# Can you intuitively explain the decreasing time-period of oscillation with increasing pendulum length in some cases?

Consider a rigid body suspended about an axis of rotation which, in general does not pass through it's center of mass (COM) and has a moment of inertia (MOI) $$0 < I_{axis}$$ about that axis. Let $$I_C$$ denote the moment of inertial of the object about an axis parallel to the one mentioned before and passing through the COM. The parallel axis theorem implies that $$I_{axis} = I_C + ml^2$$ where $$0 \leq l$$ is the distance between the two axes. From the rotational dynamics of the object $$I_{axis} \ddot{\theta} = -mgl\theta$$, where $$0 \leq \theta \rightarrow 0$$ is the (small) angular displacement of the object, we can fine the time period of oscillation as $$T = \frac{2\pi}{\sqrt{\frac{mgl}{I_{axis}}}}$$ so that $$T \propto \sqrt{\frac{I_{axis}}{ml}}$$.

In case of the simple pendulum, $$I_C \approx 0$$ so that $$I_{axis} = ml^2$$, and thus $$T \propto \sqrt{l}$$ leading to the usual conclusion (matching our intuitive physical understanding) that the time-period of small oscillations increases as the length $$l$$ increases. However, in the case of general rigid bodies, i.e. not point-masses, the algebra results in $$T \propto \sqrt{\frac{I_C}{ml} + l}$$. From the plot one can see that this expression explains the apparently counter-intuitive observation that for general rigid bodies (i.e. not point-masses), for small $$l$$, the time-period of small oscillations reduce as the length $$l$$ increases, in a mathematical sense.

In the case of a simple pendulum, it is physically intuitive that the time-period should increase with the increase in $$l$$ (distance traveled over an oscillation is increasing linearly with $$l$$, while the motivating force remains roughly of the same magnitude regardless of $$l$$). In the same sense what is the intuitive physical explanation of this apparently counter-intuitive behavior?

Consider a pendulum that consists of 2 points of mass $$m$$ connected by a massless rigid rod of length $$h$$. Suspend it at the center of mass. It doesn't oscillate.

Suspend it a small distance, $$\delta x$$, above the center of mass. Turn it 90 degrees. It oscillates slowly. The moment of intertial, $$I$$, is almost the same as if it was suspended at the center. The torque is $$\tau = mg \delta x$$. The angular acceleration is $$\alpha = \tau / I$$.

Suspend it $$2 \delta x$$ above the center of mass. $$I$$ is still almost the same. $$\tau$$ has doubled, and so has $$\alpha$$.

Consider other properties of the two cases.

• The distance the center of mass travels in a half oscillation has doubled from $$\pi \delta x$$ to $$\pi 2\delta x$$.
• The potential energy decrease when rotated upright has doubled from $$mg \delta x$$ to $$mg 2\delta x$$. So has the maximum rotational kinetic energy.
• The maximum angular velocity, $$\omega$$ has quadrupled.

These ratios hold at each angle during the half oscillations of each case. $$\omega$$ is quadrupled over a trajectory that is twice as long. The period has halved.

• The direction of the explanation seems great to me. Please allow me to point out that the potential energy is associated with the factor $(\delta x)^2$ (thus introducing $\theta^2$ associated with the spring-mass analogy of the harmonic-oscillator with spring-constant $mgl$ considered here). Further, the answer does not address the intuition behind this apparent counter-intuitive behavior only in the case of low absolute length $l$ (rather than the change in length $\delta x$ in your qualitative analysis). Commented Mar 7, 2021 at 0:48
• $U = mgh$, where $h$ is the change in height of the COM. In this case, $h = \delta x$ or $2 \delta x$. There is no $(\delta x)^2$ dependency. The period only decreases in special cases like when the support is very near the COM. For a usual pendulum, increasing the length a little would cause an increase in period. This is true for a Grandfather clock pendulum with a big block of brass as much as an ideal pendulum with a point mass. Commented Mar 7, 2021 at 1:01
• Agreed. It seems I misunderstood the variable $\delta x$. However the answer does not address the apparently counter-intuitive of behavior of decreasing time-period with increasing $l$ and why it happens only in the case of low absolute length $l$. Commented Mar 7, 2021 at 1:08
• I am not sure how much I can add to this. It is pretty much all the answer I have. It happens for low absolute $l = \delta x$ because $I$ doesn't change much and $\tau$ does. At bigger $l$, both $I$ and $\tau$ change. Commented Mar 7, 2021 at 1:16
• Ok, thanks for the very subtle and nice answer! Commented Mar 7, 2021 at 1:26

In the case of a simple pendulum, it is physically intuitive that the time-period should increase with the increase in l (distance traveled over an oscillation is increasing linearly with l). In the same sense what is the intuitive physical explanation of this apparently counter-intuitive behavior?

In my opinion, the behaviour of a simple pendulum is not so intuitive. I guess if a common visitor of the Paris Pantheon is asked about the period of the Foucault's pendulum, it would be no surprise if some answers put the mass as a variable for example, and miss the role of the length.

But, we can have an "educated" intuition of the rigid body oscillations: as the moment of inertia is $$\alpha mr^2$$, where $$\alpha$$ is some constant, increasing $$l$$ means decreasing the length parameter ($$\frac{r^2}{l}$$) inside the square root. The period is proportional to the square root of the "length" so to speak, if l is small compared to the other term. And the mass cancels.

So its behavior is intuitive in this meaning.

• Indeed, it is non-trivial at-first-glance that the time-period is independent of the mass. Thanks for the answer since it clarifies the basis of the posed question. However it does not address the apparently counter-intuitive of behavior of decreasing time-period with increasing $l$ and why it happens only in the case of low absolute length $l$. I understand why this happens mathematically, but I think there must be an intuitive explanation to it. Commented Mar 7, 2021 at 1:06