Two particles ' $A$ ' and ' $B$ ' start SHM at $t = 0$ .Their positions as a function of time are given by $X_{A} = A sin ω t$ , $X_{B} = A sin (ω t + π / 3)$

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Solution

$X_{A} = A sin (ω t)$
$X_{B} = A sin (ω t + \frac{π}{3}) = A sin (ω t + ω \frac{π}{3 ω})$
$v_{A} = A ω cos ω t$
$v_{B} = A ω cos (ω t + \frac{π}{3})$
A:
$∴ X_{B} = X_{A}$ after time $t = \frac{π}{3 ω}$
B:
$v_{A} = v_{B}$
$\Rightarrow 2\pi -\omega t = \omega t + \dfrac{\pi}{3} \Rightarrow t =\dfrac{5 \pi}{6\omega}$
C:
$v_{A} < 0$ after time $t = \frac{π}{2 ω}$
$v_{B} < 0$ after time $ω t + \frac{π}{3} > \frac{π}{2}$
i.e after time $ω t > \frac{π}{6}$ i.e time $t > \frac{π}{6 ω}$
Hence combining both, $v_{A} < 0$ and $v_{B} < 0$ after time $t = \frac{π}{2 ω}$
D:
$X_{A} < 0$ after time $ω t > π$ i.e $t > \frac{π}{ω}$
$X_{B} < 0$ after time $ω t + \frac{π}{3} > π$ i.e $t > \frac{2 π}{3 ω}$
Combining both we get, $t > \frac{π}{ω}$

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X_{A} = A sin ω t

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