Q98 / 180 · JIPMER 2019 · Physics

Maximum acceleration in SHM

  • Correct answer
    A
    $\omega^2 A$
  • B
    $\frac{\omega^2 A}{2}$
  • C
    $\omega^2 A^2$
  • D
    0

Solution

Acceleration of a particle in S.H.M is given by $a=-\omega^2 X$ Where, x is the displacement from mean position Acceleration will be maximum, When $\mathrm{x}=\mathrm{A}$, where A is amplitude $\begin{aligned} \therefore \quad & \mathrm{a}_{\max }=-\omega^2 \mathrm{~A} \\ & \left|\mathrm{a}_{\max }\right|=\omega^2 \mathrm{~A} \end{aligned}$

Hence, the correct answer is option: 'A'.

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