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(A) The maximum velocity $v_{\max}$ of a particle in simple harmonic motion is given by the formula $v_{\max} = A \omega$,where $A$ is the amplitude a

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$1.6 \pi \, m/s$

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(A) The maximum velocity $v_{\max}$ of a particle in simple harmonic motion is given by the formula $v_{\max} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.Given: Amplitude $A = 4 \, cm = 4 	imes 10^{-2} \, m$ and Time period $T = 0.05 \, s$.The angular frequency is $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{0.05} = 40 \pi \, rad/s$.Substituting these values into the formula:$v_{\max} = (4 	imes 10^{-2} \, m) 	imes (40 \pi \, rad/s)$$v_{\max} = 1.6 \pi \, m/s$.

$A$ particle executes simple harmonic motion with a periodic time of $0.05 \, s$ and an amplitude of $4 \, cm$. What is its maximum velocity?

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