QJEEM16Pha7fe: JEE Main 2016 Physics - Class 11 JEEM16Pha7fe Solution: Two particles are performing simple harmonic motion in a…

Angular displacement ( $\theta$ ₁) of particle 1. from equilibrium,

${y_1}$ = A sin $\theta$ ₁

$\Rightarrow$    A = Asin $\theta$ ₁

$\Rightarrow$    sin $\theta$ ₁ = 1 = sin ${\pi \over 2}$

$\therefore$     $\theta$ ₁ = ${\pi \over 2}$

Similarly for particle 2 angular displacement $\theta$ ₂ from equilibrium,

y₂ = Asin $\theta$ ₂

$\Rightarrow$     $-$ ${A \over 2}$ = Asin $\theta$ ₂

$\Rightarrow$    sin $\theta$ ₂ = $-$ ${1 \over 2}$ = sin $\left( { - {\pi \over 3}} \right)$

$\Rightarrow$     $\theta$ ₂ = $-$ ${{\pi \over 3}}$

Relative angular displacement of the two particle,

$\theta$ = $\theta$ ₁ $-$ $\theta$ ₂

= ${{\pi \over 2}}$ $-$ $\left( { - {\pi \over 6}} \right)$

= ${{{2\pi } \over 3}}$

Relative angular velocity $=$ $\omega - \left( { - \omega } \right)$ = $2\omega$

If they cross each other at time t

then,   t = ${\theta \over {2\omega }}$ = ${{2\pi } \over {3 \times 2\omega }}$ = ${\pi \over {3 \times {{2\pi } \over T}}}$ = ${T \over 6}$