Two particles undergo $S H M$ along the name line with the same time period $(T)$ and equal amplitudes $(A)$ . At a particular instant one particle is at $x = - A$ and the other is at $x = 0$ . They move in the same direction. They will cross each other at

t = \frac{4 T}{3}

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t = \frac{3 T}{3}

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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
B $t = \frac{3 T}{3}$
D $x = \frac{A}{\sqrt{2}}$

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

The correct options are B t=3T3 D x=A√2

Two particles undergo $S H M$ along the name line with the same time period $(T)$ and equal amplitudes $(A)$ . At a particular instant one particle is at $x = - A$ and the other is at $x = 0$ . They move in the same direction. They will cross each other at

The correct options are
B $t = \frac{3 T}{3}$
D $x = \frac{A}{\sqrt{2}}$