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(B) For an object placed on a board moving in $SHM$ not to lose contact,the maximum downward acceleration of the board must not exceed the acceleratio

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$2\pi \sqrt {\frac{\alpha}{g }} $

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(B) For an object placed on a board moving in $SHM$ not to lose contact,the maximum downward acceleration of the board must not exceed the acceleration due to gravity $g$.The maximum acceleration of the board in $SHM$ is given by $a_{max} = \alpha \omega^2$.To maintain contact,the condition is $a_{max} \le g$,which implies $\alpha \omega^2 \le g$.For the shortest time period $T$,we consider the limiting case where $\alpha \omega^2 = g$.Substituting $\omega = \frac{2\pi}{T}$,we get $\alpha (\frac{2\pi}{T})^2 = g$.Rearranging for $T$,we get $\frac{4\pi^2 \alpha}{T^2} = g$,which simplifies to $T^2 = \frac{4\pi^2 \alpha}{g}$.Taking the square root on both sides,we find $T = 2\pi \sqrt{\frac{\alpha}{g}}$.

$A$ flat horizontal board moves up and down in $SHM$ with an amplitude $\alpha$. What is the shortest permissible time period of the vibration such that an object placed on the board does not lose contact with the board?

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