Lift $A$ and $B$ of same mass $m$ starts moving from rest with acceleration $a$ in upward and downward directions respectively. Lift $B$ is stopped after time $t$ . If the strings are tight again in the interval of time $t^{^{'}}$ , find the value of $t^{^{'}}$ .

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Solution

The acceleration a here is inclusive of gravitational fore and prevolo fore.
Velocity of both A and B after time $l$
$\Rightarrow V = u + a t$
$V = a t$
If B stops then A goes upward for a while and then comes down.
Time taken by A to rise and stop.
$\Rightarrow V = g t^{''}$
$a t = g t^{''}$
$t^{''} = \frac{a t}{g}$
Times taken by A to some down and become tight again is $t^{11} = \frac{a t}{g}$ same taken to go up.
Now, $t^{'} = t " + t "$
$= \frac{2 a t}{g}$
Hence, the answer is $\frac{2 a t}{g} .$

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