MCQ
Two particles undergo $SHM$ along parallel lines with the same time period $(T)$ and equal amplitudes. At a particular instant, one particle is at its extreme position while the other is at its mean position. They move in the same direction. They will cross each other after a further time
  • A
    $T/8$
  • $3T/8$
  • C
    $T/6$
  • D
    $4T/3$

Answer

Correct option: B.
$3T/8$
b
The particle which is at mean position. Its $SHM$ can be represented by $x_{1}=-A \sin \omega t$. And the particle which is at extreme position. Its $SHM$ is represented by $x_{2}=A \cos \omega t$ When both cross each other

$x_{1}=x_{2}$

$\Rightarrow-A \sin \omega t=A \cos \omega t$

$\Rightarrow \tan \omega t=-1$

$\Rightarrow \omega t=-\frac{\pi}{4}, \frac{3 \pi}{4}$

$\omega t=-\frac{\pi}{4}$ will give the negative value of time which is not possible.

so

$\omega t=\frac{3 \pi}{4}$

$\Rightarrow \frac{2 \pi}{T} t=\frac{3 \pi}{4}$

$\Rightarrow t=\frac{3 T}{8}$

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