A particle of mass 10 text{ g} is executing s | vedclass

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(D) The mass of the particle is $m = 10 	ext{ g} = 0.01 	ext{ kg}$.The amplitude is $A = 0.5 	ext{ m}$.The time period is $T = \pi / 5 	ext{ s}$.T

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$0.5$

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(D) The mass of the particle is $m = 10 	ext{ g} = 0.01 	ext{ kg}$.The amplitude is $A = 0.5 	ext{ m}$.The time period is $T = \pi / 5 	ext{ s}$.The angular frequency is $\omega = 2\pi / T = 2\pi / (\pi / 5) = 10 	ext{ rad/s}$.The maximum force acting on a particle in simple harmonic motion is given by $F_{	ext{max}} = m \omega^2 A$.Substituting the values: $F_{	ext{max}} = 0.01 	imes (10)^2 	imes 0.5$.$F_{	ext{max}} = 0.01 	imes 100 	imes 0.5 = 1 	imes 0.5 = 0.5 	ext{ N}$.

$A$ particle of mass $10 \text{ g}$ is executing simple harmonic motion with an amplitude of $0.5 \text{ m}$ and a periodic time of $(\pi / 5) \text{ s}$. The maximum value of the force acting on the particle is ... $N$.

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