The maximum speed of a particle in S.H.M. is | vedclass

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(D) The correct option is $D$. Concept: For $S.H.M.$,the displacement is given by $x = A \sin(\omega t + \phi)$. The velocity is $v = \frac{dx}{dt} =

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$\frac{2 V}{\pi}$

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(D) The correct option is $D$. Concept: For $S.H.M.$,the displacement is given by $x = A \sin(\omega t + \phi)$. The velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi)$. The maximum speed is $V = A \omega$. In one complete time period $T$,the particle travels a total distance of $4A$. The average speed is defined as the total distance divided by the total time. $	ext{Average speed} = \frac{	ext{Total distance}}{	ext{Total time}} = \frac{4A}{T}$. Since $T = \frac{2\pi}{\omega}$ and $\omega = \frac{V}{A}$,we have $T = \frac{2\pi A}{V}$. Substituting $T$ into the average speed formula: $	ext{Average speed} = \frac{4A}{\left(\frac{2\pi A}{V}ight)} = \frac{4A \cdot V}{2\pi A} = \frac{2V}{\pi}$.

The maximum speed of a particle in $S.H.M.$ is $V$. The average speed is

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