# Period of a pendulum in free fall

Let's say I have a pendulum hanging from a bar that's fixed to the wall of an elevator. Assume that there's no air or anything inside the elevator, that the string of the pendulum is very light and that the bob of the pendulum is more or less a heavy point mass. After setting the pendulum in motion, the elevator starts going down, increasing the period of the pendulum, until the cable holding the elevator runs out and brings the whole contraption into a free fall situation.

The formula for the period of a pendulum with length $$L_0$$ where the bob experiences a gravitational acceleration of $$a_0$$ is: $$T = 2 \pi \sqrt{\frac{L_0}{a_0}}$$. In free fall, $$a_0 = 0$$ so the pendulum wouldn't swing at all.

However, in my hypothetical situation, bob of the pendulum could've had a velocity right before going into free fall, so wouldn't the pendulum transition into a uniform circular motion which gives rise to a new period?

If so, shouldn't there be a better formula to describe the period of a pendulum that also correctly predicts the period depending on how the acceleration on the bob changes with respect to time?

• Questions that can be answers as either "yes" or "no" aren't the best for this site. Can you perhaps instead ask about the physics concepts you are confused about? Commented Jun 24, 2020 at 13:58
• There, it's still a yes or no question but I hope the pendulum period thing can be discussed a bit better now. Commented Jun 24, 2020 at 14:06
• Thanks, I think the question is much better now :) I have also supplied an answer. Feel free to comment with any questions / suggestions. Commented Jun 24, 2020 at 14:39

As long as there is a net velocity on the pendulum bob at the moment the elevator goes into free fall, the pendulum will go into uniform circular motion.

The formula you have stated for the time period is only valid for a pendulum. Once the bob goes into circular motion, it is no longer a pendulum as there is no restoring force acting on the bob. The formula still makes logical sense as the bob will never reverse its direction and will hence take infinite time to come back to its starting path.

• So at what point in my hypothetical situation does my pendulum stop being a pendulum? If the bob has a sufficiently high speed and the system reaches free fall at the right time, the bob might go into circular motion much quicker than when the elevator falls very slowly. Commented Jun 24, 2020 at 14:23
• It stops being a pendulum when the acceleration on the bob becomes zero.
– Sam
Commented Jun 24, 2020 at 14:28
• It isn't a pendulum unless its angular velocity changes sign at the end of each half-cycle, and that means the bob must be accelerating. No acceleration, no pendulum. Commented Jun 24, 2020 at 15:35
• For it to be a pendulum it must have a restoring torque working on it Commented Jun 24, 2020 at 20:21

$$T=2\pi\sqrt{L_0/a_0}$$ is the period of a simple pendulum of length $$L_0$$ with small-angle oscillations. The parameter $$a_0$$, sometimes also denoted as $$g$$, is usually the acceleration due to gravity, but I suppose it is technically the acceleration due to some constant force that is proportional to the mass of the pendulum bob. So the equation of motion obtained from Newton's second law is

$$\frac{\text d\theta^2}{\text dt^2}=-\frac {a_0}{L_0}\cdot\sin\theta\approx \frac {a_0}{L_0}\cdot\theta$$

However, in free fall the equation of motion becomes $$\frac{\text d\theta^2}{\text dt^2}=0$$

And here is the issue. This second equation does not give you a unique period! You can have any period you want with $$\ddot\theta=0$$ depending on the initial conditions.

Linking this back to your period equation, note that when $$a_0=0$$ we get an undefined value, which is what we just determined above. So technically, $$T=2\pi\sqrt{L_0/a_0}$$ is still a valid equation for your free-fall scenario: it tells us the period is not defined by this equation, which makes sense. The period is instead defined by the angular velocity $$\omega_0$$ when free fall began: $$T=\frac{2\pi}{\omega_0}$$

• I think this answer best addresses the uncertainty of this hypothetical situation. Commented Jun 25, 2020 at 7:16

Look at the forces acting on bob, when elevator's acceleration is $$g$$, in the axis which is perpendicular to velocity of pendulum. Let angle between rope and $$y$$ axis be $$\theta$$. So: $$\frac{mv^2}{l}=T+ma\cos(\theta)-mg\cos(\theta)\mathrel{\stackrel{{\mbox{ a=g}}}{=}}T$$ So: $$\frac{mv^2}{l}=T$$. And there's no force in direction of velocity there's only perpendicular to it velocity doesn't change. So our equation is just for circular motion of bob. If you want to find period of this motion you'll need velocity at time that $$a(t_1)=g$$. Our period will be $$\tau=\frac{2\pi l}{v}$$And for this you'll need to find $$\theta(t)$$. And you can find it by the equation $$\ddot{\theta}=-\frac{g-a(t)}{l}\theta$$ I think it's impossible to find this without knowing $$a(t)$$. And velocity at time $$t_1$$ will be $$l\dot{\theta}(t_1)$$. And coming to better formula for period, if you look at equation for motion at time $$t_1$$ you'll see $$\ddot{\theta}=0$$ And solution for this is $$\theta=c_2t+c_1$$ from this equation you see that there's no sign of harmonic motion and there's no period of harmonic motion which is valid for our formula $$T=2\pi\sqrt{\frac{l}{a_0}}$$

• I've come to realize that this situation can lead to very chaotic outcomes, when for example the $a(t)$ doesn't monotonously approach $g$, it doesn't guarantee that the bob will stay in circular motion etc. But thanks for doing the full derivation! Commented Jun 25, 2020 at 6:59