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ejs12006
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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?