Q 11-13-060JEE MainJEE Main 2024 (29 Jan, Shift 2)Medium
A simple harmonic oscillator has an amplitude $A$ and time period $6\pi$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x = A$ to $x = \dfrac{\sqrt3}{2}A$ will be $\dfrac\pi x\ \text{s}$, where $x =$ ______.
Numerical value type. Enter your answer.
Answer: 2
$\omega = \dfrac{2\pi}{6\pi} = \dfrac13\ \text{rad s}^{-1}$. With $x = A\sin\omega t$, the particle is at $x = A$ when $\omega t = \dfrac\pi2$ and next at $x = \dfrac{\sqrt3}{2}A$ when $\omega t = \dfrac{2\pi}{3}$.
$$\Delta t = \frac{\frac{2\pi}{3} - \frac\pi2}{\omega} = \frac{\pi/6}{1/3} = \frac\pi2\ \text{s}$$
So $x = 2$.
Solution by Sreeraj P, M.Sc Physics