The equation of motion of a particle is given by $\mathrm{x}=\mathrm{a} \sin (50 \mathrm{t}+\pi / 3) \mathrm{cm}$. 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 $\_\_\_\_$.

(A) $\frac{\pi}{300} \mathrm{~s}, \frac{\pi}{75} \mathrm{~s}$

(B) $\frac{\pi}{75} \mathrm{~s}, \frac{\pi}{300} \mathrm{~s}$

(C) $\frac{\pi}{300} \mathrm{~s}, \frac{\pi}{25} \mathrm{~s}$

(D) $\frac{\pi}{50} \mathrm{~s}, \frac{\pi}{100} \mathrm{~s}$

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Elasticity & SHM

The equation of motion of a particle is given by

x = a sin (50 t + π /3) 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

____

Answer: (A)

For

x = a sin (ω t + π /3)

with

ω = 50 rad/s

, the velocity

v = aω cos (ω t + π /3)

vanishes when the particle is at an extreme, i.e.

ω t_{1} + \frac{π}{3} = \frac{π}{2}

, giving

t_{1} = \frac{π /2 - π /3}{50} = \frac{π}{300} s

. The acceleration

a = - a ω^{2} sin (ω t + π /3)

vanishes at the mean position, i.e.

ω t_{2} + \frac{π}{3} = π

, giving

t_{2} = \frac{π - π /3}{50} = \frac{π}{75} s

. The particle first comes to rest at

\frac{π}{300}

s and has zero acceleration at

\frac{π}{75}

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option