How to find period of a SHM concerning a cube and spring?

[Ari]

1. Homework Statement
The 4.00 kg cube in the figure has edge lengths d = 8.00 cm and is mounted on an axle through its center. A spring ( k = 1400 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 4.00° and released, what is the period of the resulting SHM?

m = 4 kg
d = .08 m
k = 1400 N/m
Θ = 4°
T = ?

2. Homework Equations
I = (1/6)Md²
F = ma
F = -kx = -(mω²x)
k = mω²
τ = -d(F_gsinΘ)

3. The Attempt at a Solution
Restoring force is:
F = -kx while x = dsinΘ so...
F = -kdsinΘ
F = -(1400)(.08)sin(4°)
F = -7.8127 N

Newton's 2nd Law gives:
F = ma = F/m
a = (-7.8127)/4 = -1.9532

Using k = mω:
k = mω
ω = k/m = 1400/4 = 350

Using F = -kx:
x = F/-k
x = (-7.8127)/(-1400) = .00558 m

Restoring torque is:
τ = -dmgsinΘ while τ = Iα so...
Iα = -dmgsinΘ
α = -(.08)(4)(9.81)sin(4°) = -.219

Rotational to linear:
a = αr , r = .1131 m
a = -.0248

I'm not sure where to go with this further, and to be honest, not even sure if I'm going in the right direction.
Another thing is I'm not sure whether I'm supposed to use T = 2π √(m/k) or another equation, since that equation is for SHM of springs, but I can't seem to get anywhere with it.

 

[Dr Dr news]

The equation of motion for a mass suspended by a spring is a = d^2 y/dt^2 = (k/m) y which leads to a period, T = 2 pi sqrt(m/k) What you want to set up is an equation for alpha (the angular acceleration) (the second derivative of theta with respect to time) in terms of theta. You can then find the period in terms of the coefficient of theta analogous to the simple spring-mass problem.

 

[haruspex]

x = dsinΘ

Check that.

the cube's upper corner

That does not match the picture, which shows one edge uppermost, not one corner. I would guess the diagram is as intended.

If the cube is rotated 4.00°

Is this relevant? In SHM, does the frequency depend on the amplitude?

Using k = mω:

That is for a mass moving linearly on the end of a spring. It does not apply here. What kind of motion is this? Do you know any frequency formula for this kind of motion? If not, can you derive the differential equation for the motion?

 

[Ari]

Check that.

That does not match the picture, which shows one edge uppermost, not one corner. I would guess the diagram is as intended.

Is this relevant? In SHM, does the frequency depend on the amplitude?

That is for a mass moving linearly on the end of a spring. It does not apply here. What kind of motion is this? Do you know any frequency formula for this kind of motion? If not, can you derive the differential equation for the motion?

This would be a rotational motion I believe, since you are applying a force that causes the cube to turn, which would be angular frequency ω.
The only equation I know relating to angular frequency ω would be ω = 2π/T = 2πf.

And for the diagram, the cube's upper corner would be concerning the corner attached to the spring I believe.

 

[haruspex]

The only equation I know relating to angular frequency ω would be ω = 2π/T = 2πf

Need to be careful with use of ω in rotational oscillation. The displacement is an angle,θ say. It is common to write ω for ##\dot\theta## in other contexts, but this is different from the ω in the sin(ωt) in SHM, so better not to do that here.
θ is analogous to the x in linear oscillation, so ##\dot\theta## is like ##\dot x##. Stick to the ##\dot\theta## and ##\ddot\theta## notation.
Think about torque and angular acceleration and try to write the differential equation for θ.^*

And for the diagram, the cube's upper corner would be concerning the corner attached to the spring I believe.

Yes, but the diagram does not show an upper corner of a cube so attached, it shows the middle of one edge attached, or maybe a corner of a square lamina. Suspend a cube by one corner and view it from the side. Does it look like the diagram?

^*Edit: Actually, you almost had it in post #1, but you made the mistake of plugging in the initial angle instead of leaving it as a general equation relating θ to α (##=\ddot\theta##).
And as Dr Dr news posts, you need to use the small angle approximation for sine. You have come across that with pendulums, I would think.

 

[Dr Dr news]

The point is that having solved one differential equation say d^2 x/dt^2 = -(k/m) x, x = A sin ϖt, ω = sqrt(k/m) Any other differential equation of the same form say d^ θ/dt^2 = -(τ/I) θ will have a similar solution. θ = A sin ωt , ϖ = sqrt(τ/I) , τ is the restoring torque and I is the moment of inertia. one other consideration, the restoring torque is proportional to sin θ which for small θ, can be approximated by θ.