A particle executes SHM with time period T and amplitude A. The maximum possible average speed in time $\frac{T}{4}$ is

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$\frac{2 A}{T}$

$\frac{4 A}{T}$

$\frac{8 A}{T}$

$\frac{4 \sqrt{2} A}{T}$

answer is D.

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Detailed Solution

When particle executing S.H.M moves from
$- \frac{A}{\sqrt{2}} t o + \frac{A}{\sqrt{2}}$ it takes $\frac{T}{4} S$
$\Rightarrow (V_{\max}) = \frac{(2 \frac{A}{\sqrt{2}})}{(\frac{T}{4})} = \frac{4 \sqrt{2} A}{T}$

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