Question

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 equilibium the pendulums were fastened rigifly together. What will be the frequency of small oscillations of the compound pendulum ?

Answer 1

When the two pendulums are joined rigidly and set to osciallate, each exert torques on the other, these torques are equal and opposite. We write the law of motion for the two pendulums as
`I_(1)ddot(theta)=- omega_(1)^(2)I_(1)theta+G`
`I_(2)ddot(theta)=- omega_(2)^(2)I_(2)theta-G`
where `+-G` is the torque of mutual interactions. We have written forces on each pendulum in the absence of the other as `-omega_(2)^(2)I_(1)theta` and `-omega_(2)^(2)I_(2)theta` respectively. Then
`ddot(theta)=-(I_(1)omega_(1)^(2)+I_(2)omega_(2)^(2))/(I_(1)+I_(2))theta=-omega^(2)theta`
Hence `omega=sqrt((I_(1)omega_(1)^(2)+I_(2)omega_(2)^(2))/(I_(1)+I_(2))`