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(B) For an object placed on the board not to lose contact,the downward acceleration of the board must not exceed the acceleration due to gravity $g$.T

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) For an object placed on the board not to lose contact,the downward acceleration of the board must not exceed the acceleration due to gravity $g$.The acceleration of a particle in $S.H.M.$ is given by $a = -\omega^2 y$.The maximum downward acceleration occurs at the topmost point,where $a_{\max} = \omega^2 A$.To maintain contact,the condition is $a_{\max} \leq g$,which implies $\omega^2 A \leq g$.For the shortest time period $T$,we consider the limiting case $\omega^2 A = g$.Substituting $\omega = \frac{2 \pi}{T}$,we get $\left(\frac{2 \pi}{T}\right)^2 A = g$.$\frac{4 \pi^2 A}{T^2} = g$.$T^2 = \frac{4 \pi^2 A}{g}$.$T = 2 \pi \sqrt{\frac{A}{g}}$.

$A$ flat horizontal board moves up and down in $S.H.M.$ vertically with amplitude $A$. The shortest permissible time period of the vibration such that an object placed on the board may not lose contact with the board is ..........

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