A uniform rod of length 1.8 ~m suspended by an end is made to undergo small oscillations. Find the length of the simple pendulum having the mass and time period equal to that of the rod.

Step-by-step solution

Given, length of the rod, l=1.8 ~m Time period of the oscillation of uniform rod with length l piwoted at one end is given by T=2 I m g R where, R is the distance of the point of pivot to the centre of the gravity and I is moment of inertia of the rod. R= l 2 and I= m l^2 3 From Eq. (i), T=2 m l^2 3 m g l 2 =2 2 3 l g =2 2 3 1.8 g T=2 1.2 g Time period of the simple pendulum of length l^ is given by T^ =2 l^ g ( . Given, .T^ =T ) array rlrl & & 2 l^ g & =2 1.2 g & l^ & =1.2 ~m array Hence, the length of the simple