The relation between the force (F in newton) | vedclass

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(C) Given the equation of motion: $500 F + \pi^2 y = 0$. Since $F = ma$, we can write: $500 ma + \pi^2 y = 0$. Dividing by $500 m$: $a + \left(\frac{\

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(C) Given the equation of motion: $500 F + \pi^2 y = 0$. Since $F = ma$, we can write: $500 ma + \pi^2 y = 0$. Dividing by $500 m$: $a + \left(\frac{\pi^2}{500 m}ight) y = 0$. Thus, $a = -\left(\frac{\pi^2}{500 m}ight) y$ ... $(i)$. For simple harmonic motion, the standard equation is $a = -\omega^2 y$ ... (ii). Comparing $(i)$ and (ii), we get $\omega^2 = \frac{\pi^2}{500 m}$. Substituting $m = 2 	ext{ g} = 2 	imes 10^{-3} 	ext{ kg}$: $\omega^2 = \frac{\pi^2}{500 	imes 2 	imes 10^{-3}} = \frac{\pi^2}{1} = \pi^2$. Therefore, $\omega = \pi$. The time period $T$ is given by $T = \frac{2\pi}{\omega} = \frac{2\pi}{\pi} = 2 	ext{ s}$.

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

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