Two particles undergo SHM along parallel line | vedclass

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Question

(A) Let the particle at the mean position be $1$ and the particle at the extreme position be $2$. Their displacements can be represented as: $x_1 = A

Vedclass · Accepted Answer

$T/8$

Vedclass · Answer

(A) Let the particle at the mean position be $1$ and the particle at the extreme position be $2$. Their displacements can be represented as: $x_1 = A \sin(\omega t)$ $x_2 = A \cos(\omega t)$ Since they move in the same direction,we consider the instant when they cross,i.e.,$x_1 = x_2$. $A \sin(\omega t) = A \cos(\omega t)$ $	an(\omega t) = 1$ $\omega t = \pi/4$ However,at $t=0$,particle $1$ is at mean position $(x=0)$ and particle $2$ is at extreme position $(x=A)$. To account for the initial phase,let $x_1 = A \sin(\omega t)$ and $x_2 = A \cos(\omega t)$. Setting $x_1 = x_2$ gives $	an(\omega t) = 1$,so $\omega t = \pi/4$. But the question implies they cross after starting from these positions. If we set $x_1 = A \sin(\omega t)$ and $x_2 = A \cos(\omega t)$,they cross when $\sin(\omega t) = \cos(\omega t)$,which is $\omega t = \pi/4$. Since $\omega = 2\pi/T$,we have $(2\pi/T)t = \pi/4$,which gives $t = T/8$.

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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