Q. A 0.10 $kg$ block oscillates back and forth along a horizontal surface. Its displacement from the origin is given by: $x = (10cm)\cos{\lbrack(10rad/s)t + \pi/2rad\rbrack.}$ What is the maximum acceleration experienced by the block
A) $10\ m/s^{2}$
B) $10\ \pi\ m/s^{2}$
C) $\frac{10\pi}{2}m/s^{2}$
D) $\frac{10\pi}{3}m/s^{2}$
Explanation
(a) $a = 10 \times 10^{- 2}m$ and$\ \omega = 10\ rad/s$
$A_{\max} = \omega^{2}a = 10 \times 10^{- 2} \times 10^{2} = 10\ m/s^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.