Maximum acceleration in SHM ($A = $ amplitude, $\omega = $ angular frequency) is:
- $A\omega$
- $A\omega^2$
- $A^2\omega$
- $\omega/A$
Show answer and solution
Correct answer: B. $A\omega^2$
$a = -\omega^2 x$ ⇒ $|a|_{max} = \omega^2 A$ at the extremes $x = \pm A$.
**Compare:** $v_{max} = A\omega$ at mean position.