Q 11-13-131JEE MainJEE Main 2019 (10 Jan, Shift 2)Medium
A particle executes simple harmonic motion with an amplitude of $5\ \text{cm}$. When the particle is at $4\ \text{cm}$ from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is:
Answer: (A) $\dfrac{8\pi}{3}$
$v = \omega\sqrt{A^2-x^2} = \omega\sqrt{0.05^2-0.04^2} = 0.03\,\omega$ and $a = \omega^2x = 0.04\,\omega^2$.
Equal magnitudes: $0.03\,\omega = 0.04\,\omega^2$, so $\omega = \dfrac34\ \text{rad/s}$ and
$$T = \frac{2\pi}{\omega} = \frac{8\pi}{3}\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics