For a particle performing linear S.H.M.,its a | vedclass

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(A) The total distance covered by a particle in one complete oscillation of linear $S.H.M.$ is equal to $4A$,where $A$ is the amplitude.The time perio

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$4$

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(A) The total distance covered by a particle in one complete oscillation of linear $S.H.M.$ is equal to $4A$,where $A$ is the amplitude.The time period $T$ of the oscillation is related to the frequency $n$ by the formula $T = \frac{1}{n}$.Average speed is defined as the total distance divided by the total time taken.Therefore,average speed $v_{av} = \frac{	ext{Total Distance}}{	ext{Total Time}} = \frac{4A}{T} = \frac{4A}{1/n} = 4nA$.

For a particle performing linear $S.H.M.$,its average speed over one oscillation is ($A=$ amplitude of $S.H.M.$,$n=$ frequency of oscillation) (in $nA$)

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