Velocity & Acceleration in SHM question from MHT CET 2026 (17 Apr, Shift 1)

Answer: (d)

1. $v = \omega\sqrt{A^2 - x^2}$, $v_{max} = \omega A$
2. $\omega\sqrt{A^2 - x^2} = \frac{\omega A}{4} \Rightarrow A^2 - x^2 = \frac{A^2}{16}$
3. $x^2 = \frac{15A^2}{16} \Rightarrow x = \frac{\sqrt{15}A}{4}$
4. So the answer is (d) $\frac{A\sqrt{15}}{4}$.