The 3.80 kg cube in Fig. 15-45 has edge lengths d = 5.50 cm and is mounted on an axle through its center. A spring (k = 1300 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 2.00° and released, what is the period of the resulting SHM?
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To find the period of the resulting Simple Harmonic Motion (SHM) of the cube, we need to use the equation:
T = 2π√(m/k)
where T is the period, m is the mass of the cube, and k is the spring constant.
In this case, the cube has a mass of 3.80 kg and the spring constant is 1300 N/m. To find the mass, we can use the equation:
m = ρ * V
where ρ is the density of the cube and V is the volume.
Since we know the edge length of the cube (d = 5.50 cm = 0.055 m), we can calculate the volume:
V = d^3
V = (0.055 m)^3 = 0.000166 m^3
Now we can find the mass using the density. However, the density is not given in the question, so we need more information to proceed.