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Question

A particle of mass $$m$$ is executing oscillations about the origin on the $$x$$-axis. It's potential energy is $$U(x)=k\left | x \right |^{3}$$, where $$k$$ is a positive constant. If the amplitude of oscillation is $$a$$, then its time period $$T$$ is:


A
proportional to 1a
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B
independent of a
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C
proportional to a
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D
proportional to a3/2
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Solution

The correct option is A proportional to $$\dfrac{1}{\sqrt{a}} $$
$$U=k |x|^3 \Rightarrow F=-\dfrac{dU}{dx}=-3k|x|^2 ...(i)$$
Also, for SHM $$x=a \sin \omega t$$ and $$\dfrac{d^2x}{dt^2}+\omega ^2x=0$$
$$\Rightarrow acceleration\ a=\dfrac{d^2x}{dt^2}=-\omega ^2x \Rightarrow  F=ma$$
$$=m \dfrac{d^2x}{dt^2}= - m \omega ^2x ...(ii)$$
From equation $$(i)$$ & $$ (ii)$$ we get $$\omega =\sqrt{\dfrac{3kx}{m}} $$
$$ \Rightarrow T=\dfrac{2 \pi}{\omega}= 2 \pi \sqrt{\dfrac{m}{3kx}}= 2 \pi \sqrt{\dfrac{m}{3k(a \sin \omega t)}} \Rightarrow T \propto \dfrac{1}{\sqrt a}$$

Physics

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