Question

Pendulum A is a physical pendulum made from a thin, rigid, and uniform rod whose length is d. One end of this rod is attached to the ceiling by a frictionless hinge, so the rod is free to swing back and forth. Pendulum B is a simple pendulum whose length is also d. Obtain the ratio TA/TB of their periods for small-angle oscillations.

Answer 1

Period TA = 2 (pi)v [ I / mgL ]

Where I = moment of inertia = ( 1/ 3) md^ 2

L = d / 2 Since the rod is uniform

So, TA = 2(pi) v[ 2md^ 2 / ( 3mgd) ]

= 2(pi) v[ 2d / 3g ]

Period TB = 2(pi) v[ d / g ]

TA / TB = v[ 2/3] = 0.8164

Answer 2

Period TA = 2 (?)? [ I / mgL ]
I = moment of inertia = ( 1/ 3) md^ 2
L = d / 2 (since the rod is uniform)
TA = 2(?) ?[ 2md^ 2 / ( 3mgd) ]
= 2(?) ?[ 2d / 3g ]
Period TB = 2(?) ?[ d / g ]
TA / TB = ?[ 2/3] = 0.8164