Calculating the Period of SHM for a Suspended Cubical Box

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The discussion focuses on calculating the period of simple harmonic motion (SHM) for a suspended cubical box. The box, with edge length ##a##, is suspended from a hinge and undergoes SHM when displaced. The total energy equation includes rotational inertia and gravitational potential energy, leading to the differential equation for angular displacement. A correction to the moment of inertia is noted, which impacts the final formula for the period. The correct period of oscillation is ultimately expressed as ## T = 2\pi \sqrt{\frac{7\sqrt{2}a}{9g}} ##.
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Homework Statement


We have a cubical hollow box, edge length ##a## suspended horizontally from a frictionless hinge along one of its edges. The box is displaced slightly and undergoes SHM. Show that the period of the oscillation is given by ## T = 2\pi \sqrt{\frac{7\sqrt{2}a}{9g}} ##

Homework Equations

The Attempt at a Solution



##E_{tot} = \frac{1}{2}I\omega^2 + mg\left(\frac{\sqrt{2}}{2a}\cos\theta\right) ##Then apply taylor expansion and differentiate to get: ## \ddot\theta = \frac{mga}{\sqrt{2}I} \theta ## and using ##I = \frac{2}{3}ma^2## gives ##T = 2\pi \sqrt{\frac{2\sqrt{2}a}{3g}}##

Help is much appreciated
 
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Ah, turns out the moment of inertia was incorrect...
 
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