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A cube of side ‘a’ and mass ‘m’ is suspended vertically from one of its edges as shown in the figure. The time period of small oscillations is

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$T = 2 π \sqrt{\frac{\sqrt{2} a}{3 g}}$

$T = 2 π \sqrt{\frac{3 \sqrt{2} a}{2 g}}$

$T = 2 π \sqrt{\frac{2 a}{3 g}}$

$T = 2 π \sqrt{\frac{2 \sqrt{2} a}{3 g}}$

answer is D.

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Detailed Solution

MI of cube about O is $\frac{2}{3} m a^{2}$

$\Leftrightarrow \frac{2}{3} m a^{2} α = - m g \frac{a}{\sqrt{2}} \sin θ = \frac{- m g a}{\sqrt{2}} θ$

$\Rightarrow α = \frac{- 3 g}{2 \sqrt{2} a} θ \Rightarrow T = 2 π \sqrt{\frac{2 \sqrt{2} a}{3 g}}$

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