A physical pendulum performs small oscillations about the horizontal axis with frequency `omega_(1)=15.0 s^(-1)`. When a small body of mass `m=50 g` is fixed to the pendulum at a distance `l=20 cm` below the axis, the oscillation frequency becomes equal to `omega_(2)=10.0 s^(-1)`. Find the moment of inertia of the pendulum relative to the oscillation axis.

Explanation

Let, moment of inertia of the pendulum, about the axis, concerned is `I`, then writing `N_(z)=Ibeta_(z)`, for th ependulum
`-mgx sin a theta=I dot(theta) `or `dot(theta)=-(mgx)/(I)theta` `(` For small `theta)`
which is the required equation for S.H.M. So, the frequency of oscillation,
`omega_(1)=sqrt((Mgx)/(I))` or `x=(I)/(Mg)sqrt(omega_(1)^(2)) ...(1)`
Now, when the mass `m` is attached to the pendulum, at a distance `l` below the oscillating axis,
`-M g x sin theta^(')-mg l sin theta^(')=(I+mg^(2))(d^(2) theta^('))/(dt^(2))`
or `-(g(Mx+ml))/((I+ml^(2)))theta^(')=(d^(2)theta)/(dt^(2))`, (For small `theta)`
which is again the equation of S.H.M.,So, the new frequency,
`omega_(2)=sqrt((g(Mx+ml))/((I+ml^(2)))) ...(2)`
Solving Eqns `(1)` and `(2)`
`omega_(2)=sqrt((g((I//g)omega_(1)^(2)+ml))/((I+ml^(2)))`
or `omega_(2)^(2)=(Iomega_(1)^(2)+mgl)/(I+ml^(2))`
or, `I(omega_(2)^(2)-omega_(1)^(2))=mgl-m omega_(2)^(2)l^(2)`
and hence, `I=ml^(2)(omega_(2)^(2)-g//l)//(omega_(1)^(2)-omega_(2)^(2))=0.8g.m^(2)`