Q130 / 160 · TS EAMCET 2024 (09 May Shift 2) · Physics

A particle of mass 4 mg is executing simple harmonic motion along $x$-axis with an angular frequency of 40 rad $\mathrm{s}^{-1}$. If the potential energy of the particle is $V(x)=a+$ $b x^2$, where $V(x)$ is in joule and $x$ is in metre, then the value of $b$ is

  • A
    $800 \times 10^{-6} \mathrm{Jm}^{-2}$
  • B
    $1600 \times 10^{-6} \mathrm{Jm}^{-2}$
  • Correct answer
    C
    $3200 \times 10^{-6} \mathrm{Jm}^{-2}$
  • D
    $6400 \times 10^{-6} \mathrm{Jm}^{-2}$

Solution

For particle executing SHM, $\mathrm{m}=4 \times 10^{-3} \times 10^{-3} \mathrm{~kg}=4 \times 10^{-6} \mathrm{~kg}, \omega=40 \mathrm{rad} / \mathrm{s}$ potential energy, $\mathrm{V}(\mathrm{x})=\mathrm{a}+\mathrm{bx} \mathrm{x}^2$ $\therefore$ Restoring force, $F=-\frac{d V}{d x}=-2 b x \qquad ...\mathrm{(i)}$ Also, $\mathrm{F}=-\mathrm{m} \omega^2 \mathrm{x} \qquad ...\mathrm{(ii)}$ From eq(i) and (ii), we get $\begin{aligned} & \mathrm{mw}^2=2 \mathrm{~b} \Rightarrow \mathrm{~b}=\frac{\mathrm{Mw}^2}{2}=\frac{4 \times 10^{-6} \times(40)^2}{2} \\ & =3200 \times 10^{-6} \mathrm{Jm}^{-2} \end{aligned}$

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

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