A particle performs SHM with a period $T$ and amplitude $A$ . The mean velocity of the particle over the time interval during which it travels a distance $\frac{A}{2}$ from the extreme position is

\frac{A}{T}

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\frac{2 A}{T}

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\frac{3 A}{T}

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\frac{A}{2 T}

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Solution

The correct option is C $\frac{3 A}{T}$
Displacement of particle when starting from extreme position,
$x = A cos ω t$
At $A / 2$ from extreme position,
$\frac{A}{2} = A cos ω t_{o}$
$cos \frac{π}{3} = cos ω t_{o} \frac{π}{3} = ω t_{o} ∴ t_{o} = \frac{π}{3 ω} = \frac{T}{6}$
Mean velocity, $¯ v = \frac{Total displacement}{Total time taken}$
$¯ v = \frac{A / 2}{T / 6} = \frac{3 A}{T}$

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