Freq. of small oscillations in two pendulums

[ejs12006]

Homework Statement

Consider two pendulums, I and II. I consists of a bob of mass 2m at the end of a rod of length L. II consists of one bob of mass m at the end of a rod of length L and another bob of mass m halfway up the road, at L/2. What is the ratio of the frequency of small oscillations of pendulum II to that of pendulum I?

Homework Equations

As far as I know, for small oscillations, the frequency is given by sqrt(g/L), where g is the acceleration of gravity, and L is the length of the pendulum. Therefore, the frequency of small oscillations is proportional to sqrt(1/L).

The Attempt at a Solution

if wII is the frequency for pendulum II and LII is the length of pendulum II, and wI is the frequency for pendulum I and LI is the length of pendulum I:

wII/wI = sqrt(1/LII)/sqrt(1/LI)

As I understand it, for an ideal pendulum, L refers to the position of the bob. The way I saw it, the "Bob" of pendulum II is located at the center of mass of the two Bobs. since one is at L and the other is at L/2, the center of mass is at 3L/4. Therefore:

wII/wI = sqrt(4/3)/sqrt(1/1) = sqrt(4/3)

However, the answer is actually sqrt(6/5) ! Any Ideas?

 

[LowlyPion]

Maybe consider the frequency of a compound pendulum.

f = (m*g*L / I)^1/2/ (2 * π)

Aren't you going to want to be interested in the moment of inertia and not just the center of mass?

Edit: as davieddy points out the L in this case is the distance to the center of mass

 

[davieddy]

L (in your formula) is the distance of the center of mass from the pivot,
not the L specified in the question.

 

[LowlyPion]

L (in your formula) is the distance of the center of mass from the pivot,
not the L specified in the question.

Thanks for the clarification.