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(D) The maximum acceleration $(a_{\max})$ of a particle executing simple harmonic motion is given by the formula: $a_{\max} = \omega^2 A$,where $\omeg

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$200 \pi^2 \ m/s^2$

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(D) The maximum acceleration $(a_{\max})$ of a particle executing simple harmonic motion is given by the formula: $a_{\max} = \omega^2 A$,where $\omega = 2\pi f$.Given: Amplitude $(A)$ = $0.02 \ m$,Frequency $(f)$ = $50 \ Hz$.First,calculate the angular frequency: $\omega = 2 	imes \pi 	imes 50 = 100\pi \ rad/s$.Now,calculate the maximum acceleration: $a_{\max} = (100\pi)^2 	imes 0.02$.$a_{\max} = 10000 \pi^2 	imes 0.02 = 200 \pi^2 \ m/s^2$.

$A$ particle is executing simple harmonic motion with an amplitude of $0.02 \ m$ and frequency $50 \ Hz$. The maximum acceleration of the particle is

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