Two physical pendulums perform small oscillations about the same horizontal axis with frequencies $ω_{1} and ω_{2}$ 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 angular frequency of small oscillations of the compound pendulum?

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Solution

When the two pendulum are joint rigidly and set to oscillate, each exert torques on the other, these torques are equal and opposite . We write the l;aw of motion for the two pendulum $I_{1} ¨ θ = - ω_{1}^{2} I_{1} θ + G$
$I_{2} ¨ θ = - ω_{1}^{2} I_{2} θ - G$
where $\pm G$ is the torque of mutual interaction. We have writtten the restoring forces on each pendulum in the absence of the other as $- ω_{1}^{2} I_{1} θ - G$ and $- ω_{2}^{2} I_{2} θ - G$ respectively. Then
$¨ θ = - \frac{- ω_{1}^{2} I_{1} θ + ω_{2}^{2} I_{2} θ}{I_{1} + I_{1}} θ = - ω^{2} θ$
Hence $ω = \sqrt{\frac{- ω_{1}^{2} I_{1} θ + ω_{2}^{2} I_{2} θ}{I_{1} + I_{1}} θ}$

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