The force (F in newton) acting on a particle | vedclass

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(D) The equation of motion for a particle in simple harmonic motion is $F = -ky$. Given $F + 0.04 \pi^2 y = 0$, we have $F = -0.04 \pi^2 y$. Comparing

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$4$

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(D) The equation of motion for a particle in simple harmonic motion is $F = -ky$. Given $F + 0.04 \pi^2 y = 0$, we have $F = -0.04 \pi^2 y$. Comparing this with $F = -ky$, the force constant $k = 0.04 \pi^2 	ext{ N/m}$.The mass of the particle is $m = 90 	ext{ g} = 0.09 	ext{ kg}$.The angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{0.04 \pi^2}{0.09}} = \sqrt{\frac{4 \pi^2}{9}} = \frac{2 \pi}{3} 	ext{ rad/s}$.The amplitude $A = \frac{6}{\pi} 	ext{ m}$.The maximum velocity $v_{	ext{max}}$ is given by $v_{	ext{max}} = A \omega$.Substituting the values, $v_{	ext{max}} = \left( \frac{6}{\pi} ight) 	imes \left( \frac{2 \pi}{3} ight) = 4 	ext{ m/s}$.

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