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A body is in S.H.M with time period T=0.5 s and amplitude A=1 cm. Find the average velocity in the interval in which it moves from equilibrium to half of its amplitude.

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12 cm/s

16 cm/s

4 cm/s

6 cm/s

answer is C.

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Detailed Solution

$V_{a v} = \frac{(A / 2)}{(T / 12)} = \frac{6 A}{T} = \frac{6 x 1}{0.5} = 12 c m / s$

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