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Mechanical Oscillations

Problem 4.50

Two physical pendulums perform small oscillations about the same horizontal axis with frequencies \(\omega_{1}\) and \(\omega_{2} .\) Their moments of inertia relative to the given axis are equal to \(I_{1}\) and \(I_{2}\) respectively. In a state of stable equilibrium the pendulums were fastened rigidly together. What will be the frequency of small oscillations of the compound pendulum?

Reveal Answer
ω=(I1ω12+I2ω22)/(I1+I2)\omega=\sqrt{\left(I_{1} \omega_{1}^{2}+I_{2} \omega_{2}^{2}\right) /\left(I_{1}+I_{2}\right)}
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