GRE 61, small oscillation problem

[quantumworld]

A long, straight, and massless rod pivots about one end in a vertical plane. In configuration I, two small identical masses are attached to the free end; in configuration II, one mass is moved to the center of the rod. What is the ratio of the frequency of small oscillations of configuration II to that of configuration I?
(A) (6/5)^1/2
(B) (3/2)^1/2
(C) 6/5
(D) 3/2
(E) 5/3

ok, here is my problem: now we have a pendulum that oscillates normally ( configuration I), and another one ( configuration II) that I have no idea how to calculate its frequency of oscillation. First I thought that the second configuration will have two normal modes, but I messed up, because this is a rod, so it is a rigid body problem :grumpy: .
the correct answer is A. :rolleyes:
thank u so much

[Tide]

Using the small angle approximation
I \frac {d^2 \theta}{dt^2} = -g \theta \Sigma_i m_i L_i
where I is the moment of inertia.

[quantumworld]

Thank u Tide
I did solve it! :biggrin:
I used the formula (Mgd/I)^.5
where I = I of the first mass + I of the second mass
and Mgd is like the reduced mass times its distance from the pivot.
so the key thing to remember, is that, when we calculate I, we do it
separately for each mass, and when we calculate Mgd, we find the center of mass of the system, and the reduced mass. :biggrin:
I am so happy, that it turned out to be easy.