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
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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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