A particle is executing S.H.M. If its amplitu | vedclass

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(C) The maximum velocity $(v_{\max})$ of a particle executing $S.H.M.$ is given by the formula: $v_{\max} = \omega a$.Here,$\omega$ is the angular fre

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$2\pi \, m/s$

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(C) The maximum velocity $(v_{\max})$ of a particle executing $S.H.M.$ is given by the formula: $v_{\max} = \omega a$.Here,$\omega$ is the angular frequency,which is defined as $\omega = \frac{2\pi}{T}$.Given: Amplitude $a = 2 \, m$ and periodic time $T = 2 \, s$.Substituting these values into the formula:$v_{\max} = \left( \frac{2\pi}{T} ight) 	imes a$$v_{\max} = \left( \frac{2\pi}{2} ight) 	imes 2$$v_{\max} = \pi 	imes 2 = 2\pi \, m/s$.Therefore,the maximum velocity is $2\pi \, m/s$.

$A$ particle is executing $S.H.M.$ If its amplitude is $2 \, m$ and periodic time is $2 \, s$,then the maximum velocity of the particle will be:

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