Solve Coupled Oscillator Homework: Find 2 Eigenfrequencies

[roeb]

Homework Statement

A thin hoop of radius R and mass M oscillates in its own plane with one point of the hoop fixed. Attached to the hoop is a small mass M constrained to move (in a frictionless manner) along the hoop. Consider only small oscillations, and show that the eigenfrequencies are blah blah blah (two eigenfrequencies).

Homework Equations

The Attempt at a Solution

My difficulties are in setting up this problem. I believe that I am picturing the system correctly, but I can't quite figure out how to do it. I need to find the Lagrangian of the system first, but I am having a hard time with the kinetic energy part.The oscillation of the hoop if I am not mistaken will be like that of a pendulum.

Hoop: T = 1/2*Iw^2 = 1/2 m * R^2 * $$\omega ^2$$
Small Mass: T = 1/2 mR^2 $$\theta '$$

I have set up theta as the angle between the center of the hoop and the position of the small mass. The problem is that I don't quite know how to get the second generalized coordinate -- I am assuming there are two generalized coordinates because this is a coupled oscillator problem and I am given two eigenfrequencies.

I am tempted to say that $$\omega = R \theta '$$ but that doesn't yield a correct answer and I don't think it's right to begin with...

Any help?

 

[Jhero]

First, let's define the two generalized coordinates for this system. Let x be the displacement of the small mass along the hoop, and let θ be the angle of the hoop with respect to its equilibrium position. Now, we can write the Lagrangian as:

L = T - V

where T is the kinetic energy and V is the potential energy. Let's start with the kinetic energy:

T = 1/2m(dx/dt)^2 + 1/2I(dθ/dt)^2

where m is the mass of the small mass and I is the moment of inertia of the hoop. Now, we need to express dx/dt and dθ/dt in terms of our generalized coordinates. We can do this by using the chain rule:

dx/dt = R(dθ/dt)

dθ/dt = θ'

where θ' is the angular velocity of the hoop. Substituting these expressions into our kinetic energy equation, we get:

T = 1/2mR^2(θ')^2 + 1/2I(θ')^2

Now, let's move on to the potential energy. The small mass has a potential energy due to gravity, given by mgh, where h is the height of the small mass above its equilibrium position. Since the hoop is fixed at one point, the height h is given by Rθ. The hoop also has a potential energy due to its rotation, given by 1/2kθ^2, where k is the spring constant of the hoop. Putting this all together, we get:

V = mgRθ + 1/2kθ^2

Now, we can write the Lagrangian as:

L = 1/2mR^2(θ')^2 + 1/2I(θ')^2 - mgRθ - 1/2kθ^2

To find the equations of motion, we can use the Euler-Lagrange equations:

d/dt(∂L/∂θ') - ∂L/∂θ = 0

d/dt(∂L/∂x') - ∂L/∂x = 0

Solving these equations will give us the eigenfrequencies of the system, which will be two oscillatory modes. I hope this helps!