# Time period of a rigid rod pendulum has same value at two different point of suspension, why?

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.

• It is not really clear what you're asking: what is it that you don't get? The period of a simple pendulum is what it is, it must be proportional to $\sqrt{l/g}$ because that would be the only combination of units giving back a unit of time, so it cannot be otherwise. – gented Aug 24 '17 at 8:32

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:

1. the value of $mgx$ increases because $x$ increases

2. 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.

• I got your answer, but what's my doubt is, as I move along x axis in your graph, I see the w(x) value increases, and it's logical, as the torque acting on COM of both section, have different value and certainly the one with larger portion of rod, has greater torque that other. And in similar manner this should go on, but what am not getting is why there is a $\displaystyle\text{w(x)}_{max}$ at x not equal to $\dfrac{l}{2}$. – Vivekanand Mohapatra Aug 24 '17 at 13:26
• @VivekanandMohapatra: the torque is proportional to $x$ and the change in the moment of inertia is proportional to $x^2$. So the two changes have different functional forms. There is no reason to suppose the frequency would peak halfway from the centre to the end of the rod. – John Rennie Aug 24 '17 at 14:28
• okay got it, its all because of the x and $x^2$ variation of distance and the moment of inertia. – Vivekanand Mohapatra Aug 25 '17 at 8:25