Q83 / 110 · VITEEE 2024 · Physics

The potential energy of a particle of mass 0.1 kg moving along the $X$-axis, is given by $U=5 x(x-4) \mathrm{J}$, where $x$ is in metres. Choose the wrong option.

  • A
    The speed of the particle is maximum at $x=2 \mathrm{~m}$
  • B
    The particle executes simple harmonic motion
  • C
    The period of oscillation of the particle is $\frac{\pi}{5} \mathrm{~s}$
  • Correct answer
    D
    None of the above

Solution

$\begin{aligned} & U=5 x^2-20 x, \frac{d U}{d x}=10 x-20 \\ & F=-\frac{d U}{d x}=(-10 x+20)\end{aligned}$ Let us suppose, $x=(X+2)$. Then, $F=-10 X$, $F=0$ at $X=0$ or $x=2$, i.e. $x=2 \mathrm{~m}$ is the mean position about which particle is in SHM. $k=10$ $\therefore T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{0.1}{10}}=\frac{\pi}{5} \mathrm{~s}$ At mean position kinetic energy is maximum and force acting on particle in SHM is variable.

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

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