A particle of mass 2 , kg executing SHM has a | vedclass

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Question

(C) The maximum speed $(V_{\max})$ of a particle executing $SHM$ is given by the formula $V_{\max} = A \omega$,where $A$ is the amplitude and $\omega$

Vedclass · Accepted Answer

$1.256$

Vedclass · Answer

(C) The maximum speed $(V_{\max})$ of a particle executing $SHM$ is given by the formula $V_{\max} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.Given:Amplitude $A = 20 \, cm = 0.2 \, m$Time period $T = 1 \, s$The angular frequency is $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{1} = 2 \pi \, rad/s$.Substituting the values into the formula:$V_{\max} = A \omega$$V_{\max} = 0.2 \, m 	imes 2 \pi \, rad/s$$V_{\max} = 0.4 \pi \, m/s$Using $\pi \approx 3.14$:$V_{\max} = 0.4 	imes 3.14 = 1.256 \, m/s$.Therefore,the correct option is $C$.

$A$ particle of mass $2 \, kg$ executing $SHM$ has an amplitude of $20 \, cm$ and a time period of $1 \, s$. Its maximum speed is ......... $m/s$.

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