Question

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.

Answer 1

Explanation:

Let us consider A to be the suspension point and B to be the centre of gravity. We know that the moment of inertia about A is $A \text { is } 1=I_{c \cdot g}+m h^{2}=>I=\left(\frac{m l^{2}}{3}\right)$ where l is the length of the uniform rod and $h=\frac{i}{2}$.  Thus,  $T=2 \pi \sqrt{\left(\frac{t}{m g\left(\frac{l}{3}\right)}\right)}=2 \pi \sqrt{\frac{2 l}{3 g}}$ . If the time period of simple pendulum of length x is equal to the time period T,  $T=2 \pi \sqrt{\frac{x}{g}}$ where x is the length of the pendulum. Therefore,  $x=\frac{2 l}{3}$. Hence the length of the pendulum is  $\frac{2 l}{3}$.