For a particle performing linear SHM,its aver | vedclass

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(B) In one complete oscillation,a particle in $SHM$ travels from the mean position to one extreme $(A)$,back to the mean position $(A)$,to the other e

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$4$

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(B) In one complete oscillation,a particle in $SHM$ travels from the mean position to one extreme $(A)$,back to the mean position $(A)$,to the other extreme $(A)$,and back to the mean position $(A)$.Total distance travelled in $1$ oscillation $= A + A + A + A = 4 \ A$.The time taken for one complete oscillation is the time period $T$.The frequency of oscillation is $n = \frac{1}{T}$,which implies $T = \frac{1}{n}$.Average speed is defined as the total distance divided by the total time.$	ext{Average speed} = \frac{	ext{Total distance}}{	ext{Total time}} = \frac{4 \ A}{T}$.Substituting $T = \frac{1}{n}$,we get $	ext{Average speed} = 4 \ A \ n$.

For a particle performing linear $SHM$,its average speed over $1$ oscillation is ($A =$ amplitude of $SHM$,$n =$ frequency of oscillation). (in $A \ n$)

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