While solving the problem of the rigid rod pendulum I figured out that when the point of suspension is at the end of the rod and and at $x = \frac{l}{3}$ from same end, time period of oscillation has same value.
I don't get it why does it actually have same value, as the length decreases the time period must increase. Because at COM time period is $\infty$, so why it decrease and then increases? I don't understand it physically.
Best Answer
For any rigid pendulum the angular frequency is given by:
$$ \omega = \sqrt{\frac{mgx}{I}} $$
where $m$ is the centre of mass of the rigid object and $x$ is the distance of the pivot from the centre of mass. Suppose we start with the pivot passing through the centre of mass, i.e. $x=0$, then obviously the angular frequency is zero (the period is infinite).
As we move the pivot away from the centre of mass two things happen:
the value of $mgx$ increases because $x$ increases
the moment of inertia $I$ increases as described by the parallel axis theorem
Since the frequency is proportional to the ratio of these two, $\sqrt{mgx/I}$, how the frequency changes depends on how the two quantities change.
If the moment of inertia about an axis through the centre of mass is $I_0$ then the parallel axis theorem tells us the the moment of inertia about an axis a distance $x$ from the centre of mass is:
$$ I(x) = I_0 + mx^2 $$
and we can substitute this in our equation for the angular frequency to get:
$$ \omega(x) = \sqrt{\frac{mgx}{I_0 + mx^2}} $$
Now consider the limits where $x$ is very small and $x$ is very large. For small $x$ we have $I_0 \gg mx^2$ so the frequency is approximately:
$$ \omega(x) = \sqrt{\frac{mgx}{I_0}} \propto x $$
So as we increase $x$ away from zero the frequency increases linearly with $x$. However for large $x$ we have $I_0 \ll mx^2$ so the frequency is approximately:
$$ \omega(x) = \sqrt{\frac{mgx}{mx^2}} \propto \frac{1}{x} $$
So as $x$ gets large the frequency decreases with increasing $x$.
The end result is that as we move the axis away from the centre of mass the angular frequency increases at first, but then reaches a maximum and for large $x$ it starts decreasing again. And this is what you are seeing in your calculation. If you put in the value of $I_0$ for a rigid rod of length $2\ell$:
$$ I_0 = \frac{m(2\ell)^2}{12} $$
Then you'll find $\omega(x)$ looks like:
So there value of $\omega(x)$ is indeed the same one third of the way along as it is at the end of the rod.