Period of Oscillations near equator

[Decadohedron]

Homework Statement

A bead slides along a frictionless wire which lies in the N/S direction, midpoint at the equator. All points along the wire are the same distance from the center of the earth. The bead is initially at rest then released a small distance, δ, to the north of the equator. Because of effective g doesn't point directly towards the Earth's center, there is a small component along the wire that always points back towards the equator. This means that when released, the bead will oscillate back and forth just like a mass on a spring. What is the period of these oscillations?

Homework Equations

Coriolis force would be 0 as the bead is fixed on a wire, and the centrifugal force would be the force always pointing it back towards the equator - thus the formula being

1.
-GMm/r² + mω²r = 0

2.
T = 2π√(m/k); ω²=m/k

The Attempt at a Solution

I was going to solve for ω and then just plug into the Period formula of the mass-spring system. But it seems overly simple and feels like I'm missing something.

Am I on the right track or should I be thinking about this differently??

Thanks.

 

[haruspex]

-GMm/r² + mω²r = 0

It isn't in orbit.

 

[Decadohedron]

It isn't in orbit.

So I should be using mr'' = -GMm/r² + mω²r instead and solve the PDE and go from there?

 

[haruspex]

So I should be using mr'' = -GMm/r² + mω²r instead and solve the PDE and go from there?

What about the forces from the wire?

 

[Decadohedron]

What about the forces from the wire?

There would be the parallel component in the r direction mω²rcos²(Φ) + the perpendicular component mω²rsin(Φ)cos(Φ)

 

[haruspex]

There would be the parallel component in the r direction mω²rcos²(Φ) + the perpendicular component mω²rsin(Φ)cos(Φ)

You don't need to calculate the forces from the wire, just take them into account in your analysis.

Draw a diagram in the NS, up-down plane showing the bead at latitude Φ. What forces act (centrifugal being a fictitious force)? What is the net force?