What Is the Period of SHM for a Rotated Cube Attached to a Spring?

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The discussion revolves around calculating the period of simple harmonic motion (SHM) for a 3.00 kg cube attached to a spring, with the spring constant of 950 N/m. The cube, with edge lengths of 6.30 cm, is rotated 3.00° before being released. To solve the problem, one must convert the rotational motion into a linear distance that the spring stretches, considering the geometry involved. Establishing a relationship between the angle of rotation and the torque on the cube is essential, alongside determining the moment of inertia. This approach allows for the derivation of the period of SHM for small angular displacements.
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The 3.00 kg cube has edge lengths d = 6.30 cm and is mounted on an axle through its center. A spring (k = 950 N/m) connects the cube's upper corner to a rigid wall. Initially the spring is at its rest length. If the cube is rotated 3.00° and released, what is the period of the resulting SHM?

Please provide a solution rather than just an answer. I really appreciate any help:)
 
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You are going to find that you won't get much of a response by just throwing questions into a template and posting them. You need to show that you've made at least some attempt at solving the problem on your own.
 
what if i don't get it at all :(
 
brownniegyal said:
what if i don't get it at all :(

Then you should probably start with a simpler question where you are only partially lost. For this problem convert the rotational motion of the cube into a distance the spring is stretched (a picture would really help here - I'm not sure what the geometry is). Then determine an approximate linear relation between the angle of rotation and the torque on the cube which, together with knowing the moment of inertial of the cube, gives you a SHO for small angular displacements.
 
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