Q132 / 160 · AP EAMCET 2024 (21 May Shift 1) · Physics

The potential energy of a particle of mass 10 g as a function of displacement $x$ is $\left(50 x^2+100\right) \mathrm{J}$. The frequency of oscillation is

  • A
    $\frac{10}{\pi} s^{-1}$
  • B
    $\frac{5}{\pi} s^{-1}$
  • C
    $\frac{100}{\pi} s^{-1}$
  • Correct answer
    D
    $\frac{50}{\pi} s^{-1}$

Solution

$\mathrm{m}=10 \mathrm{~gm} =10 \times 10^{-3} \mathrm{~kg}, \mathrm{U}=\left(50 \mathrm{x}^2+100\right) \mathrm{J}$ $\therefore$ Restoring force, $\mathrm{F}=-\frac{\mathrm{du}}{\mathrm{dx}}$ $\begin{aligned} & \Rightarrow \mathrm{F}=-\frac{\mathrm{d}}{\mathrm{dx}}\left(50 \mathrm{x}^2+100\right) \\ & \Rightarrow \mathrm{ma}=-(100 \mathrm{x}) \end{aligned}$ $\begin{aligned} & \Rightarrow \quad \mathrm{a}=-\frac{100}{10 \times 10^{-3}} \mathrm{x}=-10^4 \mathrm{x}=-\omega^2 \mathrm{x} \\ & \Rightarrow \omega^2=10^4 \Rightarrow \omega=100 \\ & \therefore \quad \text { Frequency, } \mathrm{f}=\frac{\omega}{2 \pi}=\frac{100}{2 \pi}=\frac{50}{\pi} \mathrm{~s}^{-1}\end{aligned}$

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

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