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As shown in figure a horizontal platform with a mass m placed on it is executing SHM along y-axis. If the amplitude of oscillation is 2.5 cm, the minimum period of the motion for the mass not to be detached from the platform is . $(g = 10 m / \sec^{2} = π^{2})$

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$\frac{10}{π} \sec$

$\frac{1}{\sqrt{10}} \sec$

$\frac{π}{10} \sec$

$\frac{π}{\sqrt{10}} \sec$

answer is B, D.

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Detailed Solution

According to the FBD,

$N = m g - m w^{2} A - - - (1)$

According to question $N \geq 0$

$\Rightarrow m g - m w^{2} A \geq 0$

$\Rightarrow g \geq w^{2} A$

$\Rightarrow w \leq \sqrt{\frac{g}{A}}$

$w_{\max} = \sqrt{\frac{g}{A}}$

$\begin{aligned} T = 2 π \sqrt{\frac{A}{g}} \\ T = 2 π \sqrt{\frac{2.5 \times 10^{- 2}}{10}} \\ T = \frac{π}{10} \sec (or) \frac{1}{\sqrt{10}} \sec \end{aligned}$

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