Q92 / 180 · JIPMER 2017 · Physics

A particle of mass $m$ is executing oscillation about the origin on $X$-axis. Its potential energy is $V(x)=K|x|^3$. Where $K$ is a positive constant. If the amplitude of oscillation is $a$, then its time period $T$ is proportional to

  • Correct answer
    A
    $1 / \sqrt{a}$
  • B
    a
  • C
    $\sqrt{a}$
  • D
    $a^{3 / 2}$

Solution

Given, $V=k x^3$ Total energy $(E)=$ Maximum potential energy $=k a^3$ From conservation of energy $\begin{aligned} \frac{1}{2} m\left(\frac{d x}{d t}\right)^2+k x^2 & =k a^3 \\ \Rightarrow \frac{1}{2} m\left(\frac{d x}{d t}\right)^2 & =k\left(a^3-x^3\right) \\ \Rightarrow \frac{d x}{d t} & =\sqrt{\frac{2 k}{m}\left(a^3-x^3\right)} \\ \frac{d x}{\sqrt{a^3-x^3}} & =\sqrt{\frac{2 k}{m}} d t\end{aligned}$ or Integrating $\int_0^a \frac{d x}{\sqrt{a^3-x^3}}=\sqrt{\frac{2 k}{m}} \int_0^{T / 4} d t$ or $\int_0^a \frac{d x}{\sqrt{a^3-x^3}}=\sqrt{\frac{2 k}{m}} \frac{T}{4}$ Let $x=a \sin \theta \Rightarrow d x=a \cos \theta d \theta$ $\therefore \int_0^{\pi / 2} \frac{a \cos \theta d \theta}{\sqrt{a^3-(a \sin \theta)^3}}=\sqrt{\frac{2 k}{m}} \frac{T}{4}$

Hence, the correct answer is option: 'A'.

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