How Does a Bar Magnet Oscillate?

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SUMMARY

The discussion focuses on the oscillation of a bar magnet when displaced from its equilibrium position, demonstrating that it oscillates with a period T defined by the formula T = 2π√(I/mB). Here, I represents the moment of inertia of the bar magnet, calculated as I = M(L² + a²)/12, where M is the mass, L is the length, and a is the width of the magnet. The Earth's horizontal component of the magnetic field is denoted as B. The discussion also suggests using torque (τ = Iα) to derive the equation for simple harmonic motion (SHM) analogous to a physical pendulum.

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Marcstylzzz
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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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