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) $5/8$
Concept. Use the reference circle. From the extreme ($x=A$, phase $\pi/2$) to $x=A/\sqrt2$ (phase $\pi/4$) the radius turns through $\pi/4$.
Step 1. Fraction of a period. $\dfrac{\pi/4}{2\pi}=\dfrac18$.
Step 2. Time. $t=\dfrac T8=\dfrac58$ s.
Answer. $5/8$ s (Option C).
Solution by Sreeraj P, M.Sc Physics
