Two particles undergo SHM along the same line with the same time period ( $T$ ) and equal amplitude ( $A$ ). At a particular instant one is at $x = A$ and the other is at $x = 0$ . They move in the opposite direction they will cross each other at

t = \frac{4 T}{3}

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t = \frac{T}{8}

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x = \frac{A}{2}

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x = \frac{A}{\sqrt{2}}

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Solution

The correct options are
C $t = \frac{T}{8}$
D $x = \frac{A}{\sqrt{2}}$
Let they cross each other at time t and position x
first particle $x_{1} = A sin w t$
Second particle $x_{2} = A cos w t$
When they cross each other
$x_{1} = x_{2}$
$A sin w t = A cos w t$
Then $w t = \frac{π}{4}$
$\frac{2 π}{T} t = \frac{π}{4}; t = \frac{T}{8}$
At $\frac{T}{8}$ time position is $x = \frac{A}{\sqrt{2}}$

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