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?
To find the period of the resulting Simple Harmonic Motion (SHM), we can use the equation for the period of a mass-spring system:
T = 2π * √(m/k)
where T is the period, m is the mass of the object, and k is the spring constant.
First, let's find the mass of the cube. Given that the density of solid objects is given by ρ = m/V, where ρ is the density, m is the mass, and V is the volume, we can calculate the mass:
m = ρ * V
Since the cube has edge lengths d = 5.50 cm, its volume can be calculated as V = d³. Given that the density ρ for the cube is not provided in the question, we will assume a typical density for the material from which the cube is made. Let's use the density of water, which is approximately ρ = 1000 kg/m³.
Converting the edge length to meters, we have:
d = 5.50 cm = 0.055 m.
Now we can calculate the volume:
V = d³ = (0.055 m)³.
Substituting the values we obtained into the equation for mass, we can find the mass of the cube:
m = ρ * V = (1000 kg/m³) * [(0.055 m)³].
Next, we can calculate the period using the given spring constant k of 1300 N/m and the mass we just found. Substituting these values into the equation for the period:
T = 2π * √(m/k) = 2π * √[(m) / (1300 N/m)].
Evaluating this expression will give us the period of the resulting SHM.
Note: Make sure to convert all the units to the correct ones before performing the calculations to get the final answer in the appropriate units.