• # question_answer 7) The potential energy of a particle of mass 0.1 kg moving along the $x-$axis is given by $U=5x(x-4)\,J,$where$x$is in metre. It can be concluded that: A)  the particle is acted upon by a constant force.B)  the speed of the particle is maximum at x = 2 m.C)  the particle cannot execute simple harmonic motion.D)  the period of oscillation of the particle is $\frac{\pi }{20}s$

$F=-\frac{dU}{dx}=-\frac{d}{dx}[5x(x-4)]=-(10x-20)$ $\therefore$$a=\frac{F}{m}=-\left( \frac{10x-20}{0.1} \right)=-100(x-2)$ or $a=-100X$ where $X=x-2$ Hence, the motion is simple harmonic. In SHM, $a=-{{\omega }^{2}}\times$$\therefore$$\omega =10$ or $\frac{2\pi }{T}=10\,or\,T=\frac{\pi }{5}\,\sec .$ In the case of SHM. The speed is maximum at mean position. This position is given by $X=0$or $x-2=0$or $x=2\,\,m.$ Hence the correction option is [b].