}^{-1}}{{T}^{-2}} \right]$
Now this time period will be dependent upon mass, amplitude and $K$.
Therefore we can assume that the time period is proportional to the mass of the particle raise to $x$, amplitude raise to $y$ and $K$ raise to $z$. This can be expressed as,
$T\propto {{m}^{x}}{{A}^{y}}{{K}^{z}}$
Taking the dimensional analysis,
$\left[ {{M}^{0}}{{L}^{0}}T \right]={{\left[ M \right]}^{x}}{{\left[ L \right]}^{y}}{{\left[ M{{L}^{-1}}{{T}^{-2}} \right]}^{z}}$
Equating the powers will be given as,

& -2z=1 \\\ & z=\dfrac{-1}{2} \\\ \end{aligned}$$ Or, $$y-z=0$$ Substituting the value of $$z$$, $$\begin{aligned} & y+\dfrac{1}{2}=0 \\\ & \Rightarrow y=-\dfrac{1}{2} \\\ \end{aligned}$$ As we already mentioned the time period will be proportional to $$y$$. That is, $$\begin{aligned} & T\propto {{\left( a \right)}^{\dfrac{-1}{2}}} \\\ & \therefore T\propto \dfrac{1}{\sqrt{a}} \\\ \end{aligned}$$ Therefore the time period has been found to be proportional to $$\dfrac{1}{\sqrt{a}}$$. **So, the correct answer is “Option A”.** **Note:** The time period of oscillation can be defined as the number of oscillations completed in a unit time. The time period can be defined as the reciprocal of the frequency of the oscillation. The unit of the time period will be in seconds.