The relation between the force ( F is newton) acting on a particle executing simple harmonic motion and the…

The relation between the force ( F is newton) acting on a particle executing simple harmonic motion and the displacement of the particle ( $y$ in metre) is $500 \mathrm{~F}+\pi^2 y=0$. If the mass of the particle is 2 g , the time period of oscillation of the particle is
  1. 8 s
  2. 6 s
  3. 2 s
  4. 4 s

Solution

$500 F+\pi^2 y=0$ $\Rightarrow \frac{\mathrm{F}}{\mathrm{m}}+\left(\frac{\pi^2}{500 \mathrm{~m}}\right) \mathrm{y}=0$ $\therefore \quad a=-\left(\frac{\pi^2}{500 m}\right) y$ $\qquad ...\mathrm{(i)}$ For SHM, $a=-\omega^2 y$ $\qquad ...\mathrm{(ii)}$ Comparing eq(i) and (ii), we get $\begin{aligned} & \omega^2=\frac{\pi^2}{500 \mathrm{~m}} \Rightarrow \frac{2 \pi}{\mathrm{~T}}=\sqrt{\frac{\pi^2}{500 \mathrm{~m}}} \\ & \therefore \quad \mathrm{~T}=2 \pi \sqrt{\frac{500 \mathrm{~m}}{\pi^2}}=2 \sqrt{500 \times 2 \times 10^{-3}}=2 \mathrm{~s} \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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