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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
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$1 / \sqrt{a}$
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 $x=a \sin \theta \Rightarrow d x=a \cos \theta d \theta$
$\therefore \int_0^{\pi / 2} \frac{a \cos \theta d \theta}{\sqrt{a^3-(a \sin \theta)^3}}=\sqrt{\frac{2 k}{m}} \frac{T}{4}$
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 $x=a \sin \theta \Rightarrow d x=a \cos \theta d \theta$
$\therefore \int_0^{\pi / 2} \frac{a \cos \theta d \theta}{\sqrt{a^3-(a \sin \theta)^3}}=\sqrt{\frac{2 k}{m}} \frac{T}{4}$
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