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A 0.10 kg block oscillates back and forth along a horizontal surface. Its displacement from the origin is given by: $x = (10 cm) \cos [(10 rad / s) t + π / 2 rad]$ . What is the maximum acceleration experienced by the block

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$\frac{10 π}{2} m / s^{2}$

$\frac{10 π}{3} m / s^{2}$

$10 m / s^{2}$

$10 π m / s^{2}$

answer is A.

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Detailed Solution

$a = 10 \times 10^{- 2} m$ and $ω = 10 rad / \sec$

$A_{\max} = ω^{2} a = 10 \times 10^{- 2} \times 10^{2} = 10 m / \sec^{2}$

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