A small wooden cube is placed on a plank. The plank performs a vertical S.H.M of frequency $\frac{3}{\pi}…

A small wooden cube is placed on a plank. The plank performs a vertical S.H.M of frequency $\frac{3}{\pi} \mathrm{Hz}$. The maximum amplitude of the plank so that the wooden block does not leave the plank is [take $g=10 \mathrm{~m} / \mathrm{s}^2$ ]
  1. $\frac{7}{12} \mathrm{~m}$
  2. $\frac{5}{2} m$
  3. $\frac{5}{18} \mathrm{~m}$
  4. $\frac{11}{18} \mathrm{~m}$

Solution

Concept: wooden cube can leave the plank at the top extreme location when the plank just begins to move towards the equilibrium position. See the figure below: For cube not to leave plank the weight of the block should be more than the pseudo force on the cube: $\text { ma } \leq \mathrm{mg}$ Therefore, $\mathrm{a} \leq \mathrm{g}$. $|a|=\omega^2 x$ when, $x=A$ the amplitude $\therefore \omega^2 \mathrm{~A} \leq \mathrm{g} \text { or } \mathrm{A} \leq \frac{\mathrm{g}}{\omega^2}$ Given, $\mathrm{f}=\frac{3}{\pi} \mathrm{Hz} \frac{\omega}{2 \pi}=\frac{1}{\mathrm{~T}}$ $\Rightarrow \omega=6 \mathrm{~Hz}$ $A \leq \frac{10}{6^2} \mathrm{~m}$ $\mathrm{A} \leq \frac{5}{18} \mathrm{~m}$ :

Asked in: MHT CET 2022 (08 Aug Shift 2)

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