alleryMaximum acceleration in SHM ($A = $ amplitude, $\omega = $ angular frequency) is:
A$A\omega$
B$A\omega^2$
C$A^2\omega$
D$\omega/A$
Answer & 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.
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