Velocity & Acceleration in SHM question from MHT CET 2026 (14 May, Shift 2)

Answer: (d)

1. $\omega = \frac{2\pi}{T} = \frac{2\pi}{16} = \frac{\pi}{8}$ rad/s.
2. Particle is at the mean position at $t = 2$ s, so $x = A\sin\omega(t - 2)$ and $v = A\omega\cos\omega(t-2)$.
3. At $t = 4$ s: $\omega(t-2) = \frac{\pi}{4}$, so $4 = A\cdot\frac{\pi}{8}\cdot\frac{1}{\sqrt{2}}$.
4. $A = \frac{32\sqrt{2}}{\pi}$ m.
5. So the answer is (d) $\frac{32\sqrt{2}}{\pi}$.