Physics · Simple Harmonic Motion

JEE Main 2026 — 2 April, Evening Shift — Question 17

The equation of motion of a particle is given by $x = a\sin(50t + \pi/3)\mathrm{cm}$ . The particle will come to rest at time $t_1$ and it will have zero acceleration at time $t_2$ . The $t_1$ and $t_2$ respectively are

Option A:
$\frac{\pi}{300}\mathrm{s},\frac{\pi}{75}\mathrm{s}$
Correct
Option B:
$\frac{\pi}{75}\mathrm{s},\frac{\pi}{300}\mathrm{s}$
Option C:
$\frac{\pi}{300}\mathrm{s},\frac{\pi}{25}\mathrm{s}$
Option D:
$\frac{\pi}{50}\mathrm{s},\frac{\pi}{100}\mathrm{s}$

Answer: A

Step-by-step solution

$v=0$ when $\cos(50t+\pi/3)=0$ → $50t+\pi/3=\pi/2$ → $t=\pi/300$ $a=0$ when $\sin(50t+\pi/3)=0$ → $50t+\pi/3=\pi$ → $t=\pi/75$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Kinematics of SHM, Phase and Energy in SHM