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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.
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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