How Does a Bar Magnet Oscillate?

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    Magnet Oscillation
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A bar magnet displaced from its equilibrium position will oscillate with a period T defined by the formula T = 2*pi*sqrt(I/mB), where I is the moment of inertia, M is mass, L is length, and a is width. The moment of inertia for the bar magnet is calculated as I = M(L^2 + a^2)/12. The discussion compares the magnet's oscillation to that of a compass needle, suggesting the use of simple harmonic motion (SHM) principles. To derive the oscillation period, torque (tau) can be expressed as tau = I*@, linking moment of inertia to angular acceleration. This approach aligns with the behavior of physical pendulums under small angle oscillations.
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Please help, if a magnet is displaced from its equilibrium position by a small angle (theta), show that it will oscillate with a period T given by T= 2*pi*sqrt(I/mB)
I = moment of inertia of the bar magnet and equals M(L^2 + a^2)/12
B = Earth's horizontal component of magnetic field
M is the mass
L is the length and a its width
 
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Is the magnet you're referring to oscillating like a compass needle does (the period they gave looks a lot like the period for the physical pendulum)?

If that is the case, then suitable method would be to find a way to write out an equation displaying SHM (much like the pendulum with oscillations of small angles). A good way of deriving it would probably be using torque (tau) as tau = I*@, in which I is moment of inertia and @ is angular acceleration.
 
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