A small wooden cube is placed on a plank. The | vedclass

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(C) Concept: The wooden cube will leave the plank at the top extreme position if the downward acceleration of the plank exceeds the acceleration due t

Vedclass · Accepted Answer

$\frac{5}{18} 	ext{ m}$

Vedclass · Answer

(C) Concept: The wooden cube will leave the plank at the top extreme position if the downward acceleration of the plank exceeds the acceleration due to gravity.For the block not to leave the plank,the normal force $N$ must be greater than or equal to zero.From the free body diagram of the block,the equation of motion is: $mg - N = ma$.For the block to remain in contact,$N \geq 0$,which implies $mg \geq ma$,or $a \leq g$.In $S$.$H$.$M$.,the maximum acceleration is given by $a_{max} = \omega^2 A$.Therefore,the condition for the block not to leave the plank is $\omega^2 A \leq g$,or $A \leq \frac{g}{\omega^2}$.Given frequency $f = \frac{3}{\pi} 	ext{ Hz}$.Angular frequency $\omega = 2\pi f = 2\pi \left( \frac{3}{\pi} ight) = 6 	ext{ rad/s}$.Substituting the values: $A \leq \frac{10}{6^2} = \frac{10}{36} = \frac{5}{18} 	ext{ m}$.Thus,the maximum amplitude is $\frac{5}{18} 	ext{ m}$.

$A$ small wooden cube is placed on a plank. The plank performs a vertical $S$.$H$.$M$. of frequency $\frac{3}{\pi} \text{ Hz}$. The maximum amplitude of the plank so that the wooden block does not leave the plank is [take $g = 10 \text{ m/s}^2$].

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