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Mechanical Oscillations

Problem 4.49

A physical pendulum performs small oscillations about the horizontal axis with frequency ω1=15.0 s−1.\omega_{1}=15.0 \mathrm{~s}^{-1} . When a small body of mass m=50 gm=50 \mathrm{~g} is fixed to the pendulum at a distance l=20 cml=20 \mathrm{~cm} below the axis, the oscillation frequency becomes equal to ω2=\omega_{2}= =10.0 s−1.=10.0 \mathrm{~s}^{-1} . Find the moment of inertia of the pendulum relative to the oscillation axis.

Reveal Answer
I=ml2(ω22−g/l)/(ω12−ω22)=0.8 g⋅m2I=m l^{2}\left(\omega_{2}^{2}-g / l\right) /\left(\omega_{1}^{2}-\omega_{2}^{2}\right)=0.8 \mathrm{~g} \cdot \mathrm{m}^{2}
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