Question

A 0.10$kg$ block oscillates back and forth along a horizontal surface. Its displacement from the origin is given by: $x=\left(10cm\right)\cos \left[\right(10rad/s)t+\pi /2rad]$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. $\frac{10\pi }{3}m/s^2$

Answer 1

The correct option is A 10$m/s^2$

The amplitude of the oscillation is $a=10\times {10}^{-2}m$

and the angular frequency is $\omega =10rad/sec$

The maximum acceleration of the particle is:

$A_{max}={\omega }^{2}a$

$=10\times {10}^{-2}\times {10}^2$

$=10m/se{c}^2$