A 0.10 kg block oscillates back and forth along a horizontal surf… - Vidyadip

MCQ

A $0.10 kg$ block oscillates back and forth along a horizontal surface. Its displacement from the origin is given by: $x=(10 cm ) \cos [(10 rad / s ) t+\pi / 2 rad ]$. What is the maximum acceleration experienced by the block [AMU (Engg.) 2000]

✓
$10 m / s ^2$
B
$10 \pi m / s ^2$
C
$\frac{10 \pi}{2} m / s ^2$
D
$\frac{10 \pi}{3} m / s ^2$

✓

Answer

Correct option: A.

$10 m / s ^2$

$ a=10 \times 10^{-2} \mathrm{~m} \text { and } \omega=10 \mathrm{rad} / \mathrm{sec}$
$ A_{\max }=\omega^2 a=10 \times 10^{-2} \times 10^2=10 \mathrm{~m} / \mathrm{sec}^2$

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