Hooke's law and circular motion.

[negation]

Part of the challenge is to come up with your own illustration.You are mixing up two versions of the formula for centripetal acceleration:
a_c = v²/r = ω²r

That was costly. Can't trust memory.
S = Θ/r = (dΘ.r - Θdr)/r² = vr/r² = v/r
ω=v/r
∴v = ω²r²
a_c = v₂/r = ω²r²/r = ω²r
F_C = m[ω²r] = 2kg[(2πf)²r] = 2kg(39.4784176 rad s^-1)(0.7m)
Good to go?

Edit: I tried working backwards from the answers I was provided.

In my final equation:
F_C = m[ω²r] = 2kg[(2πf)²r] = 2kg(39.4784176 rad s^-1)(0.7m), this gives
55.26978465 N
55.26978465 N = |-k|(0.4m) = 138.1744616

But should I divide 55.26978465 N by 2, and equating this to |-k|(0.4m) I get ~69 which is correct.
I can only rationalize that dividing the value 55.26978465 N by 2 is such that only the force acting on one mass can be determined.

 

[Doc Al]

That was costly. Can't trust memory.
S = Θ/r = (dΘ.r - Θdr)/r² = vr/r² = v/r
ω=v/r
∴v = ω²r²
a_c = v₂/r = ω²r²/r = ω²r
F_C = m[ω²r] = 2kg[(2πf)²r] = 2kg(39.4784176 rad s^-1)(0.7m)
Good to go?

Except that you added the masses together for some reason. Think in terms of Newton's 2nd law, which you are applying to each mass separately.

 

[ehild]

Why do you use 2kg for the mass? Consider one of the masses. It moves along a circular orbit, with given radius and time period. The spring exerts force on the mass and that force provides the centripetal force.
The same force acts on the other end of the spring on the other mass, and it moves along the same orbit.

ehild

 

[negation]

Except that you added the masses together for some reason. Think in terms of Newton's 2nd law, which you are applying to each mass separately.

I did an edit in the previous post

 

[negation]

Why do you use 2kg for the mass? Consider one of the masses. It moves along a circular orbit, with given radius and time period. The spring exerts force on the mass and that force provides the centripetal force.
The same force acts on the other end of the spring on the other mass, and it moves along the same orbit.

ehild

It hit me that since we considered the total change in length of the spring due to the 2 masses, we give consideration to the total mass.

 

[ehild]

The force acts in a point. If you speak about an extended body, with forces acting at different points, you can sum up the individual equations m_ia_i=F_i, and arrive at the equation for the acceleration of the centre of mass: a_CM∑m_i=∑Fi . If there are only forces of interaction (no external force) the sum of forces is zero. The system, which has 2 kg mass altogether, has zero acceleration. But the individual masses feel the force of the spring and have centripetal acceleration.

ehild

 

[ehild]

It hit me that since we considered the total change in length of the spring due to the 2 masses, we give consideration to the total mass.

A stretched spring has tension. To keep it stretched there must be forces equal to that tension acting at both ends. You can not exert force on a spring if the other end is free. Hook's law refers to the tension which is equal to kΔL, and it is equal to the force acting on one end of the spring (but force of the same magnitude and opposite direction has to act on the other end).
Again, the net force on the stretched spring is zero, as the forces acting at the ends are opposite and of equal magnitude.

ehild