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प्रश्न
A uniform rod of length l is suspended by an end and is made to undergo small oscillations. Find the length of the simple pendulum having the time period equal to that of the road.
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उत्तर
It is given that the length of the rod is l.
Let point A be the suspension point and point B be the centre of gravity.
Separation between the point of suspension and the centre of mass, l' = \[\frac{l}{2}\]
Also, h =\[\frac{l}{2}\]
Using parallel axis theorem, the moment of inertia about A is given as,
\[I = I_{CG} + m h^2 \]
\[ = \frac{m l^2}{12} + \frac{m l^2}{4} = \frac{m l^2}{3}\]
\[\text { the time period }\left( T \right) \text { is given by, }\]
\[T = 2\pi\sqrt{\frac{I}{mgl'}} = 2\pi\sqrt{\frac{I}{mg\frac{l}{2}}}\]
\[ = 2\pi\sqrt{\frac{2m l^2}{3mgl}} = 2\pi\sqrt{\frac{2l}{3g}}\]
Let T' be the time period of simple pendulum of length x.
Time Period \[(T')\] is given by ,\[T' = 2\pi\sqrt{\left( \frac{x}{g} \right)}\] \[\text { As the time period of the simple pendulum is equal to the time period of the rod,} \] \[T' = T\]
\[ \Rightarrow \frac{2l}{3g} = \frac{x}{g}\]
\[ \Rightarrow x = \frac{2l}{3}\]
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