The potential energy of a particle of mass $0.1\,kg,$ moving along $x-$ axis, is given by $U = 5x(x-4)\,J$ where $x$ is in metres. It can be concluded that
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$F=-\frac{d U}{d x}=-\frac{d}{d x}\left(5 x^{2}-20 x\right)=-10 x+20$
Here we see that force is directly proportional to the displacement and opposite to the direction of displacement. So it satisfies the condition of $SHM$
Speed of the particle is maximum in mean position. i.e. where force is zero. i.e. $F=-10 x+20=0 \Rightarrow x=2 m$
Speed of the particle is maximum at $x=2 m . \Rightarrow B$ is correct. since $F=-10 x+20$ comparing it with general equation of $\operatorname{SHM} F=-k x,$ we have $k=10 N / m$
$\Rightarrow T=2 \pi \sqrt{\frac{m}{k}}$
$\Rightarrow T=2 \pi \sqrt{\frac{0.1}{10}}=2 \pi \sqrt{\frac{1}{100}}=\frac{2 \pi}{10}=\frac{\pi}{5} s$
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