Q 11-13-038JEE MainJEE Main 2026 (6 Apr, Shift 1)Easy
A particle is executing simple harmonic motion. Its amplitude is $A$ and time period is $5$ sec. The time required by it to move from $x = A$ to $x = \dfrac{A}{\sqrt{2}}$ is ______ sec.
Answer: (C) $5/8$
Starting from the extreme, $x = A\cos\omega t$. For $x = \dfrac{A}{\sqrt{2}}$: $\omega t = \dfrac{\pi}{4}$.
$t = \dfrac{\pi/4}{2\pi/T} = \dfrac{T}{8} = \dfrac{5}{8}$ s.
Solution by Sreeraj P, M.Sc Physics