Pendulum $A$ is a physical pendulum made from a thin, rigid, uniform rod whose length is $d$ . One end of this rod is attached to the ceiling by a frictionless hinge so that the rod is free to swing back and forth. Pendulum $B$ is a simple pendulum whose length is also $d$ . If the ratio of time periods of their small angle oscillations is found to be $\frac{T_{A}}{T_{B}} = \sqrt{\frac{x}{3}},$ the value of $x$ is

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Solution

$T_{A} = 2 π \sqrt{\frac{I}{m g l}} = 2 π \sqrt{\frac{m d^{2} / 3}{m g (\frac{d}{2})}} = \sqrt{\frac{2}{3}} [2 π \sqrt{\frac{d}{g}}]$

$T_{B} = 2 π \sqrt{\frac{d}{g}}$

$\Rightarrow \frac{T_{A}}{T_{B}} = \sqrt{\frac{2}{3}}$

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TA=2π√Imgl=2π   ⎷md2/3mg(d2)=√23[2π√dg ] TB=2π√dg ⇒TATB=√23

$T_{A} = 2 π \sqrt{\frac{I}{m g l}} = 2 π \sqrt{\frac{m d^{2} / 3}{m g (\frac{d}{2})}} = \sqrt{\frac{2}{3}} [2 π \sqrt{\frac{d}{g}}]$

$T_{B} = 2 π \sqrt{\frac{d}{g}}$

$\Rightarrow \frac{T_{A}}{T_{B}} = \sqrt{\frac{2}{3}}$