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Q.

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

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a

$\frac{A}{2 T}$

b

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

c

$\frac{2 A}{T}$

d

$\frac{8 (\sqrt{2} - 1) A}{\sqrt{2} T}$

answer is D.

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Detailed Solution

$x = A \cos ω t \frac{A}{\sqrt{2}} = A \cos ω t ∴ \cos ω t = \frac{1}{\sqrt{2}} \Rightarrow ω t = \frac{π}{4} t = \frac{π}{4 ω} ∴ v_{a v} = \frac{(A - \frac{A}{\sqrt{2}})}{t} = \frac{A}{2 π} \frac{\sqrt{2} - 1}{\sqrt{2}} \times 4 ω = \frac{8 (\sqrt{2} - 1) A}{\sqrt{2} T}$

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