A particle executes simple harmonic motion along a straight line according to \( x(t) = A \sin(\omega t + \phi) \) with angular frequency \( \omega = 2\pi\text{ rad s}^{-1} \). If the relation between acceleration and displacement is \( a = -\omega^2 x \), what is the magnitude of acceleration of the particle when its displacement is \( x = -0.5\text{ m} \)?

Correct answer · Option 2

\( 2\pi^2\text{ m s}^{-2} \)

Acceleration magnitude \( |a| = \omega^2 |x| = (2\pi)^2 \times 0.5 = 4\pi^2 \times 0.5 = 2\pi^2\text{ m s}^{-2} \).