Q153 / 155 · WBJEE 2021 · Physics

The potential energy of a particle of mass $0.02 \mathrm{~kg}$ moving along $x$-axis is given by $V=A x(x-4) J$ where $x$ is in metres and $\mathrm{A}$ is a constant. Which of the following is/are correct statement(s)?

  • A
    The particle is acted upon by a constant force
  • Correct answer
    B
    The particle executes simple harmonic motion
  • Correct answer
    C
    The speed of the particle is maximum at $x=2 m$
  • D
    The period of oscillation of the particle is $\frac{\pi}{5} \mathrm{sec}$

Solution

$V=A x(x-4) J m=0.02=A x^{2}-4 A x$ $$ \begin{array}{l} \mathrm{F}=-\frac{\mathrm{d} \mathrm{V}}{\mathrm{dx}}=-2 \mathrm{Ax}+4 \mathrm{~A} \text { (not constant) } \\ \mathrm{F}=-2 \mathrm{~A}(\mathrm{x}-2) \end{array} $$ Hence position execute S.H.M mean position is when force $=0$ i.e. $x=2$ and speed is maximum at mean position $$ \omega^{2}=\frac{2 \mathrm{~A}}{\mathrm{~m}} \mathrm{~T}=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{~A}}}=2 \pi \sqrt{\frac{1}{100 \mathrm{~A}}}=\frac{\pi}{5} \sqrt{\frac{1}{\mathrm{~A}}} $$

Hence, the correct answer is options: 'B', 'C'.

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